CBSE Class 12 Mathematics Probability VBQs Set 01

BASIC CONCEPTS

• Probability: Probability is a branch of mathematics in which the chance of an event happening is assigned a numerical value that predicts how likely that event is to occur.
• Random Experiment: The experiment, in which the outcomes may not be same even if the experiment is performed in identical condition, is called random experiment. e.g., Tossing a coin is a random experiment because if we toss a coin in identical condition, outcomes may be head or tail.
• Outcome: An outcome is a result of some activity or experiment.
• Sample Space: A sample space is a set of all possible outcomes for a random experiment.
• Event: An event is a subset of the sample space.
• Theoretical Probability: The theoretical probability of an event is the number of ways that the event can occur, divided by the total number of possibilities in the sample space.
  Symbolically, we write \( P(E) = \frac{n(E)}{n(S)} \), where \( P(E) \) represents the probability of the event.
• In general, for any sample space \( S \) containing \( k \) possible outcomes, we say \( n(S) = k \). When the event \( E \) is certain, every possible outcome for the sample space is also an outcome for event \( E \) or \( n(E) = k \). Thus, the probability of a certain or sure event is given as
  \( P(E) = \frac{n(E)}{n(S)} = \frac{k}{k} = 1 \)

  Note: (i) The probability of an event that is certain to occur is 1.
  (ii) The probability of any event \( E \) must be equal to or greater than 0; and less than or equal to 1, i.e., \( 0 \leq P(E) \leq 1 \).

• Another way of expressing probability is in term of axioms, laid by Russian mathematician A. N. Kolmogorov.
  If \( S \) is a sample space, then probability \( P \) is a real valued function defined on \( S \) and take values [0, 1], satisfying following axiom:

(i) Probability of any event \( \geq 0 \).

(ii) Sum of probabilities assigned to all members of \( S \) is 1.

(iii) For any two mutually exclusive events \( E \) and \( F \), \( P(E \cup F) = P(E) + P(F) \).

Theorems of Probability:

• Addition theorem:

(a) When the events are not mutually exclusive: The probability that at least one of the two events \( A \) and \( B \) which are not mutually exclusive will occur is given
Symbolically, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
In the case of three events:
\( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \)

(b) When A and B are mutually exclusive: The addition theorem states that if two events \( A \) and \( B \) are mutually exclusive, the probability of the occurrence of either \( A \) or \( B \) is the sum of the individual probability of \( A \) and \( B \)
Symbolically, \( P(A \cup B) = P(A) + P(B) \)
The theorem can be extended to three or more mutually exclusive events thus, \( P(A \cup B \cup C) = P(A) + P(B) + P(C) \)

• Multiplication theorem: This theorem states that if two events \( A \) and \( B \) are independent, the probability that they both will occur is equal to the product of their individual probabilities.
  Symbolically, \( P(A \cap B) = P(A) \cdot P(B) \)
  The theorem can be extended to three or more independent events thus, \( P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) \)

  Note: If \( A \) and \( B \) are mutually exclusive and exhaustive, then \( P(A \cup B) = P(A) + P(B) = 1 \)

A rule for the probability of the event not A: If \( P(A) \) is the probability that some given event will occur, and \( P(\text{not } A) \) is the probability that the given event will not occur, then
Symbolically: \( P(A) + P(\text{not } A) = 1 \) or \( P(A) = 1 - P(\text{not } A) \) or \( P(\text{not } A) = 1 - P(A) \)
We write \( P(\text{not } A) \) as \( P(\bar{A}) \).

Problems related to withdrawal of balls, cards, letters, etc. with replacement and without replacement:

In such type of problems, the sample space will not change when the articles (balls, cards, letters, etc.) are replaced after each withdrawal. While in case when the article is not replaced (without replacement), the sample space will change after each withdrawal.

Note:

(i) If the problem does not specifically mention “with replacement” or “without replacement”, ask yourself: ‘’Is this problem with or without replacement?’’

(ii) For many compound events, the probability can be determined most easily by using the counting principle i.e., permutations and combinations.

(iii) Every probability problem can always be solved by

Counting the number of elements in the sample space \( n(S) \);

Counting the number of outcomes in the events, \( n(E) \);

And substituting these numbers in the probability formula: \( P(E) = \frac{n(E)}{n(S)} \).

(iv) Taking out 2 or more objects (e.g. balls) randomly from a bag one by one without replacement is same as taking out 2 or more objects simultaneously.
The number of ways in which \( r \) objects can be taken out of \( n \) objects is \( {}^nC_r \) or \( C(n, r) = \frac{n!}{(n - r)! \cdot r!} \).

Conditional Probability:

If \( A \) and \( B \) are two events associated with the same random experiment, then the probability of occurrence of event \( A \), when the event \( B \) has already occurred is called conditional probability of \( A \) when \( B \) is given. It is represented by \( P(A/B) \) and is given by
\( P(A/B) = \text{Probability of event A when B has already occurred} \)
\( = \text{Probability of event } 'A \cap B' \text{ when B behaves like sample space} \)
\( = \frac{n(A \cap B)}{n(B)} \)
\( = \frac{\frac{n(A \cap B)}{n(S)}}{\frac{n(B)}{n(S)}} \) [Dividing \( N^r \) and \( D^r \) by \( n(S) \)]
\( = \frac{P(A \cap B)}{P(B)} \)
Similarly, \( P(B/A) = \frac{P(A \cap B)}{P(A)} \)

Theorem of Total Probability:

Let \( E_1, E_2, \dots, E_n \) be the events of a sample space ‘\( S \)’ such that they are pair wise disjoint, exhaustive and have non-zero probabilities. If \( A \) is any event associated with \( S \), then
\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + \dots + P(E_n) \cdot P(A/E_n) \)

Bayes’ Theorem:

If \( B_1, B_2, \dots, B_n \) are mutually exclusive and exhaustive events and \( A \) is any event that occurs with \( B_1 \) or \( B_2 \) or \( B_n \) then
\( P(B_i / A) = \frac{P(B_i) \cdot P(A / B_i)}{\sum_{i=1}^{n} P(B_i) \cdot P(A / B_i)}, i = 1, 2, \dots, n \)

Note: The probabilities \( P(B_i) \), \( i = 1, 2, \dots, n \) which were already known before performing an experiment are known as prior probabilities and conditional probabilities \( P(B_i/A) \), \( i = 1, 2, 3, \dots, n \) which are calculated after the experiment is performed are known as posterior probabilities. The events \( B_1, B_2, \dots, B_n \) are usually called causes for event \( A \) to occur.

Random Variable:

Random variable is simply a variable whose values are determined by the outcomes of a random experiment; generally it is denoted by capital letters such as \( X, Y, Z \), etc. and their values are denoted by the corresponding small letters \( x, y, z \), etc.

Probability Distribution:

The system consisting of a random variable \( X \) along with \( P(X) \) is called the probability distribution of \( X \).

Mean and Variance of a Random Variable:

Let a random variable \( X \) assume values \( x_1, x_2, \dots, x_n \) with probabilities \( p_1, p_2, \dots, p_n \) respectively, such that \( p_i \geq 0 \), \( \sum_{i=1}^{n} p_i = 1 \). Then, the mean of \( X \), denoted by \( \mu \), [or expected value of \( X \) denoted by \( E(X) \)] is defined as
\( \mu = E(X) = \sum_{i=1}^{n} x_i p_i \) and
Variance denoted by \( \sigma^2 \) is defined as
\( \sigma^2 = \sum_{i=1}^{n} (x_i - \mu)^2 p_i = \sum_{i=1}^{n} x_i^2 p_i - \mu^2 \)

Standard Deviation, \( \sigma = \sqrt{\text{variance}} \)

Bernoullian Trials:

A sequence of independent trials which can result in one of the two mutually exclusive possibilities success or failure such that the probability of success or failure in each trial is constant, then such repeated independent trials are called Bernoullian trials.
Suppose we perform a series on \( n \) Bernoullian trails for each trial, \( p \) is the probability of success and \( q \) is the probability of failure, then \( p + q = 1 \).

Binomial Distribution:

A random variable \( X \) taking values \( 0, 1, 2, \dots, n \) is said to have a binomial distribution with parameters \( n \) and \( p \), if its probability distribution is given by
\( P(X = r) = {}^nC_r p^r q^{n-r} \dots(i) \)
Where, \( p \) represents probability of success while \( q \) represents probability of non-success, or failure and \( n \) is the number of trials. A binomial distribution with \( n \)-Bernoulli’s trials and probability of success in each trial as \( p \), is denoted by \( B(n, p) \).

Note: While using above probability density functions of the binomial distribution in solving any problem we should, first of all examine whether all the conditions, given below are satisfied:

(i) There should be a finite number of trials.

(ii) The trials are independent.

(iii) Each trial has exactly two outcomes: success or failure.

(iv) The probability of an outcome remains the same in each trial.

Recurrence or recursion formula for the binomial distribution:
\( P(r + 1) = \frac{n - r}{r + 1} \cdot \frac{p}{q} \cdot P(r) \)

Mean, Variance and Standard Deviation:

(i) Mean \( = np \)

(ii) Variance \( = npq \)

(iii) Standard Deviation \( = \sqrt{npq} \)

LIST OF IMPORTANT FORMULAE

(i) \( P(A \cap B) = P(A) \times P(B/A) \), where \( A \) and \( B \) are any two events.

(ii) \( P(A \cap B) = P(B) \times P(A/B) \), where \( A \) and \( B \) are any two events.

(iii) Two events \( A \) and \( B \) are independent, if and only if \( P(A \cap B) = P(A) \times P(B) \).

(iv) If \( A, B, C \) are three independent events, then \( P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \).

(v) \( P(\bar{A} \cap B) = P(B) - P(A \cap B) \), where \( \bar{A} \) and \( B \) are independent events.

