Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.2

 

Question 1. A family has two children. What is the probability that both children are girls, given that at least one of them is a girl?
Answer: Let B represent a boy and G represent a girl. The possible outcomes for two children are \( S = \{BG, GB, BB, GG\} \). So, the total number of outcomes is \( n(S) = 4 \).
Let E be the event that both children are girls. Then \( E = \{GG\} \), so \( n(E) = 1 \).
Let F be the event that at least one of them is a girl. Then \( F = \{BG, GB, GG\} \), so \( n(F) = 3 \).
The probability of event F is \( P(F) = \frac{n(F)}{n(S)} = \frac{3}{4} \).
The intersection of events E and F, \( E \cap F \), is when both children are girls AND at least one is a girl, which is just \( \{GG\} \). So, \( n(E \cap F) = 1 \).
The probability of this intersection is \( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{4} \).
We need to find the conditional probability that both children are girls, given that at least one is a girl, which is \( P(E/F) \).
Using the formula for conditional probability: \( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{1}{4}}{\frac{3}{4}} \)
\( \implies P(E/F) = \frac{1}{3} \)
In simple words: First, list all the possible pairs of children (like boy-girl). Then, find the number of ways to have both girls and the number of ways to have at least one girl. Divide the probability of having both girls (which also means at least one girl) by the probability of having at least one girl.

🎯 Exam Tip: Clearly define your sample space and events before applying conditional probability formulas to avoid confusion between \( P(A/B) \) and \( P(B/A) \).

 

Question 2. A die is thrown twice and the sum of the number appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
Answer: When a die is thrown twice, the total number of outcomes (sample space S) is \( n(S) = 36 \). Here is the list of all possible outcomes:
\( S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \)
\( \quad (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \)
\( \quad (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \)
\( \quad (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \)
\( \quad (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \)
\( \quad (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} \)
Let A be the event that the sum of the numbers appearing is 6. The outcomes for A are:
\( A = \{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)\} \). So, \( n(A) = 5 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{5}{36} \).
Let B be the event that the number 4 has appeared at least once. The outcomes for B are:
\( B = \{(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)\} \). So, \( n(B) = 11 \).
The intersection of events A and B, \( A \cap B \), is when the sum is 6 AND 4 has appeared at least once. The outcomes are:
\( A \cap B = \{(2, 4), (4, 2)\} \). So, \( n(A \cap B) = 2 \).
The probability of this intersection is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} \).
We need to find the conditional probability of B given A, which is \( P(B/A) \). This means, what is the probability that 4 appeared at least once, given that the sum is 6.
Using the formula for conditional probability: \( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{\frac{2}{36}}{\frac{5}{36}} \)
\( \implies P(B/A) = \frac{2}{5} \)
In simple words: First, list all the ways two dice can sum to 6. Then, from that list, pick out only the ones that have a '4' in them. Divide how many pairs have a '4' by the total number of pairs that sum to 6.

🎯 Exam Tip: When dealing with conditional probability, always clearly define the reduced sample space (the "given" event) before calculating probabilities for the desired event.

 

Question 3. An unbiased die is thrown twice. Let the event A be an odd number on the first throw and B the event odd number on the second throw. Check whether A and B events are independent.
Answer: When an unbiased die is thrown twice, the total number of outcomes (sample space S) is \( n(S) = 36 \).
Let A be the event of getting an odd number on the first throw. The outcomes for A are:
\( A = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \)
\( \quad (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \)
\( \quad (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \). So, \( n(A) = 18 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{18}{36} = \frac{1}{2} \).
Let B be the event of getting an odd number on the second throw. The outcomes for B are:
\( B = \{(1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), \)
\( \quad (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5), \)
\( \quad (5, 1), (5, 3), (5, 5), (6, 1), (6, 3), (6, 5)\} \). So, \( n(B) = 18 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{18}{36} = \frac{1}{2} \).
The intersection of events A and B, \( A \cap B \), is when both the first and second throws are odd numbers. The outcomes are:
\( A \cap B = \{(1, 1), (1, 3), (1, 5), \)
\( \quad (3, 1), (3, 3), (3, 5), \)
\( \quad (5, 1), (5, 3), (5, 5)\} \). So, \( n(A \cap B) = 9 \).
The probability of this intersection is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{9}{36} = \frac{1}{4} \).
To check for independence, we compare \( P(A \cap B) \) with \( P(A) \cdot P(B) \).
\( P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \).
Since \( P(A \cap B) = P(A) \cdot P(B) \) (both equal \( \frac{1}{4} \)), events A and B are independent.
In simple words: We find the chance of getting an odd number on the first roll and the chance of getting an odd number on the second roll. Then, we find the chance of both happening. If the chance of both happening is the same as multiplying their individual chances, then the two rolls don't affect each other.

