CBSE Class 12 Mathematics Probability VBQs Set A

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, \( P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{8/35}{3/14 + 8/35} = \frac{8/35}{31/70} = \frac{16}{31} \).