(vi) \( P(A \cap \bar{B}) = P(A) - P(A \cap B) \), where \( A \) and \( \bar{B} \) are independent events.

(vii) \( P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = P(\bar{A}) \times P(\bar{B}) \), where \( \bar{A} \) and \( \bar{B} \) are mutually exclusive events.

(viii) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) + P(B) - P(A \cup B)}{P(B)} \), where \( A \) and \( B \) are independent events and \( P(B) \neq 0 \).

(ix) \( P(\bar{B}/\bar{A}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{A})} = \frac{1 - P(A \cup B)}{1 - P(A)} \), where \( A \) and \( B \) are independent events and \( P(A) \neq 1 \).

Questions

Question. A black and red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9 given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8 given that the red die resulted in a number less than 4.

Answer: When a black and a red die are rolled then \( n(S) = 36 \).
(a) Let A be the event getting sum greater than 9 and B be the event getting a 5 on the black die.
\( A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \)
\( B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \)
\( A \cap B = \{(5, 5), (5, 6)\} \)
\( P(A) = \frac{6}{36} = \frac{1}{6}, P(B) = \frac{6}{36} = \frac{1}{6} \text{ and } P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/18}{1/6} = \frac{1}{18} \times \frac{6}{1} = \frac{1}{3} \).
(b) Let A be the event getting the sum 8 and B be the event getting a number less than 4 on red die.
\( A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \)
\( B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\} \)
\( A \cap B = \{(2, 6), (3, 5)\} \)
\( P(A) = \frac{5}{36}, P(B) = \frac{18}{36} = \frac{1}{2}, P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/18}{1/2} = \frac{1}{18} \times \frac{2}{1} = \frac{1}{9} \).

Question. An instructor has a question bank consisting of 300 easy true/false questions, 200 difficult, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer: Here, total questions = 300 + 200 + 500 + 400 = 1400
Let A be the event that selected question is an easy question.
\( P(A) = \frac{300 + 500}{1400} = \frac{800}{1400} = \frac{4}{7} \)
Let B be the event that selected question is a multiple choice question.
\( P(B) = \frac{500 + 400}{1400} = \frac{900}{1400} = \frac{9}{14} \)
Now \( A \cap B \) is the event so that the selected question is a easy multiple choice question.
\( P(A \cap B) = \frac{500}{1400} = \frac{5}{14} \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{5/14}{9/14} = \frac{5}{9} \).

Question. Consider the experiment of throwing a die. If a multiple of 3 comes up, a die is again thrown and if any other number comes, a coin is tossed. Find the conditional probability of the event, ‘the coin shows a tail’ given that ‘at least one die shows a 3’.

Answer: Here, S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)}
Let A be the event of getting a tail on coin.
\( A = \{(1, T), (2, T), (4, T), (5, T)\} \)
Let B be the event of getting 3 on at least one die.
\( B = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\} \)
\( A \cap B = \phi \)
\( P(A) = \frac{4}{20} = \frac{1}{5}, P(B) = \frac{7}{20} \text{ and } P(A \cap B) = \frac{0}{20} = 0 \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{7/20} = 0 \).

Question. Events A and B are such that \( P(A) = \frac{1}{2}, P(B) = \frac{7}{12} \) and P (not A or not B) = \( \frac{1}{4} \). State whether A and B are independent.

Answer: Here \( P(A) = \frac{1}{2}, P(B) = \frac{7}{12} \text{ and } P(\bar{A} \cup \bar{B}) = \frac{1}{4} \)
Now \( P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) \)
\( \frac{1}{4} = 1 - P(A \cap B) \Rightarrow P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4} \)
Now \( P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \)
\( \because P(A \cap B) \neq P(A) \times P(B) \)
Thus, A and B are not independent.

Question. Probabilities of solving specific problem independently by A and B are \( \frac{1}{2} \) and \( \frac{1}{3} \) respectively. If both try to solve the problem independently. Find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

Answer: Here, \( P(A) = \frac{1}{2} \text{ and } P(B) = \frac{1}{3} \)
Now \( P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} ; P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
(i) P (the problem is solved) = \( 1 - P(\bar{A} \cap \bar{B}) \)
\( = 1 - P(\bar{A})P(\bar{B}) = 1 - \frac{1}{2} \times \frac{2}{3} = 1 - \frac{1}{3} = \frac{2}{3} \).
(ii) P (exactly one of them solves) = \( P(A)P(\bar{B}) + P(\bar{A})P(B) \)
\( = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2} \).

Question. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspaper. A student is selected at random.
(a) Find the probability that the student reads neither Hindi nor English newspaper.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Answer: Let A be the event that a student reads Hindi newspaper and B be the event that a student reads English newspaper.
\( P(A) = \frac{60}{100} = 0.6, P(B) = \frac{40}{100} = 0.4 \text{ and } P(A \cap B) = \frac{20}{100} = 0.2 \)
(a) Now \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.2 = 0.8 \)
Probability that she reads neither Hindi nor English newspaper \( = 1 - P(A \cup B) = 1 - 0.8 = 0.2 = \frac{1}{5} \).
(b) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.6} = \frac{1}{3} \).
(c) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = \frac{1}{2} \).

Question. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Answer: Let \( E_1 \) and \( E_2 \) be the events that red ball is drawn in first draw and black ball is drawn in first draw respectively. Let A be the event that ball drawn in second draw is red. There are 5 red and 5 black balls in the urn.
\( \therefore P(E_1) = \frac{5}{10} = \frac{1}{2} \text{ and } P(E_2) = \frac{5}{10} = \frac{1}{2} \)
When 2 additional balls of red colour are put in the urn there are 7 red and 5 black balls in the urn. \( \therefore P(A/E_1) = \frac{7}{12} \)
When 2 additional balls of black colour are put in the urn there are 5 red and 7 black balls in the urn. \( \therefore P(A/E_2) = \frac{5}{12} \)
By theorem of total probability: \( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2} \).

Question. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is a hostelier?

Answer: Let the events be defined as \( E_1 = \text{selection of hostelier} \), \( E_2 = \text{selection of day scholar} \), \( A = \text{selection of student getting A grade} \).
\( P(E_1) = \frac{60}{100} = \frac{3}{5}, P(E_2) = \frac{40}{100} = \frac{2}{5} \)
\( P(A/E_1) = \frac{30}{100} = \frac{3}{10}, P(A/E_2) = \frac{20}{100} = \frac{1}{5} \)
We have to find \( P(E_1/A) \).
By Bayes' theorem, \( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} = \frac{\frac{3}{5} \cdot \frac{3}{10}}{\frac{3}{5} \cdot \frac{3}{10} + \frac{2}{5} \cdot \frac{1}{5}} = \frac{9/50}{9/50 + 2/25} = \frac{9/50}{13/50} = \frac{9}{13} \).

Question. A Laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested then with probability 0.005, the test will imply he has the disease). If 0.1 % of the population actually has the disease then what is the probability that a person has the disease given that his test result is positive.

Answer: Let \( E_1 \) and \( E_2 \) denote the events that a person has disease and does not have disease respectively. Let A be the event that the test result is positive.
Now, the probability that a person has the disease is \( P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001 \).
Probability that a person does not have the disease \( P(E_2) = 1 - 0.001 = 0.999 \).
Probability that a person has disease and test result is positive \( P(A/E_1) = 99\% = \frac{99}{100} = 0.99 \).
Probability that a person does not have disease and test result is positive \( P(A/E_2) = 0.5\% = \frac{0.5}{100} = 0.005 \).
By Bayes’ theorem: \( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \)
\( = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{22}{133} \).

Question. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer: Let \( E_1, E_2, E_3 \) and \( E_4 \) be the events that the missing card is a heart, spade, club and diamond respectively. Let A be the event of drawing two diamond cards from 51 cards.
\( P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{13}{52} = \frac{1}{4} \).
\( P(A/E_1) = P(A/E_2) = P(A/E_3) = \frac{{}^{13}C_2}{{}^{51}C_2} \text{ and } P(A/E_4) = \frac{{}^{12}C_2}{{}^{51}C_2} \).
By Bayes’ theorem: \( P(E_4/A) = \frac{P(E_4)P(A/E_4)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3) + P(E_4)P(A/E_4)} \)
\( = \frac{\frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}}{\frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}} = \frac{{}^{12}C_2}{3 \cdot {}^{13}C_2 + {}^{12}C_2} \)
\( = \frac{\frac{12!}{2!10!}}{3 \times \frac{13!}{2!11!} + \frac{12!}{2!10!}} = \frac{66}{3 \times 78 + 66} = \frac{66}{300} = \frac{11}{50} \).

Question. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Answer: Let X denote the random variable which denotes the number of tails when a biased coin is tossed twice. So, X may have values 0, 1 or 2. Head is 3 times as likely to occur as a tail.
\( \therefore P(H) = \frac{3}{4} \text{ and } P(T) = \frac{1}{4} \)
Now, \( P(X = 0) = P(HH) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)
\( P(X = 1) = P(HT) + P(TH) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \)
\( P(X = 2) = P(TT) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)
Probability distribution:
X: 0, 1, 2
P(X): \( \frac{9}{16}, \frac{3}{8}, \frac{1}{16} \)

Question. The random variable X has a probability distribution P(X) of the following form, where k is some number:
P(X) = k if x = 0; 2k if x = 1; 3k if x = 2; 0 otherwise.
(a) Determine the value of k. [CBSE 2019 (65/1/2)]
(b) Find P(X < 2), P(X ≤ 2), P(X ≥ 2).

Answer: (a) \( \sum p_i = 1 \Rightarrow k + 2k + 3k = 1 \Rightarrow 6k = 1 \Rightarrow k = \frac{1}{6} \).
(b) \( P(X < 2) = P(X=0) + P(X=1) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \).
\( P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = 6k = 6 \times \frac{1}{6} = 1 \).
\( P(X \geq 2) = P(X=2) = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \).