🎯 Exam Tip: Remember that two events A and B are independent if and only if \( P(A \cap B) = P(A) \cdot P(B) \). Always calculate both sides of this equation to verify independence.

 

Question 4. Probability of solving a 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 the problem is (i) solved, (ii) exactly one of them solves the problem.
Answer: Given probabilities are \( P(A) = \frac{1}{2} \) (A solves the problem) and \( P(B) = \frac{1}{3} \) (B solves the problem).
Since A and B work independently, the probability that A does NOT solve the problem is \( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \).
Similarly, the probability that B does NOT solve the problem is \( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \).

(i) The probability that the problem is solved:
The problem is solved if at least one of them solves it. This is \( P(A \cup B) \).
It can also be found by subtracting the probability that *none* of them solve it from 1.
\( P(\text{problem is solved}) = 1 - P(\text{none of them solve the problem}) \)
\( \implies = 1 - P(\overline{A} \cap \overline{B}) \)
Since A and B are independent, \( \overline{A} \) and \( \overline{B} \) are also independent.
\( \implies = 1 - P(\overline{A}) \cdot P(\overline{B}) \)
\( \implies = 1 - (\frac{1}{2}) (\frac{2}{3}) \)
\( \implies = 1 - \frac{2}{6} \)
\( \implies = 1 - \frac{1}{3} \)
\( \implies = \frac{2}{3} \)

(ii) The probability that exactly one of them solves the problem:
This means either A solves it and B doesn't, OR B solves it and A doesn't.
\( P(\text{exactly one solves}) = P(A \cap \overline{B}) + P(\overline{A} \cap B) \)
Due to independence, this becomes:
\( \implies = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B) \)
\( \implies = (\frac{1}{2}) (\frac{2}{3}) + (\frac{1}{2}) (\frac{1}{3}) \)
\( \implies = \frac{2}{6} + \frac{1}{6} \)
\( \implies = \frac{3}{6} \)
\( \implies = \frac{1}{2} \)
In simple words: For part (i), we figure out the chance that neither person solves the problem and take that away from 1. For part (ii), we add up two separate chances: first, the chance that A solves it and B does not, and second, the chance that B solves it and A does not.

🎯 Exam Tip: When events are independent, the probability of their intersection is the product of their individual probabilities. This simplifies calculations for "none" or "exactly one" scenarios.

 

Question 5. Suppose one person is selected at random from a group of 100 persons as given in the following table. What is the probability that the man selected is a Psychologist?

Answer: We are given a table showing the distribution of psychologists, socialists, and democrats among men and women.
The total number of men in the group is 50.
The number of men who are psychologists is 15.
We need to find the probability that a man selected is a psychologist, which is a conditional probability. This means we are only considering the group of men.
The probability of selecting a psychologist, given that the selected person is a man, is given by:
\( P(\text{Psychologist} / \text{Man}) = \frac{\text{Number of men who are Psychologists}}{\text{Total number of men}} \)
\( \implies = \frac{15}{50} \)
\( \implies = \frac{3}{10} \)
In simple words: To find the chance that a chosen man is a psychologist, just look at the 'Men' row in the table. Divide the number of men who are psychologists by the total number of men.

🎯 Exam Tip: For conditional probability problems involving tables, first identify the relevant subset (the "given" condition) and then calculate the probability within that subset.

