NCERT Solutions Class-12-Chapter-13-Probability
Excercise-13.2
Note:
Multiplication Theorem: \( P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F) \).
Independent Events: Two events E and F are independent if \( P(E \cap F) = P(E) \cdot P(F) \).
If E and F are independent, then \( P(F|E) = P(F) \) and \( P(E|F) = P(E) \).
Q1
If \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{1}{5} \), find \( P(A \cap B) \) if A and B are independent events.
Solution:
Since A and B are independent events:
\( P(A \cap B) = P(A) \cdot P(B) \).
\( P(A \cap B) = \frac{3}{5} \cdot \frac{1}{5} = \frac{3}{25} \).
Q2
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution:
Total cards = 52. Total Black cards = 26.
Let A be the event that the first card is black.
\( P(A) = \frac{26}{52} = \frac{1}{2} \).
Let B be the event that the second card is black. Since the card is not replaced, there are now 51 cards total and 25 black cards left.
\( P(B|A) = \frac{25}{51} \).
Probability that both are black \( P(A \cap B) = P(A) \cdot P(B|A) \).
\( = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102} \).
Q3
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
Total oranges = 15. Good = 12. Bad = 3.
The box is approved if all 3 drawn oranges are good.
P(1st Good) = \( \frac{12}{15} \).
P(2nd Good | 1st Good) = \( \frac{11}{14} \) (since drawn without replacement).
P(3rd Good | 1st & 2nd Good) = \( \frac{10}{13} \).
Required Probability = \( \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13} \).
\( = \frac{4 \times 11 \times 10}{5 \times 14 \times 13} = \frac{44}{91} \).
Q4
A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.
Solution:
Sample space S = { (H,1), (H,2)...(H,6), (T,1)...(T,6) }. Total = 12 outcomes.
Event A (Head) = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6) }. \( P(A) = \frac{6}{12} = \frac{1}{2} \).
Event B (3 on die) = { (H,3), (T,3) }. \( P(B) = \frac{2}{12} = \frac{1}{6} \).
\( A \cap B \) (Head and 3) = { (H,3) }. \( P(A \cap B) = \frac{1}{12} \).
Check Independence: \( P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \).
Since \( P(A \cap B) = P(A) \cdot P(B) \), events A and B are independent.
Q5
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, 'the number is even,' and B be the event, 'the number is red'. Are A and B independent?
Solution:
Sample space S = {1, 2, 3, 4, 5, 6}.
A (Even) = {2, 4, 6}. \( P(A) = \frac{3}{6} = \frac{1}{2} \).
B (Red) = {1, 2, 3}. \( P(B) = \frac{3}{6} = \frac{1}{2} \).
\( A \cap B \) (Even and Red) = {2}. \( P(A \cap B) = \frac{1}{6} \).
Check: \( P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
Since \( \frac{1}{6} \neq \frac{1}{4} \), A and B are not independent.
Q6
Let E and F be events with \( P(E) = \frac{3}{5}, P(F) = \frac{3}{10} \) and \( P(E \cap F) = \frac{1}{5} \). Are E and F independent?
Solution:
Calculate \( P(E) \cdot P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50} \).
Given \( P(E \cap F) = \frac{1}{5} = \frac{10}{50} \).
Since \( \frac{9}{50} \neq \frac{10}{50} \), E and F are not independent.
Q7
Given that the events A and B are such that \( P(A) = \frac{1}{2}, P(A \cup B) = \frac{3}{5} \) and \( P(B) = p \). Find \( p \) if they are (i) mutually exclusive (ii) independent.
Solution:
We know \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
(i) Mutually Exclusive:
\( P(A \cap B) = 0 \).
\( \frac{3}{5} = \frac{1}{2} + p - 0 \Rightarrow p = \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10} \).
(ii) Independent:
\( P(A \cap B) = P(A)P(B) = \frac{1}{2}p \).
\( \frac{3}{5} = \frac{1}{2} + p - \frac{1}{2}p \Rightarrow \frac{3}{5} - \frac{1}{2} = \frac{p}{2} \).
\( \frac{1}{10} = \frac{p}{2} \Rightarrow p = \frac{2}{10} = \frac{1}{5} \).
Q8
Let A and B be independent events with \( P(A) = 0.3 \) and \( P(B) = 0.4 \). Find
(i) \( P(A \cap B) \)
(ii) \( P(A \cup B) \)
(iii) \( P(A|B) \)
(iv) \( P(B|A) \)
(i) \( P(A \cap B) \)
(ii) \( P(A \cup B) \)
(iii) \( P(A|B) \)
(iv) \( P(B|A) \)
Solution:
(i) \( P(A \cap B) = P(A)P(B) = 0.3 \times 0.4 = 0.12 \).
(ii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.12 = 0.58 \).
(iii) \( P(A|B) = P(A) = 0.3 \) (Since independent).
(iv) \( P(B|A) = P(B) = 0.4 \) (Since independent).
Q9
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).
Solution:
P(not A and not B) = \( P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \).
First find \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8} \).
Therefore, \( P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8} \).
Q10
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?
Solution:
Given \( P(A' \cup B') = \frac{1}{4} \).
By De Morgan's Law, \( A' \cup B' = (A \cap B)' \).
So, \( P((A \cap B)') = \frac{1}{4} \Rightarrow P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4} \).
Check Independence: \( P(A) \cdot P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \).
Since \( \frac{3}{4} \neq \frac{7}{24} \), A and B are not independent.