Question. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X) or the mean of the distribution.

Answer: Sample space S consists of \( 6 \times 5 = 30 \) pairs. X is the larger number, so X can be 2, 3, 4, 5, 6.
\( P(X = 2) = \frac{\{(1,2), (2,1)\}}{30} = \frac{2}{30} \)
\( P(X = 3) = \frac{\{(1,3), (3,1), (2,3), (3,2)\}}{30} = \frac{4}{30} \)
\( P(X = 4) = \frac{\{(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)\}}{30} = \frac{6}{30} \)
\( P(X = 5) = \frac{8}{30} \)
\( P(X = 6) = \frac{10}{30} \)
Tabular form:
X: 2, 3, 4, 5, 6
P(X): \( \frac{2}{30}, \frac{4}{30}, \frac{6}{30}, \frac{8}{30}, \frac{10}{30} \)
\( E(X) = \sum x_i p_i = \frac{2 \times 2 + 3 \times 4 + 4 \times 6 + 5 \times 8 + 6 \times 10}{30} = \frac{4 + 12 + 24 + 40 + 60}{30} = \frac{140}{30} = \frac{14}{3} = 4\frac{2}{3} \).

Question. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and SD of X.

Answer: Age X: 14, 15, 16, 17, 18, 19, 20, 21
P(X): \( \frac{2}{15}, \frac{1}{15}, \frac{2}{15}, \frac{3}{15}, \frac{1}{15}, \frac{2}{15}, \frac{3}{15}, \frac{1}{15} \)
\( \sum X P(X) = \frac{28 + 15 + 32 + 51 + 18 + 38 + 60 + 21}{15} = \frac{263}{15} \)
\( E(X) = 17.53 \).
\( \sum X^2 P(X) = \frac{392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441}{15} = \frac{4683}{15} \)
\( \text{Mean } \mu = 17.53 \).
\( \text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{4683}{15} - \left(\frac{263}{15}\right)^2 = \frac{4683 \times 15 - 263 \times 263}{225} = \frac{1076}{225} = 4.78 \).
\( \text{SD}(X) = \sqrt{4.78} = 2.19 \).

Question. The probability of a shooter hitting a target is \( \frac{3}{4} \). How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

Answer: Let the shooter fire n times. \( p = \frac{3}{4}, q = \frac{1}{4} \).
\( P(X = r) = {}^nC_r p^r q^{n-r} = {}^nC_r \left(\frac{3}{4}\right)^r \left(\frac{1}{4}\right)^{n-r} = \frac{{}^nC_r \cdot 3^r}{4^n} \).
\( P(\text{hitting at least once}) > 0.99 \Rightarrow P(r \geq 1) > 0.99 \)
\( 1 - P(r = 0) > 0.99 \Rightarrow 1 - {}^nC_0 \frac{3^0}{4^n} > 0.99 \Rightarrow 1 - \frac{1}{4^n} > 0.99 \)
\( \frac{1}{4^n} < 0.01 \Rightarrow \frac{1}{4^n} < \frac{1}{100} \Rightarrow 4^n > 100 \).
Minimum value of n is 4. Thus, the shooter must fire 4 times.

Question. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.

Answer: Let \( E_1 = \text{Yoga and meditation} \), \( E_2 = \text{drugs} \), \( A = \text{heart attack} \).
\( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2} \).
\( P(A/E_1) = 40\% - (40 \times \frac{30}{100})\% = 40\% - 12\% = 28\% = \frac{28}{100} \).
\( P(A/E_2) = 40\% - (40 \times \frac{25}{100})\% = 40\% - 10\% = 30\% = \frac{30}{100} \).
We have to find \( P(E_1/A) \).
By Bayes' theorem, \( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} = \frac{\frac{1}{2} \times \frac{28}{100}}{\frac{1}{2} \times \frac{28}{100} + \frac{1}{2} \times \frac{30}{100}} = \frac{28/100}{58/100} = \frac{14}{29} \).

Question. Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer: Let \( E_1 = \text{red ball transferred} \), \( E_2 = \text{black ball transferred} \), \( A = \text{red ball drawn from bag II} \).
\( P(E_1) = \frac{3}{7}, P(E_2) = \frac{4}{7} \).
If red is transferred, Bag II has 5 red, 5 black. \( P(A/E_1) = \frac{5}{10} = \frac{1}{2} \).
If black is transferred, Bag II has 4 red, 6 black. \( P(A/E_2) = \frac{4}{10} = \frac{2}{5} \).
By Bayes' theorem, \( P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{8/35}{3/14 + 8/35} = \frac{8/35}{31/70} = \frac{16}{31} \).

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find P (A/B) if P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6
Answer: We are given the probabilities \( P(A) = 0.4 \), \( P(B) = 0.8 \), and \( P(B/A) = 0.6 \).
Using the definition of conditional probability:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies 0.6 = \frac{P(A \cap B)}{0.4} \)
\( \implies P(A \cap B) = 0.6 \times 0.4 = 0.24 \)
Now, we calculate the required conditional probability \( P(A/B) \):
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{0.24}{0.8} = 0.3 \).
In simple words: Find the shared probability of both events occurring first by multiplying the probability of event A with the conditional probability of B given A. Then, divide this shared probability by the probability of B.

Exam Tip: Be careful not to confuse the formulas for \( P(A/B) \) and \( P(B/A) \) - the denominator is always the probability of the event that has already occurred.

Question 2. Find P(A  B) if A and B are two events such that P(A) = 0.5, P(B) = 0.6 and P(A  B) = 0.8
Answer: We are given \( P(A) = 0.5 \), \( P(B) = 0.6 \), and \( P(A \cup B) = 0.8 \).
According to the addition theorem of probability:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substituting the given values into the equation:
\( 0.8 = 0.5 + 0.6 - P(A \cap B) \)
\( \implies 0.8 = 1.1 - P(A \cap B) \)
\( \implies P(A \cap B) = 1.1 - 0.8 = 0.3 \) (or \( \frac{3}{10} \)).
In simple words: To find the probability of both events happening together, add their individual probabilities and subtract the probability of either event happening.

Exam Tip: This addition formula is a fundamental concept in probability. Keep in mind that for mutually exclusive events, \( P(A \cap B) = 0 \).

Question 3. A soldier fires three bullets on enemy. The probability that the enemy will be killed by one bullet is 0.7. What is the probability that the enemy is still alive?
Answer: The probability that the enemy is killed by a single bullet is \( 0.7 \).
Therefore, the probability that the enemy survives a single bullet is:
\( 1 - 0.7 = 0.3 \)
Since the soldier fires three bullets and each shot is independent, the probability that the enemy survives all three shots is the product of the survival probabilities for each shot:
\( P(\text{still alive}) = 0.3 \times 0.3 \times 0.3 = (0.3)^3 = 0.027 \).
In simple words: First find the chance of the enemy surviving a single shot, which is 0.3. Then multiply this survival chance by itself three times for the three independent shots.

Exam Tip: When dealing with "at least one" or "survival" type questions, calculating the probability of the complementary event (failure in all trials) is often much simpler.

Question 4. What is the probability that a leap year has 53 Sundays?
Answer: A leap year contains 366 days.
Dividing this by 7 to find the number of weeks:
\( 366 = 52 \text{ weeks} \times 7 \text{ days/week} + 2 \text{ days} \)
Thus, a leap year has 52 complete weeks (which guarantees 52 Sundays) and 2 remaining days.
These 2 remaining days can form any of the following 7 consecutive pairs:
1. {Sunday, Monday}
2. {Monday, Tuesday}
3. {Tuesday, Wednesday}
4. {Wednesday, Thursday}
5. {Thursday, Friday}
6. {Friday, Saturday}
7. {Saturday, Sunday}
For the leap year to have 53 Sundays, one of these two remaining days must be a Sunday. The favorable pairs are {Saturday, Sunday} and {Sunday, Monday} (2 out of 7 possibilities).
\( \implies P(53 \text{ Sundays}) = \frac{2}{7} \).
In simple words: A leap year has 52 full weeks and 2 extra days. These 2 extra days can fall in 7 different ways, and 2 of those options contain a Sunday, giving a probability of 2 out of 7.

Exam Tip: Remember that a non-leap year has 365 days, leaving only 1 extra day, which results in a \( \frac{1}{7} \) probability of having 53 Sundays.

Question 5. 20 cards are numbered 1 to 20. One card is drawn at random. What is the probability that the number on the card will be a multiple of 4?
Answer: The total number of outcomes is 20, as the cards are numbered from 1 to 20.
Let \( E \) be the event that the selected card is a multiple of 4. The multiples of 4 in the range 1 to 20 are:
\( \{4, 8, 12, 16, 20\} \)
Number of favorable outcomes \( n(E) = 5 \).
The probability is given by:
\( P(E) = \frac{n(E)}{\text{Total outcomes}} = \frac{5}{20} = \frac{1}{4} \).
In simple words: List the multiples of 4 up to 20, which gives 5 numbers. Divide this count by the total number of cards, which is 20.

Exam Tip: Always list the favorable outcomes explicitly in your working to make it clear how you arrived at your numerator.

Question 6. Three coins are tossed once. Find the probability of getting at least one head.
Answer: When three coins are tossed once, the total number of possible outcomes in the sample space is:
\( 2^3 = 8 \)
These outcomes are: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
The only outcome where no heads are obtained is {TTT} (1 outcome).
Therefore, the number of outcomes with at least one head is:
\( 8 - 1 = 7 \)
The probability of getting at least one head is:
\( P(\text{at least one head}) = \frac{7}{8} \).
In simple words: There are 8 total combinations when flipping three coins. Only 1 combination has no heads at all, meaning the remaining 7 combinations have at least one head.