 

Question 6. Two urns contain the set of balls as given in the following table. One ball is drawn from each urn and find the probability that (i) both balls are red, (ii) both balls are of the same colour.

Answer: First, let's find the total number of balls in each urn.
For Urn 1: Total balls = \( 10 (\text{White}) + 6 (\text{Red}) + 9 (\text{Black}) = 25 \) balls.
For Urn 2: Total balls = \( 3 (\text{White}) + 7 (\text{Red}) + 15 (\text{Black}) = 25 \) balls.
Since one ball is drawn from each urn, the events are independent.

(i) Probability that both balls are red:
Let A be the event of drawing a red ball from Urn 1. \( P(A) = \frac{\text{Number of Red balls in Urn 1}}{\text{Total balls in Urn 1}} = \frac{6}{25} \).
Let B be the event of drawing a red ball from Urn 2. \( P(B) = \frac{\text{Number of Red balls in Urn 2}}{\text{Total balls in Urn 2}} = \frac{7}{25} \).
Since the events are independent, the probability that both balls are red is \( P(A \cap B) = P(A) \cdot P(B) \).
\( \implies = \frac{6}{25} \times \frac{7}{25} \)
\( \implies = \frac{42}{625} \)

(ii) Probability that both balls are of the same colour:
This means either both are white, or both are red, or both are black. These are mutually exclusive events.
Let \( W_1, R_1, B_1 \) be drawing white, red, black from Urn 1, and \( W_2, R_2, B_2 \) be drawing white, red, black from Urn 2.
\( P(W_1) = \frac{10}{25} \), \( P(R_1) = \frac{6}{25} \), \( P(B_1) = \frac{9}{25} \).
\( P(W_2) = \frac{3}{25} \), \( P(R_2) = \frac{7}{25} \), \( P(B_2) = \frac{15}{25} \).
\( P(\text{same colour}) = P(W_1 \cap W_2) + P(R_1 \cap R_2) + P(B_1 \cap B_2) \).
Since draws from different urns are independent:
\( \implies = P(W_1) \cdot P(W_2) + P(R_1) \cdot P(R_2) + P(B_1) \cdot P(B_2) \)
\( \implies = (\frac{10}{25}) (\frac{3}{25}) + (\frac{6}{25}) (\frac{7}{25}) + (\frac{9}{25}) (\frac{15}{25}) \)
\( \implies = \frac{30}{625} + \frac{42}{625} + \frac{135}{625} \)
\( \implies = \frac{30 + 42 + 135}{625} \)
\( \implies = \frac{207}{625} \)
In simple words: For part (i), multiply the chance of getting a red ball from the first urn by the chance of getting a red ball from the second urn. For part (ii), calculate the chance for each color (both white, both red, both black) and then add those chances together.

🎯 Exam Tip: Remember to calculate the total number of items in each category first. When events are independent (like drawing from separate urns), probabilities multiply. When events are mutually exclusive (like "both white" or "both red"), probabilities add.

 

Question 7. Bag I contains 3 Red and 4 Black balls while Bag II contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag I.