Q11
Given two independent events A and B such that \( P(A) = 0.3, P(B) = 0.6 \). Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Solution:
(i) \( P(A \cap B) = P(A)P(B) = 0.3 \times 0.6 = 0.18 \).
(ii) \( P(A \cap B') = P(A)P(B') = 0.3 \times (1 - 0.6) = 0.3 \times 0.4 = 0.12 \).
(iii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.6 - 0.18 = 0.72 \).
(iv) \( P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.72 = 0.28 \).
Q12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
Probability of getting an odd number in one toss = \( \frac{3}{6} = \frac{1}{2} \).
Probability of getting an even number (not odd) = \( \frac{1}{2} \).
P(At least one odd) = 1 - P(None is odd) = 1 - P(All are even).
\( = 1 - (\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}) = 1 - \frac{1}{8} = \frac{7}{8} \).
Q13
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution:
Total balls = 18. P(Black) = 10/18 = 5/9. P(Red) = 8/18 = 4/9.
(i) Both Red: \( P(R) \times P(R) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81} \).
(ii) First Black, Second Red: \( P(B) \times P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81} \).
(iii) One Black, One Red: Case 1 (BR) + Case 2 (RB).
\( = (\frac{5}{9} \times \frac{4}{9}) + (\frac{4}{9} \times \frac{5}{9}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81} \).
Q14
Probability 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.
(i) the problem is solved
(ii) exactly one of them solves the problem.
Solution:
\( P(A) = \frac{1}{2} \), \( P(B) = \frac{1}{3} \). Events are independent.
(i) Problem is solved: \( P(A \cup B) = 1 - P(A' \cap B') \).
\( P(A') = 1 - \frac{1}{2} = \frac{1}{2} \), \( P(B') = 1 - \frac{1}{3} = \frac{2}{3} \).
P(Solved) = \( 1 - (\frac{1}{2} \times \frac{2}{3}) = 1 - \frac{1}{3} = \frac{2}{3} \).
(ii) Exactly one solves: \( P(A \cap B') + P(A' \cap B) \).
\( = (\frac{1}{2} \times \frac{2}{3}) + (\frac{1}{2} \times \frac{1}{3}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \).
Q15
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: 'the card drawn is a spade', F: 'the card drawn is an ace'
(ii) E: 'the card drawn is black', F: 'the card drawn is a king'
(iii) E: 'the card drawn is a king or queen', F: 'the card drawn is a queen or jack'
(i) E: 'the card drawn is a spade', F: 'the card drawn is an ace'
(ii) E: 'the card drawn is black', F: 'the card drawn is a king'
(iii) E: 'the card drawn is a king or queen', F: 'the card drawn is a queen or jack'
Solution:
(i) \( P(E) = \frac{13}{52} = \frac{1}{4} \), \( P(F) = \frac{4}{52} = \frac{1}{13} \).
\( E \cap F \) is Ace of Spades (1 card). \( P(E \cap F) = \frac{1}{52} \).
\( P(E)P(F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52} \). Independent.
(ii) \( P(E) = \frac{26}{52} = \frac{1}{2} \), \( P(F) = \frac{4}{52} = \frac{1}{13} \).
\( E \cap F \) is Black King (2 cards). \( P(E \cap F) = \frac{2}{52} = \frac{1}{26} \).
\( P(E)P(F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26} \). Independent.
(iii) \( P(E) = \frac{8}{52} = \frac{2}{13} \), \( P(F) = \frac{8}{52} = \frac{2}{13} \).
\( E \cap F \) is Queen (4 cards). \( P(E \cap F) = \frac{4}{52} = \frac{1}{13} \).
\( P(E)P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169} \).
\( \frac{1}{13} \neq \frac{4}{169} \). Not Independent.
Q16
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(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.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(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.
Solution:
H = reads Hindi, E = reads English. \( P(H) = 0.6, P(E) = 0.4, P(H \cap E) = 0.2 \).
(a) Neither = \( 1 - P(H \cup E) = 1 - (P(H) + P(E) - P(H \cap E)) \).
\( = 1 - (0.6 + 0.4 - 0.2) = 1 - 0.8 = 0.2 \).
(b) \( P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{0.2}{0.6} = \frac{1}{3} \).
(c) \( P(H|E) = \frac{P(H \cap E)}{P(E)} = \frac{0.2}{0.4} = \frac{1}{2} \).
Q17
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0 (B) \( \frac{1}{3} \) (C) \( \frac{1}{12} \) (D) \( \frac{1}{36} \)
(A) 0 (B) \( \frac{1}{3} \) (C) \( \frac{1}{12} \) (D) \( \frac{1}{36} \)
Answer: (D)
Solution:
The only even prime number is 2.
Probability of getting 2 on one die = \( \frac{1}{6} \).
Probability of getting 2 on both dice = \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \).
Q18
Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) \( P(A'B') = [1 - P(A)] [1 - P(B)] \)
(C) \( P(A) = P(B) \)
(D) \( P(A) + P(B) = 1 \)
(A) A and B are mutually exclusive
(B) \( P(A'B') = [1 - P(A)] [1 - P(B)] \)
(C) \( P(A) = P(B) \)
(D) \( P(A) + P(B) = 1 \)
Answer: (B)
Solution:
If A and B are independent, then A' and B' are also independent.
So, \( P(A' \cap B') = P(A') \cdot P(B') \).
Since \( P(A') = 1 - P(A) \) and \( P(B') = 1 - P(B) \), option (B) is correct.