Exam Tip: Using the complement rule, \( P(\text{at least one}) = 1 - P(\text{none}) \), is the quickest and most reliable method to solve these types of problems.

Question 7. The probability that a student is not a swimmer is \( \frac{1}{5} \). Find the probability that out of 5 students, 4 are swimmers.
Answer: Let success be defined as a student being a swimmer.
The probability of failure (not a swimmer) is \( q = \frac{1}{5} \).
Therefore, the probability of success (being a swimmer) is:
\( p = 1 - q = 1 - \frac{1}{5} = \frac{4}{5} \)
Here, the number of independent trials \( n = 5 \). We want to find the probability of exactly \( r = 4 \) successes.
Using the Binomial Distribution formula \( P(X = r) = \binom{n}{r} p^r q^{n-r} \):
\( P(X = 4) = \binom{5}{4} \left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^{5-4} \)
\( \implies P(X = 4) = 5 \times \left(\frac{4}{5}\right)^4 \times \frac{1}{5} = \left(\frac{4}{5}\right)^4 \).
In simple words: The chance of a student being a swimmer is 4 out of 5. Using the binomial formula for choosing 4 swimmers out of 5, the calculation simplifies to 4 over 5 raised to the power of 4.

Exam Tip: Remember to simplify the combination term \( \binom{5}{4} = 5 \) and cancel it with the denominator of \( q = \frac{1}{5} \) to reach the final simplified answer.

Question 8. Find P(A/B), if P(B) = 0.5 and P(A  B) = 0.32
Answer: We are given \( P(B) = 0.5 \) and \( P(A \cap B) = 0.32 \).
The conditional probability of event A occurring given that event B has occurred is defined as:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substituting the given values:
\( P(A/B) = \frac{0.32}{0.5} = 0.64 \) (or \( \frac{16}{25} \)).
In simple words: Divide the probability of both events happening together (0.32) by the probability of event B happening (0.5).

Exam Tip: When dividing by 0.5, it is equivalent to multiplying the numerator by 2, which helps in quick mental calculations.

Question 9. A random variable X has the following probability distribution.

Find the value of k.
Answer: We know that for any valid probability distribution, the sum of all probabilities must equal 1:
\( \sum P(X) = 1 \)
\( \implies \frac{1}{15} + k + \frac{15k-2}{15} + k + \frac{15k-1}{15} + \frac{1}{15} = 1 \)
Multiplying the entire equation by 15 to clear the denominators:
\( 1 + 15k + (15k - 2) + 15k + (15k - 1) + 1 = 15 \)
Combining the like terms:
\( 60k - 1 = 15 \)
\( \implies 60k = 16 \)
\( \implies k = \frac{16}{60} = \frac{4}{15} \) (Note: The value \( k = \frac{1}{5} \) printed in the textbook key represents an alternate version of the problem; based on the printed table, the mathematically correct answer is \( \frac{4}{15} \)).
In simple words: The sum of all probabilities in a table must always equal 1. Set up an equation adding all values together, multiply by the common denominator, and solve for \( k \).

Exam Tip: Always make sure to check if individual probabilities are positive after solving for \( k \). For example, with \( k = \frac{4}{15} \), \( P(2) = \frac{2}{15} > 0 \), confirming the solution is valid.

Question 10. A random variable X, taking values 0, 1, 2 has the following probability distribution for some number k.
\[ P(X) = \begin{cases} k & \text{if } X = 0 \\ 2k & \text{if } X = 1 \\ 3k & \text{if } X = 2 \end{cases} \]
find k.
Answer: Since the total probability for any random variable is equal to 1:
\( P(X=0) + P(X=1) + P(X=2) = 1 \)
Substituting the given expressions:
\( k + 2k + 3k = 1 \)
\( \implies 6k = 1 \)
\( \implies k = \frac{1}{6} \).
In simple words: Add the three probabilities \( k \), \( 2k \), and \( 3k \) together and set the sum to 1, then solve for \( k \).

Exam Tip: This is a very common type of 1-mark question. Just sum the coefficients of \( k \) and equate to 1 to find the answer quickly.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 11. A problem in Mathematics is given to three students whose chances of solving it are \( \frac{1}{2} \), \( \frac{1}{3} \) and \( \frac{1}{4} \). What is the probability that the problem is solved.
Answer: Let \( A \), \( B \), and \( C \) be the independent events that the first, second, and third student solve the problem respectively.
We are given:
\( P(A) = \frac{1}{2} \implies P(A') = 1 - \frac{1}{2} = \frac{1}{2} \)
\( P(B) = \frac{1}{3} \implies P(B') = 1 - \frac{1}{3} = \frac{2}{3} \)
\( P(C) = \frac{1}{4} \implies P(C') = 1 - \frac{1}{4} = \frac{3}{4} \)
The problem is solved if at least one of the three students solves it. The complement of this event is that none of the students solve the problem:
\( P(\text{none solves}) = P(A' \cap B' \cap C') \)
Since the events are independent:
\( P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4} \)
Thus, the probability that the problem is solved is:
\( P(\text{problem is solved}) = 1 - P(\text{none solves}) = 1 - \frac{1}{4} = \frac{3}{4} \).
In simple words: Find the probability that no one is able to solve the problem by multiplying their individual failure rates. Subtract this value from 1 to find the chance that at least one person solves it.

Exam Tip: Be sure to explicitly mention that since \( A \), \( B \), and \( C \) are independent, their complements \( A' \), \( B' \), and \( C' \) are also independent events.

Question 12. A die is rolled. If the outcome is an even number, what is the probability that it is a prime?
Answer: Let \( S \) be the sample space of rolling a single die: \( S = \{1, 2, 3, 4, 5, 6\} \).
Let \( A \) be the event that the outcome is an even number: \( A = \{2, 4, 6\} \).
Let \( B \) be the event that the outcome is a prime number: \( B = \{2, 3, 5\} \).
The intersection event \( A \cap B \) is the set of even prime numbers: \( A \cap B = \{2\} \).
We want to find the conditional probability \( P(B/A) \):
\( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{n(A \cap B)}{n(A)} \)
Since there is 1 even prime number and 3 even numbers in total:
\( P(B/A) = \frac{1}{3} \ fielding \).
In simple words: The even outcomes on a die are 2, 4, and 6. Out of these three numbers, only the number 2 is prime, giving a probability of 1 out of 3.

Exam Tip: For conditional probability on simple spaces, you can find the answer directly by restricting the sample space to the given condition (even numbers) and counting the favorable outcomes.

Question 13. If A and B are two events such that \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{2} \) and \( P(A \cap B) = \frac{1}{8} \). Find P (not A and not B).
Answer: We need to find \( P(\text{not } A \text{ and not } B) \), which is \( P(A' \cap B') \).
According to De Morgan's Law:
\( A' \cap B' = (A \cup B)' \)
Therefore:
\( P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \)
First, we calculate \( P(A \cup B) \) using the addition theorem:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2 + 4 - 1}{8} = \frac{5}{8} \)
Now, substituting this value back:
\( P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8} \).
In simple words: Calculate the chance of either event happening, which is 5/8. Subtract this from 1 to find the probability of neither event happening, which is 3/8.

Exam Tip: Always state De Morgan's Law explicitly in your proof as it shows a clear logical progression to the examiner.

Question 14. In a class of 25 students with roll numbers 1 to 25, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of 5 or of 7.
Answer: The total number of students is 25, so the sample space size is \( n(S) = 25 \).
Let \( A \) be the event that the selected roll number is a multiple of 5. The multiples of 5 between 1 and 25 are:
\( A = \{5, 10, 15, 20, 25\} \implies n(A) = 5 \)
Let \( B \) be the event that the selected roll number is a multiple of 7. The multiples of 7 between 1 and 25 are:
\( B = \{7, 14, 21\} \implies n(B) = 3 \)
There is no common multiple of 5 and 7 in the range 1 to 25 (the smallest positive common multiple is 35). Thus:
\( A \cap B = \phi \implies n(A \cap B) = 0 \)
The probability that the roll number is a multiple of 5 or 7 is:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{25} + \frac{3}{25} - 0 = \frac{8}{25} \).
In simple words: Count the multiples of 5 (five numbers) and multiples of 7 (three numbers) up to 25. Since there are no overlapping numbers, add them together to get 8 favorable outcomes out of 25.

Exam Tip: Confirm that the intersection of the two sets is empty before adding the individual probabilities, to ensure the events are mutually exclusive.

Question 15. A can hit a target 4 times in 5 shots B three times in 4 shots and C twice in 3 shots. They fire a volley. What is the probability that atleast two shots hit.
Answer: Let \( A \), \( B \), and \( C \) be the independent events that \( A \), \( B \), and \( C \) hit the target respectively.
We are given:
\( P(A) = \frac{4}{5} \implies P(A') = 1 - \frac{4}{5} = \frac{1}{5} \)
\( P(B) = \frac{3}{4} \implies P(B') = 1 - \frac{3}{4} = \frac{1}{4} \)
\( P(C) = \frac{2}{3} \implies P(C') = 1 - \frac{2}{3} = \frac{1}{3} \)
We need the probability that at least two shots hit:
\( P(\text{at least two hits}) = P(\text{exactly two hits}) + P(\text{all three hit}) \)
Calculating the probability of exactly two hits:
- \( A \) and \( B \) hit, \( C \) misses: \( P(A \cap B \cap C') = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{12}{60} \)
- \( A \) and \( C \) hit, \( B \) misses: \( P(A \cap B' \cap C) = \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} = \frac{8}{60} \)
- \( B \) and \( C \) hit, \( A \) misses: \( P(A' \cap B \cap C) = \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{6}{60} \)
Sum of exactly two hits:
\( P(\text{exactly two hits}) = \frac{12}{60} + \frac{8}{60} + \frac{6}{60} = \frac{26}{60} \)
Calculating the probability that all three hit:
\( P(\text{all three hit}) = P(A \cap B \cap C) = \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{24}{60} \)
Thus:
\( P(\text{at least two hits}) = \frac{26}{60} + \frac{24}{60} = \frac{50}{60} = \frac{5}{6} \).
In simple words: Work out the chances of exactly two people hitting the target, and add the chance of all three hitting the target to find the total probability of at least two hits.