Answer: This is a problem using Bayes' Theorem.
Let \( E_1 \) be the event of choosing Bag I, and \( E_2 \) be the event of choosing Bag II.
Since a bag is chosen at random, \( P(E_1) = P(E_2) = \frac{1}{2} \).
Let A be the event of drawing a red ball.
From Bag I, there are 3 red balls out of a total of \( 3+4=7 \) balls. So, the probability of drawing a red ball given that Bag I was chosen is \( P(A/E_1) = \frac{3}{7} \).
From Bag II, there are 5 red balls out of a total of \( 5+6=11 \) balls. So, the probability of drawing a red ball given that Bag II was chosen is \( P(A/E_2) = \frac{5}{11} \).
We want to find the probability that the ball was drawn from Bag I, given that it is red. This is \( P(E_1/A) \).
Using Bayes' Theorem:
\( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \)
\( \implies = \frac{\frac{1}{2} \times \frac{3}{7}}{(\frac{1}{2} \times \frac{3}{7}) + (\frac{1}{2} \times \frac{5}{11})} \)
\( \implies = \frac{\frac{3}{14}}{\frac{3}{14} + \frac{5}{22}} \)
To add the fractions in the denominator, find a common denominator, which is \( 154 \).
\( \implies = \frac{\frac{3}{14}}{\frac{3 \times 11}{14 \times 11} + \frac{5 \times 7}{22 \times 7}} \)
\( \implies = \frac{\frac{3}{14}}{\frac{33}{154} + \frac{35}{154}} \)
\( \implies = \frac{\frac{3}{14}}{\frac{33+35}{154}} \)
\( \implies = \frac{\frac{3}{14}}{\frac{68}{154}} \)
Now, invert the denominator and multiply:
\( \implies = \frac{3}{14} \times \frac{154}{68} \)
Since \( 154 = 14 \times 11 \), we can simplify:
\( \implies = \frac{3 \times 11}{68} \)
\( \implies = \frac{33}{68} \)
In simple words: Imagine you pick a bag and then a ball. We want to find the chance that you picked Bag I, knowing that the ball you found was red. We use a formula that combines the chance of picking each bag with the chance of getting a red ball from that bag.

🎯 Exam Tip: Bayes' Theorem is crucial for these types of inverse probability problems. Always clearly define the prior probabilities (choosing a bag) and the conditional probabilities (drawing a specific color from each bag) before applying the formula.

 

Question 8. The first of three urns contains 7 White and 10 Black balls, the second contains 5 White and 12 Black balls, and the third contains 17 White balls and no Blackball. A person chooses an urn at random and draws a ball from it. And the ball is found to be white. Find the probabilities that the ball comes from (i) the first urn, (ii) the second urn, (iii) the third urn.

Answer: Let \( E_1, E_2, E_3 \) be the events of choosing the first, second, and third urns, respectively.
Since an urn is chosen at random, \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
Let A be the event of drawing a white ball.
From Urn I: 7 white balls out of 17 total. So, \( P(A/E_1) = \frac{7}{17} \).
From Urn II: 5 white balls out of 17 total. So, \( P(A/E_2) = \frac{5}{17} \).
From Urn III: 17 white balls out of 17 total. So, \( P(A/E_3) = \frac{17}{17} = 1 \).
We need to find the probability that the ball came from a specific urn, given that it is white. This requires Bayes' Theorem.

(i) Probability that the ball comes from the first urn, given it is white \( P(E_1/A) \):
\( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \)
\( \implies = \frac{\frac{1}{3} \times \frac{7}{17}}{(\frac{1}{3} \times \frac{7}{17}) + (\frac{1}{3} \times \frac{5}{17}) + (\frac{1}{3} \times \frac{17}{17})} \)
\( \implies = \frac{\frac{7}{51}}{\frac{7}{51} + \frac{5}{51} + \frac{17}{51}} \)
\( \implies = \frac{\frac{7}{51}}{\frac{7+5+17}{51}} \)
\( \implies = \frac{\frac{7}{51}}{\frac{29}{51}} \)
\( \implies = \frac{7}{29} \)

(ii) Probability that the ball comes from the second urn, given it is white \( P(E_2/A) \):
\( P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \)
The denominator is the same as in part (i), which is \( \frac{29}{51} \).
\( \implies = \frac{\frac{1}{3} \times \frac{5}{17}}{\frac{29}{51}} \)
\( \implies = \frac{\frac{5}{51}}{\frac{29}{51}} \)
\( \implies = \frac{5}{29} \)

(iii) Probability that the ball comes from the third urn, given it is white \( P(E_3/A) \):
\( P(E_3/A) = \frac{P(E_3)P(A/E_3)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \)
The denominator is again \( \frac{29}{51} \).
\( \implies = \frac{\frac{1}{3} \times \frac{17}{17}}{\frac{29}{51}} \)
\( \implies = \frac{\frac{17}{51}}{\frac{29}{51}} \)
\( \implies = \frac{17}{29} \)
In simple words: This problem asks us to find the chance that a white ball came from Urn 1, Urn 2, or Urn 3, after we already know it's white. We use Bayes' Theorem for each urn. It compares the chance of picking that urn and getting a white ball, to the total chance of getting a white ball from any urn.