Exam Tip: Keeping a common denominator of 60 throughout your calculations makes adding the different independent cases together much simpler at the end.

Question 16. Two dice are thrown once. Find the probability of getting an even number on the first die or a total of 8.
Answer: When two dice are thrown, the total number of outcomes in the sample space is \( n(S) = 36 \).
Let \( A \) be the event of getting an even number on the first die. The first die can be 2, 4, or 6, and the second can be 1 to 6.
\( n(A) = 3 \text{ choices} \times 6 \text{ choices} = 18 \)
Let \( B \) be the event of getting a total sum of 8. The favorable outcomes are:
\( B = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \implies n(B) = 5 \)
The intersection \( A \cap B \) represents outcomes with an even number on the first die AND a sum of 8:
\( A \cap B = \{(2, 6), (4, 4), (6, 2)\} \implies n(A \cap B) = 3 \)
Using the addition theorem of probability:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} = \frac{20}{36} = \frac{5}{9} \).
In simple words: Add the 18 combinations starting with an even number to the 5 combinations summing to 8. Subtract the 3 overlapping combinations that were counted twice to get the final probability of 5/9.

Exam Tip: Always double check for any overlap between your two events to avoid double-counting favorable outcomes.

Question 17. A and B throw a die alternatively till one of them throws a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts the game.
Answer: Let \( S \) denote getting a '6' (success) and \( F \) denote not getting a '6' (failure) in a single throw:
\( P(S) = \frac{1}{6} \), and \( P(F) = \frac{5}{6} \)
Since A starts first, A can win on the 1st, 3rd, 5th, etc., throws.
\( P(\text{A wins}) = P(S) + P(FFS) + P(FFFFS) + \dots \)
\( \implies P(\text{A wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4 \left(\frac{1}{6}\right) + \dots \)
This is an infinite geometric progression with first term \( a = \frac{1}{6} \) and common ratio \( r = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \).
Using the sum formula \( S_{\infty} = \frac{a}{1 - r} \):
\( P(\text{A wins}) = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11} \)
Since the game must eventually be won by either player, the probability that B wins is:
\( P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{5}{11} \).
In simple words: Player A can win on any of their turn cycles. Calculating this sum as an infinite geometric series shows that A has a 6/11 chance of winning, while B has a 5/11 chance.

Exam Tip: Going first in an alternating game always provides an advantage. Ensure your calculated probability for the first player is greater than \( \frac{1}{2} \).

Question 18. If A and B are events such that \( P(A) = \frac{1}{2} \), \( P(A \cup B) = \frac{3}{5} \) and P(B) = p find p if events
(i) are mutually exclusive,
(ii) are independent.
Answer: We are given \( P(A) = \frac{1}{2} \), \( P(A \cup B) = \frac{3}{5} \), and \( P(B) = p \).
We use the relation:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
(i) If the events are mutually exclusive, then \( P(A \cap B) = 0 \):
\( \frac{3}{5} = \frac{1}{2} + p - 0 \)
\( \implies p = \frac{3}{5} - \frac{1}{2} = \frac{6 - 5}{10} = \frac{1}{10} \)
(ii) If the events are independent, then \( P(A \cap B) = P(A) \times P(B) = \frac{1}{2}p \):
\( \frac{3}{5} = \frac{1}{2} + p - \frac{1}{2}p \)
\( \implies \frac{3}{5} - \frac{1}{2} = \frac{1}{2}p \)
\( \implies \frac{1}{10} = \frac{1}{2}p \implies p = \frac{2}{10} = \frac{1}{5} \).
In simple words: If the events cannot happen together, the probability is 1/10. If they are completely independent of each other, the probability is 1/5.

Exam Tip: Clearly state the defining mathematical condition for both mutually exclusive and independent events before carrying out the algebraic substitutions.

Question 19. A man takes a step forward with probability 0.4 and backward with probability 0.6. Find the probability that at the end of eleven steps he is one step away from the starting point.
Answer: Let \( X \) be the number of forward steps, and \( Y \) be the number of backward steps.
The total number of steps is:
\( X + Y = 11 \)
Being one step away from the starting point means the difference between forward and backward steps is exactly 1:
\( |X - Y| = 1 \)
This gives us two cases:
Case 1: \( X - Y = 1 \implies (X+Y) + (X-Y) = 11 + 1 \implies 2X = 12 \implies X = 6 \text{ and } Y = 5 \)
Case 2: \( Y - X = 1 \implies X = 5 \text{ and } Y = 6 \)
Let the probability of a forward step (success) be \( p = 0.4 \) and a backward step (failure) be \( q = 0.6 \).
Using the Binomial Distribution formula:
\( P(X=6) = \binom{11}{6} (0.4)^6 (0.6)^5 \)
\( P(X=5) = \binom{11}{5} (0.4)^5 (0.6)^6 \)
Since \( \binom{11}{6} = \binom{11}{5} \):
\( P(\text{one step away}) = P(X=6) + P(X=5) = \binom{11}{5} (0.4)^5 (0.6)^5 [0.4 + 0.6] \)
\( \implies P(\text{one step away}) = \binom{11}{5} (0.24)^5 \times 1 = 462 \times (0.24)^5 \approx 0.3678 \).
In simple words: To end up one step away after 11 steps, the man must take either 6 forward steps and 5 backward steps, or 5 forward and 6 backward. Calculate the probability of both scenarios using binomial theory.

Exam Tip: Combine the common terms in both cases to simplify the calculation, as \( 0.4 + 0.6 = 1 \) makes evaluating the expression much faster.

Question 20. Two cards are drawn from a pack of well shuffled 52 cards one by one with replacement. Getting an ace or a spade is considered a success. Find the probability distribution for the number of successes.
Answer: Let us first find the probability of success (drawing an ace or a spade) in a single draw.
In a deck of 52 cards, there are 13 spade cards and 4 ace cards. Since one of these aces is the Ace of Spades (which is counted in both groups), the total number of cards that are either an ace or a spade is:
\( 13 + 4 - 1 = 16 \)
The probability of success \( p \) is:
\( p = \frac{16}{52} = \frac{4}{13} \)
The probability of failure \( q \) is:
\( q = 1 - p = \frac{9}{13} \)
Let the random variable \( X \) represent the number of successes when 2 cards are drawn. \( X \) can take values 0, 1, or 2.
- \( P(X = 0) = q \times q = \left(\frac{9}{13}\right)^2 = \frac{81}{169} \)
- \( P(X = 1) = 2 \times p \times q = 2 \left(\frac{4}{13}\right)\left(\frac{9}{13}\right) = \frac{72}{169} \)
- \( P(X = 2) = p \times p = \left(\frac{4}{13}\right)^2 = \frac{16}{169} \)
Thus, the probability distribution of \( X \) is:

In simple words: The chance of drawing a spade or an ace is 4 out of 13. When drawing two cards with replacement, calculate the probability of getting 0, 1, or 2 successful draws to form the table.

Exam Tip: Always double check that the sum of the probabilities in your final distribution table is exactly 1: \( \frac{81 + 72 + 16}{169} = \frac{169}{169} = 1 \).

Question 21. In a game, a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/looses.
Answer: Let \( W \) be the random variable representing the net amount won or lost by the player.
The probability of getting a six (success) is \( \frac{1}{6} \), and the probability of getting any other number (failure) is \( \frac{5}{6} \).
Let us determine the possible outcomes, net gains, and their probabilities:
1. **Wins on the 1st throw:** He rolls a six, wins Rs. 1 and quits.
\( W = 1 \implies P(W = 1) = \frac{1}{6} \)
2. **Wins on the 2nd throw:** He rolls a non-six on the 1st throw, and a six on the 2nd. He loses Rs. 1 first, then wins Rs. 1.
\( W = -1 + 1 = 0 \implies P(W = 0) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36} \)
3. **Wins on the 3rd throw:** Non-six on the 1st and 2nd, and a six on the 3rd. He loses Rs. 2 first, then wins Rs. 1.
\( W = -2 + 1 = -1 \implies P(W = -1) = \left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{25}{216} \)
4. **Does not roll a six in 3 throws:** Non-six on all three attempts. He loses Rs. 3 and the game ends.
\( W = -3 \implies P(W = -3) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \)
Now, we calculate the expected value \( E(W) \):
\( E(W) = \sum W_i P(W_i) \)
\( \implies E(W) = 1\left(\frac{36}{216}\right) + 0\left(\frac{30}{216}\right) - 1\left(\frac{25}{216}\right) - 3\left(\frac{125}{216}\right) \)
\( \implies E(W) = \frac{36 - 25 - 375}{216} = \frac{-364}{216} = -\frac{91}{54} \approx -1.69 \) rupees.
In simple words: Work out the net winnings and probabilities for each stage of the game. Multiply each possible winning/losing value by its probability and sum them to find the average expected outcome, which is a loss of Rs. 1.69.

Exam Tip: Convert all terms to have a common denominator (like 216) before calculating the expected value to avoid errors when working with fractions.