🎯 Exam Tip: For problems with multiple conditional events, calculate the total probability of the observed event (the denominator in Bayes' Theorem) once, and then use it for all subsequent conditional probability calculations. Check that the sum of the conditional probabilities for each urn equals 1.

 

Question 9. Three boxes \( B_1, B_2, B_3 \) contain lamp bulbs some of which are defective. The defective proportions in box \( B_1 \), box \( B_2 \) and box \( B_3 \) are respectively \( \frac{1}{2} \), \( \frac{1}{3} \) and \( \frac{1}{4} \). A box is selected at random and a bulb is drawn from it. If the selected bulb is found to be defective, what is the probability that box \( B_1 \) was selected?
Answer: Let \( B_1, B_2, B_3 \) represent the events of selecting Box 1, Box 2, and Box 3, respectively.
Since a box is selected at random, the probability of choosing each box is equal:
\( P(B_1) = P(B_2) = P(B_3) = \frac{1}{3} \).
Let A be the event of selecting a defective bulb.
The probability of drawing a defective bulb given it's from Box 1 is \( P(A/B_1) = \frac{1}{2} \).
The probability of drawing a defective bulb given it's from Box 2 is \( P(A/B_2) = \frac{1}{3} \).
The probability of drawing a defective bulb given it's from Box 3 is \( P(A/B_3) = \frac{1}{4} \).
We need to find the probability that Box 1 was selected, given that the drawn bulb is defective. This is \( P(B_1/A) \).
Using Bayes' Theorem:
\( P(B_1/A) = \frac{P(B_1)P(A/B_1)}{P(B_1)P(A/B_1) + P(B_2)P(A/B_2) + P(B_3)P(A/B_3)} \)
\( \implies = \frac{\frac{1}{3} \times \frac{1}{2}}{(\frac{1}{3} \times \frac{1}{2}) + (\frac{1}{3} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{4})} \)
\( \implies = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{9} + \frac{1}{12}} \)
To sum the fractions in the denominator, find a common denominator, which is 36.
\( \implies = \frac{\frac{1}{6}}{\frac{6}{36} + \frac{4}{36} + \frac{3}{36}} \)
\( \implies = \frac{\frac{1}{6}}{\frac{6+4+3}{36}} \)
\( \implies = \frac{\frac{1}{6}}{\frac{13}{36}} \)
Now, invert the denominator and multiply:
\( \implies = \frac{1}{6} \times \frac{36}{13} \)
\( \implies = \frac{6}{13} \)
This can also be expressed as a decimal: \( \frac{6}{13} \approx 0.4615 \).
In simple words: We want to know the chance that a faulty bulb came from Box 1. We use a formula that compares the chance of picking Box 1 and getting a faulty bulb, to the total chance of getting a faulty bulb from any of the three boxes.

🎯 Exam Tip: Carefully list all given probabilities and the event you are trying to find the probability for. The denominator in Bayes' Theorem represents the total probability of the observed event (drawing a defective bulb in this case) across all possible initial conditions.

 

Question 10. Three horses A, B, C are in the race. A is twice as likely to win as B and B is twice as likely to win as C. What are their respective probabilities of winning?
Answer: Let \( P(A), P(B), P(C) \) be the probabilities that horses A, B, and C win, respectively.
We are given that A is twice as likely to win as B: \( P(A) = 2 \cdot P(B) \).
We are also given that B is twice as likely to win as C: \( P(B) = 2 \cdot P(C) \).
From the second relation, we can write \( P(C) = \frac{1}{2} P(B) \).
Now, substitute \( P(B) \) into the first relation:
\( P(A) = 2 \cdot (2 \cdot P(C)) = 4 \cdot P(C) \).
So, we have the ratios: \( P(A) : P(B) : P(C) = 4 : 2 : 1 \).
Since these are the only three horses, their probabilities must sum to 1:
\( P(A) + P(B) + P(C) = 1 \).
Let \( P(C) = k \). Then \( P(B) = 2k \) and \( P(A) = 4k \).
Substitute these into the sum equation:
\( 4k + 2k + k = 1 \)
\( \implies 7k = 1 \)
\( \implies k = \frac{1}{7} \)
Now we can find the individual probabilities:
\( P(A) = 4k = 4 \cdot \frac{1}{7} = \frac{4}{7} \)
\( P(B) = 2k = 2 \cdot \frac{1}{7} = \frac{2}{7} \)
\( P(C) = k = \frac{1}{7} \)
So, the probabilities of winning for A, B, and C are \( \frac{4}{7}, \frac{2}{7}, \frac{1}{7} \) respectively. This shows how their chances are proportioned based on the given likelihoods.
In simple words: We are told how much more likely each horse is to win compared to the next. We use these relationships to set up ratios for their chances. Since only one horse can win, all their chances must add up to 1. We use this to find the actual probability for each horse.

🎯 Exam Tip: When given proportional likelihoods, assign a variable (e.g., k) to the smallest probability and express others in terms of k. Remember that the sum of probabilities for all possible outcomes must always equal 1.

 

Question 11. A die is thrown. Find the probability of getting (i) a prime number, (ii) a number greater than or equal to 3.
Answer: When a die is thrown, the sample space (S) is the set of all possible outcomes:
\( S = \{1, 2, 3, 4, 5, 6\} \). The total number of outcomes is \( n(S) = 6 \).

(i) Probability of getting a prime number:
Let A be the event of getting a prime number. Prime numbers on a die are numbers greater than 1 that are only divisible by 1 and themselves. These are 2, 3, and 5.
So, \( A = \{2, 3, 5\} \). The number of outcomes in A is \( n(A) = 3 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \).

(ii) Probability of getting a number greater than or equal to 3:
Let B be the event of getting a number greater than or equal to 3. These numbers include 3, 4, 5, and 6.
So, \( B = \{3, 4, 5, 6\} \). The number of outcomes in B is \( n(B) = 4 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{6} = \frac{2}{3} \).
In simple words: For part (i), count how many prime numbers are on a die (numbers like 2, 3, 5) and divide by the total numbers. For part (ii), count how many numbers are 3 or bigger (3, 4, 5, 6) and divide by the total numbers.

🎯 Exam Tip: Make sure you correctly identify prime numbers (a common mistake is including 1 or even numbers other than 2) and correctly interpret "greater than or equal to".

 

Question 12. Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly, and then one card is drawn randomly. If it is known that the number on the drawn card is more than 4. What is the probability that it is an even number?
Answer: The sample space (S) is the set of numbers on the cards: \( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \). The total number of outcomes is \( n(S) = 10 \).
Let A be the event that the number on the drawn card is more than 4. This is the "given" condition.
So, \( A = \{5, 6, 7, 8, 9, 10\} \). The number of outcomes in A is \( n(A) = 6 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{6}{10} \).
Let B be the event of getting an even number. The even numbers in the full sample space are:
\( B = \{2, 4, 6, 8, 10\} \).
We need to find the probability that the number is even, given that it is more than 4. This is \( P(B/A) \).
First, find the intersection of A and B, \( A \cap B \), which means the number is both more than 4 AND even.
\( A \cap B = \{6, 8, 10\} \). The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 3 \).
The probability of this intersection is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{10} \).
Now, apply the conditional probability formula:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{\frac{3}{10}}{\frac{6}{10}} \)
\( \implies P(B/A) = \frac{3}{6} \)
\( \implies P(B/A) = \frac{1}{2} \)
In simple words: First, only look at cards with numbers higher than 4. Then, from these cards, count how many are even numbers. The answer is the count of even numbers divided by the count of numbers higher than 4.

🎯 Exam Tip: When a condition is "given", it effectively reduces your sample space. For this problem, you can think of the new sample space as just the numbers greater than 4, and then find the even numbers within that smaller group.