Question 22. Suppose that 10% of men and 5% of women have grey hair. A grey haired person is selected at random. What is the probability that the selected person is male assuming that there are 60% males and 40% females.
Answer: Let \( M \) be the event that the selected person is a male, and \( F \) be the event that the selected person is a female.
We are given:
\( P(M) = 60\% = 0.60 \)
\( P(F) = 40\% = 0.40 \)
Let \( G \) be the event that the selected person has grey hair. The conditional probabilities are:
\( P(G/M) = 10\% = 0.10 \)
\( P(G/F) = 5\% = 0.05 \)
We need to find the probability that the selected grey-haired person is male, which is \( P(M/G) \). Using Bayes' Theorem:
\( P(M/G) = \frac{P(M) P(G/M)}{P(M) P(G/M) + P(F) P(G/F)} \)
Substituting the values:
\( P(M/G) = \frac{0.60 \times 0.10}{0.60 \times 0.10 + 0.40 \times 0.05} = \frac{0.06}{0.06 + 0.02} = \frac{0.06}{0.08} = \frac{3}{4} = 0.75 \).
In simple words: Use Bayes' Theorem to divide the probability of being a grey-haired male by the total probability of anyone being grey-haired. This simplifies to 3/4 or 75%.

Exam Tip: Bayes' Theorem questions are very common in exams. Clearly define each event and write out the theorem's formula before performing any numerical calculations.

Question 23. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn. What is the probability that they both are diamonds?
Answer: Let \( E_1 \) be the event that the lost card is a diamond, and \( E_2 \) be the event that the lost card is not a diamond.
\( P(E_1) = \frac{13}{52} = \frac{1}{4} \)
\( P(E_2) = \frac{39}{52} = \frac{3}{4} \)
Let \( A \) be the event that the two drawn cards from the remaining 51 cards are both diamonds.
- **Case 1: If the lost card was a diamond (event \( E_1 \))**, there are 12 diamonds left in the pack.
\( P(A/E_1) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550} \)
- **Case 2: If the lost card was not a diamond (event \( E_2 \))**, there are 13 diamonds left in the pack.
\( P(A/E_2) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550} \)
According to the law of total probability:
\( P(A) = P(E_1) P(A/E_1) + P(E_2) P(A/E_2) \)
\( \implies P(A) = \frac{1}{4} \left(\frac{132}{2550}\right) + \frac{3}{4} \left(\frac{156}{2550}\right) \)
\( \implies P(A) = \frac{132 + 468}{4 \times 2550} = \frac{600}{10200} = \frac{1}{17} \).
In simple words: The lost card could either be a diamond or another suit. Calculate the probability of drawing two diamonds in both scenarios, weight them by the chance of that card being lost, and sum them to get 1/17.

Exam Tip: Simplify the fraction at the very end to save time and prevent mathematical calculation errors in the intermediate steps.

Question 24. Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Answer: Let \( p \) be the probability of selecting a defective egg in a single draw:
\( p = 10\% = 0.1 \)
Therefore, the probability of selecting a non-defective egg is:
\( q = 1 - p = 0.9 \)
Let \( X \) be the random variable representing the number of defective eggs drawn in \( n = 10 \) trials.
We need to find the probability of getting at least one defective egg:
\( P(X \ge 1) = 1 - P(X = 0) \)
Using the binomial probability for \( X = 0 \):
\( P(X = 0) = \binom{10}{0} p^0 q^{10} = (0.9)^{10} \)
Thus:
\( P(X \ge 1) = 1 - (0.9)^{10} \).
In simple words: The chance of getting no defective eggs in 10 tries is \( 0.9^{10} \). Subtract this from 1 to find the chance of getting at least one defective egg.

Exam Tip: It is standard practice to leave exponents of decimals (like \( 0.9^{10} \)) unexpanded in final answers, as the exact decimal value is extremely tedious to calculate by hand.

Question 25. Find the variance of the number obtained on a throw of an unbiased die.
Answer: Let \( X \) be the discrete random variable denoting the number obtained on a throw of a die: \( X \in \{1, 2, 3, 4, 5, 6\} \).
Since the die is unbiased, each outcome has an equal probability:
\( P(X = i) = \frac{1}{6} \text{ for } i = 1, 2, \dots, 6 \)
First, find the mean (expectation) of \( X \):
\( E(X) = \sum X_i P(X_i) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2} \)
Next, find \( E(X^2) \):
\( E(X^2) = \sum X_i^2 P(X_i) = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} \)
\( \implies E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \)
Now, we calculate the variance \( \sigma^2 \):
\( \text{Var}(X) = E(X^2) - [E(X)]^2 \)
\( \implies \text{Var}(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4} \)
\( \implies \text{Var}(X) = \frac{182 - 147}{12} = \frac{35}{12} \).
In simple words: Find the average value of rolling a die, which is 3.5. Then find the average of the squared values, and subtract the square of the average to get the variance of 35/12.

Exam Tip: Memorize the standard variance formula \( \text{Var}(X) = E(X^2) - [E(X)]^2 \), as it is used in almost all variance problems.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 26. In a hurdle race, a player has to cross 8 hurdles. The probability that he will clear a hurdle is \( \frac{4}{5} \), what is the probability that he will knock down in fewer than 2 hurdles?
Answer: Let success be defined as the player knocking down a hurdle.
The probability of clearing a hurdle is \( \frac{4}{5} \), so the probability of success \( p \) is:
\( p = 1 - \frac{4}{5} = \frac{1}{5} \)
The probability of failure \( q \) is:
\( q = \frac{4}{5} \)
Here, the number of independent trials \( n = 8 \). We need to find the probability of knocking down fewer than 2 hurdles, i.e., \( P(X < 2) \):
\( P(X < 2) = P(X = 0) + P(X = 1) \)
Using the Binomial Distribution formula \( P(X = r) = \binom{n}{r} p^r q^{n-r} \):
- For \( X = 0 \):
\( P(X = 0) = \binom{8}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^8 = \left(\frac{4}{5}\right)^8 \)
- For \( X = 1 \):
\( P(X = 1) = \binom{8}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^7 = 8 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^7 = 2 \times \frac{4}{5} \times \left(\frac{4}{5}\right)^7 = 2 \left(\frac{4}{5}\right)^8 \)
Adding these together:
\( P(X < 2) = \left(\frac{4}{5}\right)^8 + 2 \left(\frac{4}{5}\right)^8 = 3 \left(\frac{4}{5}\right)^8 \).
In simple words: The chance of knocking down a hurdle is 1/5. Calculate the chance of knocking down exactly 0 or 1 hurdles in 8 trials, and add them up to find the final probability.

Exam Tip: Writing \( 8 \times \frac{1}{5} \) as \( 2 \times \frac{4}{5} \) is a useful algebra step that allows you to factor out \( \left(\frac{4}{5}\right)^8 \) directly.

Question 27. Bag A contains 4 red, 3 white and 2 black balls. Bag B contains 3 red, 2 white and 3 black balls. One ball is transferred from bag A to bag B and then a ball is drawn from bag B. The ball so drawn is found to be red. Find the probability that the transferred ball is black.
Answer: Let us define the events for the transferred ball from Bag A (which has 4 red, 3 white, 2 black, total 9 balls):
- \( E_1 \): Transferred ball is red \( \implies P(E_1) = \frac{4}{9} \)
- \( E_2 \): Transferred ball is white \( \implies P(E_2) = \frac{3}{9} \)
- \( E_3 \): Transferred ball is black \( \implies P(E_3) = \frac{2}{9} \)
Let \( R \) be the event that a red ball is drawn from Bag B (which initially has 3 red, 2 white, 3 black, total 8 balls).
We calculate the conditional probabilities:
- If a red ball is transferred, Bag B has 4 red out of 9: \( P(R/E_1) = \frac{4}{9} \)
- If a white ball is transferred, Bag B has 3 red out of 9: \( P(R/E_2) = \frac{3}{9} \)
- If a black ball is transferred, Bag B has 3 red out of 9: \( P(R/E_3) = \frac{3}{9} \)
We need to find \( P(E_3/R) \) using Bayes' Theorem:
\( P(E_3/R) = \frac{P(E_3) P(R/E_3)}{P(E_1)P(R/E_1) + P(E_2)P(R/E_2) + P(E_3)P(R/E_3)} \)
\( \implies P(E_3/R) = \frac{\frac{2}{9} \times \frac{3}{9}}{\frac{4}{9} \times \frac{4}{9} + \frac{3}{9} \times \frac{3}{9} + \frac{2}{9} \times \frac{3}{9}} \)
\( \implies P(E_3/R) = \frac{6}{16 + 9 + 6} = \frac{6}{31} \).
In simple words: The transferred ball could be red, white, or black. Find the chance of drawing a red ball in each case, and use Bayes' Theorem to determine that the probability of the transferred ball being black is 6/31.

Exam Tip: Be sure to write the updated counts of the balls in Bag B for each case clearly, as this is where most mistakes occur.

Question 28. If a fair coin is tossed 10 times, find the probability of getting.
(i) exactly six heads,
(ii) at least six heads,
(iii) at most six heads.
Answer: This is a binomial experiment with \( n = 10 \), and the probability of success (heads) is \( p = \frac{1}{2} \) and failure (tails) is \( q = \frac{1}{2} \).
(i) Exactly 6 heads:
\( P(X = 6) = \binom{10}{6} \left(\frac{1}{2}\right)^{10} = \frac{210}{1024} = \frac{105}{512} \)
(ii) At least 6 heads:
\( P(X \ge 6) = \sum_{r=6}^{10} \binom{10}{r} \left(\frac{1}{2}\right)^{10} \)
\( \implies P(X \ge 6) = \left[ \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \right] \frac{1}{1024} \)
\( \implies P(X \ge 6) = [210 + 120 + 45 + 10 + 1] \frac{1}{1024} = \frac{386}{1024} = \frac{193}{512} \)
(iii) At most 6 heads:
\( P(X \le 6) = 1 - P(X \ge 7) \)
\( \implies P(X \le 6) = 1 - \left[ \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \right] \frac{1}{1024} \)
\( \implies P(X \le 6) = 1 - \frac{120 + 45 + 10 + 1}{1024} = 1 - \frac{176}{1024} = \frac{848}{1024} = \frac{53}{64} \).
In simple words:
(i) Use the binomial formula for exactly 6 successes to get 105/512.
(ii) Sum the outcomes for 6, 7, 8, 9, and 10 heads to get 193/512.
(iii) Subtract the outcomes for 7 or more heads from 1 to find the probability of at most 6 heads, which is 53/64.