 

Question 13. There are 1000 students in a school out of which 450 are girls. It is known that out of 450, 20% of the girls studying in class XI. A student is randomly selected from 1000 students. What is the probability that the selected student is from class XI given that the selected student is a girl?
Answer: The total number of students in the school is \( n(S) = 1000 \).
The total number of girls is 450.
Let G be the event that the selected student is a girl. So, \( P(G) = \frac{\text{Number of girls}}{\text{Total students}} = \frac{450}{1000} \).
Let C be the event that the selected student is from class XI.
We are told that 20% of the girls are studying in Class XI. This means the number of girls in Class XI is \( 20\% \text{ of } 450 \).
Number of girls in Class XI \( = \frac{20}{100} \times 450 = 90 \).
Let \( C \cap G \) be the event that the selected student is a girl AND from Class XI. So, \( n(C \cap G) = 90 \).
The probability of this event is \( P(C \cap G) = \frac{n(C \cap G)}{n(S)} = \frac{90}{1000} \).
We need to find the probability that the selected student is from Class XI, given that the selected student is a girl. This is \( P(C/G) \).
Using the conditional probability formula:
\( P(C/G) = \frac{P(C \cap G)}{P(G)} \)
\( \implies = \frac{\frac{90}{1000}}{\frac{450}{1000}} \)
\( \implies = \frac{90}{450} \)
\( \implies = \frac{1}{5} \)
\( \implies = 0.2 \)
In simple words: We want to know the chance that a student is in Class 11, but only if we already know that student is a girl. We take the number of girls in Class 11 and divide it by the total number of girls.

🎯 Exam Tip: When dealing with percentages, convert them to decimal or fractional form before calculation. For "given that" problems, focus on the reduced sample space (the condition) for your denominator.

 

Question 14. From a pack of 52 cards, two cards are drawn at random. Find the probability that one is a king and the other is a queen.
Answer: A standard pack of 52 cards has 4 kings and 4 queens.
The total number of ways to draw 2 cards from 52 cards is given by the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \).
Total ways to draw 2 cards \( = C(52, 2) = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \).
We want to draw one king AND one queen.
Number of ways to choose 1 king from 4 kings \( = C(4, 1) = 4 \).
Number of ways to choose 1 queen from 4 queens \( = C(4, 1) = 4 \).
The number of ways to draw one king and one queen is the product of these combinations (since they are independent choices):
Number of favorable outcomes \( = C(4, 1) \times C(4, 1) = 4 \times 4 = 16 \).
The probability is the ratio of favorable outcomes to the total possible outcomes:
\( P(\text{1 King and 1 Queen}) = \frac{\text{Number of favorable outcomes}}{\text{Total ways to draw 2 cards}} \)
\( \implies = \frac{16}{1326} \)
\( \implies = \frac{8}{663} \)
As a decimal, this is approximately 0.012.
In simple words: First, find all the different ways you can pick any two cards from the deck. Then, find all the ways you can pick one king and one queen. Divide the second number by the first number to get the probability.

🎯 Exam Tip: Remember to use combinations (\( C(n,k) \)) when the order of selection does not matter. The "and" in probability often implies multiplication of possibilities.

 

Question 15. A card is drawn from a pack of playing cards and then another card is drawn without the first being replaced. What is the probability of drawing (i) two aces, (ii) two spades?
Answer: A standard pack has 52 cards. When a card is drawn and not replaced, the total number of cards for the second draw changes.

(i) Probability of drawing two aces:
There are 4 aces in a pack of 52 cards.
Probability of drawing the first ace \( = \frac{4}{52} \).
After drawing one ace, there are 3 aces left and 51 cards remaining in total.
Probability of drawing the second ace (given the first was an ace) \( = \frac{3}{51} \).
The probability of drawing two aces consecutively without replacement is:
\( P(\text{Two Aces}) = \frac{4}{52} \times \frac{3}{51} \)
\( \implies = \frac{12}{2652} \)
\( \implies = \frac{1}{221} \)
Alternatively, using combinations:
Total ways to choose 2 cards from 52 \( = C(52, 2) = \frac{52 \times 51}{2 \times 1} = 1326 \).
Ways to choose 2 aces from 4 aces \( = C(4, 2) = \frac{4 \times 3}{2 \times 1} = 6 \).
\( P(\text{Two Aces}) = \frac{6}{1326} = \frac{1}{221} \).

(ii) Probability of drawing two spades:
There are 13 spades in a pack of 52 cards.
Probability of drawing the first spade \( = \frac{13}{52} \).
After drawing one spade, there are 12 spades left and 51 cards remaining.
Probability of drawing the second spade (given the first was a spade) \( = \frac{12}{51} \).
The probability of drawing two spades consecutively without replacement is:
\( P(\text{Two Spades}) = \frac{13}{52} \times \frac{12}{51} \)
\( \implies = \frac{156}{2652} \)
\( \implies = \frac{13}{221} \)
Alternatively, using combinations:
Total ways to choose 2 cards from 52 \( = C(52, 2) = 1326 \).
Ways to choose 2 spades from 13 spades \( = C(13, 2) = \frac{13 \times 12}{2 \times 1} = 78 \).
\( P(\text{Two Spades}) = \frac{78}{1326} = \frac{13}{221} \).
In simple words: For both parts, first find the chance of picking the first specific card (an ace or a spade). Then, because you don't put the card back, there's one less card of that kind and one less total card. Find the chance of picking the second specific card from the smaller group, and multiply these two chances together.

🎯 Exam Tip: When drawing cards "without replacement", the denominator (total cards) and the numerator (number of specific cards) decrease by one for each subsequent draw. Both direct multiplication and combination methods are valid; choose the one you find easier.

 

Question 16. A company has three machines A, B, C which produce 20%, 30%, and 50% of the product respectively. Their respective defective percentages are 7%, 3%, and 5%. From a randomly chosen product, one item is selected and inspected. If it is found to be defective, what is the probability that it has been made by machine C?
Answer: Let \( P(A), P(B), P(C) \) be the probabilities that a product comes from machine A, B, or C, respectively.
Given: \( P(A) = 20\% = \frac{20}{100} \)
\( P(B) = 30\% = \frac{30}{100} \)
\( P(C) = 50\% = \frac{50}{100} \)
Let D be the event that a selected product is defective.
The defective percentages are conditional probabilities:
\( P(D/A) = 7\% = \frac{7}{100} \) (probability of defective given from machine A)
\( P(D/B) = 3\% = \frac{3}{100} \) (probability of defective given from machine B)
\( P(D/C) = 5\% = \frac{5}{100} \) (probability of defective given from machine C)
We need to find the probability that the defective product was made by machine C, which is \( P(C/D) \).
Using Bayes' Theorem:
\( P(C/D) = \frac{P(C)P(D/C)}{P(A)P(D/A) + P(B)P(D/B) + P(C)P(D/C)} \)
\( \implies = \frac{\frac{50}{100} \times \frac{5}{100}}{(\frac{20}{100} \times \frac{7}{100}) + (\frac{30}{100} \times \frac{3}{100}) + (\frac{50}{100} \times \frac{5}{100})} \)
\( \implies = \frac{\frac{250}{10000}}{\frac{140}{10000} + \frac{90}{10000} + \frac{250}{10000}} \)
\( \implies = \frac{\frac{250}{10000}}{\frac{140+90+250}{10000}} \)
\( \implies = \frac{250}{140+90+250} \)
\( \implies = \frac{250}{480} \)
\( \implies = \frac{25}{48} \)
As a decimal, this is approximately 0.5208.
In simple words: We want to know the chance that a faulty product came from Machine C. We use a special formula called Bayes' Theorem. It compares the chance of Machine C making a product and that product being faulty, to the total chance of any machine making a faulty product.

🎯 Exam Tip: Always convert percentages to decimals or fractions at the beginning of the calculation to avoid errors. Bayes' Theorem is used when you know the outcome (defective product) and want to find the probability of a specific cause (which machine made it).

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