Exam Tip: For "at most" calculations, using \( 1 - P(\text{unwanted events}) \) saves you from adding up 7 different positive terms directly.

Question 29. A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter by other means of transport are respectively \( \frac{3}{13} \), \( \frac{1}{5} \), \( \frac{1}{10} \) and \( \frac{2}{5} \). The probabilities that he will be late are \( \frac{1}{4} \), \( \frac{1}{3} \), and \( \frac{1}{12} \) if he comes by train, bus and scooter respectively but if comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?
Answer: Let \( E_1, E_2, E_3, \) and \( E_4 \) be the events that the doctor travels by train, bus, scooter, and other means of transport respectively:
\( P(E_1) = \frac{3}{13}, \; P(E_2) = \frac{1}{5}, \; P(E_3) = \frac{1}{10}, \; P(E_4) = \frac{2}{5} \)
Let \( L \) be the event that the doctor arrives late. The conditional probabilities are:
\( P(L/E_1) = \frac{1}{4}, \; P(L/E_2) = \frac{1}{3}, \; P(L/E_3) = \frac{1}{12}, \; P(L/E_4) = 0 \)
We need to find the probability that he came by train given that he is late, which is \( P(E_1/L) \). Using Bayes' Theorem:
\( P(E_1/L) = \frac{P(E_1) P(L/E_1)}{\sum P(E_i) P(L/E_i)} \)
Numerator:
\( P(E_1) P(L/E_1) = \frac{3}{13} \times \frac{1}{4} = \frac{3}{52} \)
Denominator:
\( P(E_1) P(L/E_1) + P(E_2) P(L/E_2) + P(E_3) P(L/E_3) + P(E_4) P(L/E_4) \)
\( = \frac{3}{52} + \left(\frac{1}{5} \times \frac{1}{3}\right) + \left(\frac{1}{10} \times \frac{1}{12}\right) + 0 = \frac{3}{52} + \frac{1}{15} + \frac{1}{120} = \frac{207}{1560} \)
Thus:
\( P(E_1/L) = \frac{3/52}{207/1560} = \frac{3}{52} \times \frac{1560}{207} = \frac{10}{23} \).
In simple words: Find the probability that the doctor is late on the train, and divide it by the total overall probability of being late across all modes of transport. This simplifies to 10/23.

Exam Tip: Notice that \( P(L/E_4) = 0 \) because the doctor is never late when using other means of transport, which simplifies your denominator calculation.

Question 30. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is six. Find the probability that it is actually a six.
Answer: Let \( E \) be the event that the die shows a six, and \( E' \) be the event that it does not show a six:
\( P(E) = \frac{1}{6} \), and \( P(E') = \frac{5}{6} \)
Let \( S \) be the event that the man reports a six.
- **He reports six when it actually is a six** (he is speaking the truth):
\( P(S/E) = \frac{3}{4} \)
- **He reports six when it is not a six** (he is telling a lie):
\( P(S/E') = 1 - \frac{3}{4} = \frac{1}{4} \)
We need to find \( P(E/S) \) using Bayes' Theorem:
\( P(E/S) = \frac{P(E) P(S/E)}{P(E) P(S/E) + P(E') P(S/E')} \)
\( \implies P(E/S) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{3 + 5} = \frac{3}{8} \).
In simple words: The man can report a six either by telling the truth when it is a six, or by lying when it is some other number. Calculating these ratios shows there is a 3/8 probability that it is indeed a six.

Exam Tip: Remember that if the die does not show a six, the probability of him lying and reporting a six is still \( \frac{1}{4} \), not any other fraction.

Question 31. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively one of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer: Let the events \( E_1, E_2, \) and \( E_3 \) represent selecting a scooter driver, a car driver, and a truck driver respectively.
Total number of drivers \( = 2000 + 4000 + 6000 = 12000 \).
\( P(E_1) = \frac{2000}{12000} = \frac{1}{6} \)
\( P(E_2) = \frac{4000}{12000} = \frac{1}{3} \)
\( P(E_3) = \frac{6000}{12000} = \frac{1}{2} \)
Let \( A \) be the event that the driver meets with an accident:
\( P(A/E_1) = 0.01, \; P(A/E_2) = 0.03, \; P(A/E_3) = 0.15 \)
We need to find the probability that the driver is a scooter driver given that they had an accident, which is \( P(E_1/A) \). Using Bayes' Theorem:
\( P(E_1/A) = \frac{P(E_1) P(A/E_1)}{P(E_1) P(A/E_1) + P(E_2) P(A/E_2) + P(E_3) P(A/E_3)} \)
\( \implies P(E_1/A) = \frac{\frac{1}{6} \times 0.01}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15} = \frac{0.01}{0.01 + 0.06 + 0.45} = \frac{0.01}{0.52} = \frac{1}{52} \).
In simple words: Find the weighted probability of a scooter accident and divide it by the total overall accident rate across all categories, yielding 1/52.

Exam Tip: Be sure to keep the denominators aligned when substituting into the fraction, as decimal expansion can lead to calculation errors.

Question 32. Two cards from a pack of 52 cards are lost. One card is drawn from the remaining cards. If drawn card is heart, find the probability that the lost cards were both hearts.
Answer: Let \( E_1 \) be the event that both lost cards were hearts, \( E_2 \) be the event that one was a heart and one was not, and \( E_3 \) be the event that neither was a heart.
Total cards = 52, Hearts = 13, Non-hearts = 39.
\( P(E_1) = \frac{\binom{13}{2}}{\binom{52}{2}} = \frac{13 \times 12}{52 \times 51} = \frac{6}{34 \times 3} = \frac{6}{102} \)
\( P(E_2) = \frac{13 \times 39}{\binom{52}{2}} = \frac{13 \times 39}{26 \times 51} = \frac{39}{102} \)
\( P(E_3) = \frac{\binom{39}{2}}{\binom{52}{2}} = \frac{39 \times 38}{52 \times 51} = \frac{57}{102} \)
Let \( H \) be the event that a heart is drawn from the remaining 50 cards.
- If two hearts are lost, 11 hearts remain: \( P(H/E_1) = \frac{11}{50} \)
- If one heart is lost, 12 hearts remain: \( P(H/E_2) = \frac{12}{50} \)
- If no hearts are lost, 13 hearts remain: \( P(H/E_3) = \frac{13}{50} \)
Using Bayes' Theorem:
\( P(E_1/H) = \frac{P(E_1) P(H/E_1)}{P(E_1)P(H/E_1) + P(E_2)P(H/E_2) + P(E_3)P(H/E_3)} \)
\( \implies P(E_1/H) = \frac{6 \times 11}{6 \times 11 + 39 \times 12 + 57 \times 13} = \frac{66}{66 + 468 + 741} = \frac{66}{1275} = \frac{22}{425} \).
In simple words: Determine the probabilities of different heart loss scenarios, weight them by the remaining hearts in each case, and use Bayes' Theorem to find that the probability is 22/425.

Exam Tip: This is a challenging Bayes' Theorem question. Write down each case of combinations \( \binom{n}{r} \) explicitly to make sure your coefficients are correct.

Question 33. A box X contains 2 white and 3 red balls and a bag Y contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag Y.
Answer: Let \( E_1 \) and \( E_2 \) be the events of choosing box X and bag Y respectively.
Since a bag is chosen at random:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( R \) be the event of drawing a red ball.
- Probability of red from Box X (3 red out of 5): \( P(R/E_1) = \frac{3}{5} \)
- Probability of red from Bag Y (5 red out of 9): \( P(R/E_2) = \frac{5}{9} \)
Using Bayes' Theorem:
\( P(E_2/R) = \frac{P(E_2) P(R/E_2)}{P(E_1) P(R/E_1) + P(E_2) P(R/E_2)} \)
\( \implies P(E_2/R) = \frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}} = \frac{5/9}{3/5 + 5/9} = \frac{5/9}{52/45} = \frac{25}{52} \).
In simple words: Find the probability of choosing bag Y and getting a red ball, and divide it by the total overall probability of getting a red ball from either container.

Exam Tip: Be sure to cancel the common factor of \( \frac{1}{2} \) from both the numerator and denominator before carrying out the final fraction addition.

Question 34. In answering a question on a multiple choice, a student either knows the answer or guesses. Let \( \frac{3}{4} \) be the probability that he knows the answer and \( \frac{1}{4} \) be the probability that he guesses. Assuming that a student who guesses at the answer will be incorrect with probability \( \frac{1}{4} \). What is the probability that the student knows the answer, given that he answered correctly?
Answer: Let \( E_1 \) be the event that the student knows the answer, and \( E_2 \) be the event that he guesses:
\( P(E_1) = \frac{3}{4} \), and \( P(E_2) = \frac{1}{4} \)
Let \( C \) be the event that the student answers correctly.
- If he knows the answer, he will definitely answer correctly: \( P(C/E_1) = 1 \)
- If he guesses, the probability of being incorrect is \( \frac{1}{4} \), so the probability of being correct is:
\( P(C/E_2) = 1 - \frac{1}{4} = \frac{3}{4} \)
We need to find \( P(E_1/C) \) using Bayes' Theorem:
\( P(E_1/C) = \frac{P(E_1) P(C/E_1)}{P(E_1) P(C/E_1) + P(E_2) P(C/E_2)} \)
\( \implies P(E_1/C) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{3}{4}} = \frac{3/4}{3/4 + 3/16} = \frac{3/4}{15/16} = \frac{4}{5} \)
(Note: If using the standard textbook variant where guessing has a success rate of \( \frac{1}{4} \), the calculation yields \( \frac{12}{13} \)).
In simple words: Find the probability that the student answered correctly by knowing the answer, and divide it by the total probability of answering correctly either by knowing or guessing.

Exam Tip: Read carefully whether the question specifies the guessing success rate or the guessing failure rate, as this is a common point of difference in exam papers.

Question 35. Suppose a girl throws a die. If she gets 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head. What is the probability that she throws 1, 2, 3 or 4 with the die?
Answer: Let \( E_1 \) be the event of getting 5 or 6 on the die, and \( E_2 \) be the event of getting 1, 2, 3, or 4 on the die.
\( P(E_1) = \frac{2}{6} = \frac{1}{3} \)
\( P(E_2) = \frac{4}{6} = \frac{2}{3} \)
Let \( H \) be the event of getting exactly one head.
- **If she gets 5 or 6**, she tosses the coin 3 times. The probability of getting exactly one head is:
\( P(H/E_1) = \binom{3}{1} \left(\frac{1}{2}\right)^3 = \frac{3}{8} \)
- **If she gets 1, 2, 3, or 4**, she tosses the coin once. The probability of getting exactly one head is:
\( P(H/E_2) = \frac{1}{2} \)
We need to find \( P(E_2/H) \) using Bayes' Theorem:
\( P(E_2/H) = \frac{P(E_2) P(H/E_2)}{P(E_1) P(H/E_1) + P(E_2) P(H/E_2)} \)
\( \implies P(E_2/H) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{1/3}{1/8 + 1/3} = \frac{1/3}{11/24} = \frac{8}{11} \).
In simple words: The coin flipping rules depend on the die roll. Find the chance of getting one head in both paths, and calculate the probability that the roll was 1, 2, 3, or 4, which is 8/11.

Exam Tip: Be sure to use the binomial expression \( \binom{3}{1} \left(\frac{1}{2}\right)^3 \) to calculate the probability of getting exactly one head in three tosses.

Question 36. In a bolt factory machines A, B and C manufacture 60%, 30% and 10% of the total bolts respectively, 2%, 5% and 10% of the bolts produced by them respectively are defective. A bolt is picked up at random from the product and is found to be defective. What is the probability that it has been manufactured by machine A?
Answer: Let \( E_1, E_2, \) and \( E_3 \) be the events that a bolt is produced by machine A, B, and C respectively:
\( P(E_1) = 0.60, \; P(E_2) = 0.30, \; P(E_3) = 0.10 \)
Let \( D \) be the event that the selected bolt is defective:
\( P(D/E_1) = 0.02, \; P(D/E_2) = 0.05, \; P(D/E_3) = 0.10 \)
We need to find the probability that a defective bolt came from machine A, which is \( P(E_1/D) \). Using Bayes' Theorem:
\( P(E_1/D) = \frac{P(E_1) P(D/E_1)}{P(E_1)P(D/E_1) + P(E_2)P(D/E_2) + P(E_3)P(D/E_3)} \)
\( \implies P(E_1/D) = \frac{0.60 \times 0.02}{0.60 \times 0.02 + 0.30 \times 0.05 + 0.10 \times 0.10} \)
\( \implies P(E_1/D) = \frac{0.012}{0.012 + 0.015 + 0.010} = \frac{0.012}{0.037} = \frac{12}{37} \).
In simple words: Use Bayes' Theorem to find the probability of a defective bolt being made by machine A. Divide the rate of defective bolts from A by the total rate of defectives, giving 12/37.

Exam Tip: Expressing the decimal probabilities as whole numbers after multiplying by 1000 makes the final fraction simplification much easier.

Question 37. Two urns A and B contain 6 black and 4 white, 4 black and 6 white balls respectively. Two balls are drawn from one of the urns. If both the balls drawn are white, find the probability that the balls are drawn from urn B.
Answer: Let \( E_1 \) be the event of selecting urn A, and \( E_2 \) be the event of selecting urn B:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( W \) be the event of drawing two white balls.
- **From Urn A** (4 white out of 10):
\( P(W/E_1) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} = \frac{2}{15} \)
- **From Urn B** (6 white out of 10):
\( P(W/E_2) = \frac{\binom{6}{2}}{\binom{10}{2}} = \frac{15}{45} = \frac{5}{15} = \frac{1}{3} \)
We need to find \( P(E_2/W) \) using Bayes' Theorem:
\( P(E_2/W) = \frac{P(E_2) P(W/E_2)}{P(E_1)P(W/E_1) + P(E_2)P(W/E_2)} \)
\( \implies P(E_2/W) = \frac{\frac{1}{2} \times \frac{5}{15}}{\frac{1}{2} \times \frac{2}{15} + \frac{1}{2} \times \frac{5}{15}} = \frac{5}{2 + 5} = \frac{5}{7} \).
In simple words: Find the probability of getting two white balls from urn B, and divide it by the total combined probability of getting two white balls from either urn.

Exam Tip: Be sure to write the formula for combinations \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to show your work for finding the probability of drawing multiple balls without replacement.

Question 38. Two cards are drawn from a well shuffled pack of 52 cards. Find the mean and variance for the number of face cards obtained.
Answer: Let \( X \) be the random variable denoting the number of face cards obtained. There are 12 face cards and 40 non-face cards in a pack of 52.
We evaluate both standard interpretations of this problem (with and without replacement) below:
Case I: Drawing without replacement (simultaneous drawing)
\( X \) can take values 0, 1, 2.
- \( P(X=0) = \frac{\binom{40}{2}}{\binom{52}{2}} = \frac{40 \times 39}{52 \times 51} = \frac{130}{221} \)
- \( P(X=1) = \frac{12 \times 40}{\binom{52}{2}} = \frac{480}{1326} = \frac{80}{221} \)
- \( P(X=2) = \frac{\binom{12}{2}}{\binom{52}{2}} = \frac{12 \times 11}{52 \times 51} = \frac{11}{221} \)
Evaluating the mean and variance:
- \( \text{Mean } E(X) = 0\left(\frac{130}{221}\right) + 1\left(\frac{80}{221}\right) + 2\left(\frac{11}{221}\right) = \frac{102}{221} = \frac{6}{13} \)
- \( E(X^2) = 0^2\left(\frac{130}{221}\right) + 1^2\left(\frac{80}{221}\right) + 2^2\left(\frac{11}{221}\right) = \frac{124}{221} \)
- \( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{124}{221} - \left(\frac{6}{13}\right)^2 = \frac{1000}{2873} \).
Case II: Drawing with replacement
Success probability is \( p = \frac{12}{52} = \frac{3}{13} \), failure is \( q = \frac{10}{13} \).
For \( n = 2 \) independent trials:
- \( \text{Mean } E(X) = np = 2 \times \frac{3}{13} = \frac{6}{13} \)
- \( \text{Variance} = npq = 2 \times \frac{3}{13} \times \frac{10}{13} = \frac{60}{169} \).
In simple words: Find the probability distribution of drawing face cards from a deck. Calculate the expectation and variance using both the independent and dependent card-drawing methods.

Exam Tip: Clearly state whether you are performing the calculations with or without replacement, as the variance value differs between the two methods.

Question 39. Write the probability distribution for the number of heads obtained when three coins are tossed together. Also, find the mean and variance of the number of heads.
Answer: Let \( X \) be the random variable representing the number of heads obtained in three tosses of a fair coin.
\( X \) can take values 0, 1, 2, or 3.
The probability of getting a head is \( p = \frac{1}{2} \), and a tail is \( q = \frac{1}{2} \).
Using the Binomial Distribution formula with \( n = 3 \):
- \( P(X = 0) = \binom{3}{0} p^0 q^3 = \frac{1}{8} \)
- \( P(X = 1) = \binom{3}{1} p^1 q^2 = \frac{3}{8} \)
- \( P(X = 2) = \binom{3}{2} p^2 q^1 = \frac{3}{8} \)
- \( P(X = 3) = \binom{3}{3} p^3 q^0 = \frac{1}{8} \)
Now, we calculate the mean and variance:
- **Mean \( E(X) \)**:
\( E(X) = np = 3 \times \frac{1}{2} = 1.5 \) (or \( \frac{3}{2} \))
- **Variance \( \text{Var}(X) \)**:
\( \text{Var}(X) = npq = 3 \times \frac{1}{2} \times \frac{1}{2} = 0.75 \) (or \( \frac{3}{4} \)).
The probability distribution of \( X \) is:

In simple words: Set up the probability table for the number of heads when tossing three coins. The average number of heads is 1.5, and the variance is 0.75.

Exam Tip: Since tossing coins constitutes Bernoulli trials, using the simplified formulas \( \text{Mean} = np \) and \( \text{Var} = npq \) is much faster than computing \( \sum X_i P(X_i) \) and \( \sum X_i^2 P(X_i) \).

Question 40. Two groups are competing for the position on the Board of Directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer: Let \( E_1 \) be the event that the first group wins, and \( E_2 \) be the event that the second group wins:
\( P(E_1) = 0.6 \), and \( P(E_2) = 0.4 \)
Let \( N \) be the event that a new product is introduced. The conditional probabilities are:
\( P(N/E_1) = 0.7 \)
\( P(N/E_2) = 0.3 \)
We need to find the probability that the new product was introduced by the second group, which is \( P(E_2/N) \). Using Bayes' Theorem:
\( P(E_2/N) = \frac{P(E_2) P(N/E_2)}{P(E_1) P(N/E_1) + P(E_2) P(N/E_2)} \)
Substituting the values:
\( P(E_2/N) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = \frac{2}{9} \).
In simple words: Use Bayes' Theorem to divide the probability of the second group winning and launching the product (0.12) by the total probability of the product launching under either group (0.54), giving 2/9.

Exam Tip: Expressing your final fraction as \( \frac{2}{9} \) is preferred over writing a repeating decimal like \( 0.222\dots \).