Exercise 13.2, Question 1
\(P(A) = \tfrac35\), \(P(B) = \tfrac15\), A and B independent. Find \(P(A \cap B)\).
Show solution
- \(\tfrac35 \times \tfrac15\).
Answer: \(\tfrac{3}{25}\)
NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2
Exercise 13.2: Multiplication rule and independent events. \(P(E \cap F) = P(E)P(F|E)\); for draws without replacement multiply the changing fractions. E and F are independent when \(P(E \cap F) = P(E)P(F)\) (then their complements are independent too), so 'at least one' \(= 1 - P(\text{none})\).
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Exercise 13.2 questions and solutions
Exercise 13.2, Question 2
Two cards are drawn without replacement from a pack of 52. Find the probability both are black.
Show solution
- \(\tfrac{26}{52} \times \tfrac{25}{51}\).
Answer: \(\tfrac{25}{102}\)
Exercise 13.2, Question 3
Three oranges are drawn without replacement from 15 (12 good, 3 bad); the box is approved only if all three are good. Find the probability of approval.
Show solution
- \[\tfrac{12}{15} \times \tfrac{11}{14} \times \tfrac{10}{13}\]
Answer: \(\tfrac{44}{91}\)
Exercise 13.2, Question 4
A coin and a die are tossed. A: head; B: 3 on the die. Are A and B independent?
Show solution
- \(P(A) = \tfrac12\), \(P(B) = \tfrac16\), \(P(A \cap B) = \tfrac{1}{12} = P(A)P(B)\).
Answer: Yes, independent
Exercise 13.2, Question 5
A die has 1, 2, 3 in red and 4, 5, 6 in green. A: even number; B: red number. Are they independent?
Show solution
- \(P(A) = \tfrac12\), \(P(B) = \tfrac12\); \(A \cap B = \{2\}\), \(P = \tfrac16 \ne \tfrac14\).
Answer: Not independent
Exercise 13.2, Question 6
\(P(E) = \tfrac35\), \(P(F) = \tfrac{3}{10}\), \(P(E \cap F) = \tfrac15\). Are E and F independent?
Show solution
- \(P(E)P(F) = \tfrac{9}{50} \ne \tfrac15\).
Answer: Not independent
Exercise 13.2, Question 7
\(P(A) = \tfrac12\), \(P(A \cup B) = \tfrac35\), \(P(B) = p\). Find p if A and B are:
(i) mutually exclusive
Show solution
- \(\tfrac35 = \tfrac12 + p\).
Answer: \(p = \tfrac{1}{10}\)
(ii) independent
Show solution
- \(\tfrac35 = \tfrac12 + p - \tfrac12p\).
Answer: \(p = \tfrac15\)
Exercise 13.2, Question 8
A, B independent with \(P(A) = 0.3\), \(P(B) = 0.4\). Find:
(i) \(P(A \cap B)\)
Show solution
- \(0.3 \times 0.4\).
Answer: \(0.12\)
(ii) \(P(A \cup B)\)
Show solution
- \(0.3 + 0.4 - 0.12\).
Answer: \(0.58\)
(iii) \(P(A|B)\)
Show solution
- Independent: \(P(A|B) = P(A)\).
Answer: \(0.3\)
(iv) \(P(B|A)\)
Show solution
- \(P(B|A) = P(B)\).
Answer: \(0.4\)
Exercise 13.2, Question 9
\(P(A) = \tfrac14\), \(P(B) = \tfrac12\), \(P(A \cap B) = \tfrac18\). Find \(P(\text{not A and not B})\).
Show solution
- \[\begin{aligned}P(A' \cap B') &= 1 - P(A \cup B) \\ &= 1 - \left(\tfrac14 + \tfrac12 - \tfrac18\right)\end{aligned}\]
Answer: \(\tfrac38\)
Exercise 13.2, Question 10
\(P(A) = \tfrac12\), \(P(B) = \tfrac{7}{12}\), \(P(\text{not A or not B}) = \tfrac14\). Are A and B independent?
Show solution
- \(P(A' \cup B') = 1 - P(A \cap B)\), so \(P(A \cap B) = \tfrac34\), while \(P(A)P(B) = \tfrac{7}{24}\).
- They differ, so A and B are not independent. (In fact these numbers cannot all hold, since \(P(A \cap B)\) cannot exceed \(P(A) = \tfrac12\); the independence test fails either way.)
Answer: Not independent
Exercise 13.2, Question 11
A, B independent with \(P(A) = 0.3\), \(P(B) = 0.6\). Find:
(i) \(P(A \text{ and } B)\)
Show solution
- \(0.3 \times 0.6\).
Answer: \(0.18\)
(ii) \(P(A \text{ and not } B)\)
Show solution
- \(0.3 \times 0.4\).
Answer: \(0.12\)
(iii) \(P(A \text{ or } B)\)
Show solution
- \(0.3 + 0.6 - 0.18\).
Answer: \(0.72\)
(iv) \(P(\text{neither A nor B})\)
Show solution
- \(0.7 \times 0.4\).
Answer: \(0.28\)
Exercise 13.2, Question 12
A die is tossed three times. Find the probability of an odd number at least once.
Show solution
- \[1 - P(\text{no odd}) = 1 - \left(\tfrac12\right)^3\]
Answer: \(\tfrac78\)
Exercise 13.2, Question 13
Two balls are drawn with replacement from 10 black and 8 red balls. Find the probability that:
(i) both are red
Show solution
- \(\left(\tfrac{8}{18}\right)^2\).
Answer: \(\tfrac{16}{81}\)
(ii) the first is black and the second red
Show solution
- \(\tfrac{10}{18} \times \tfrac{8}{18}\).
Answer: \(\tfrac{20}{81}\)
(iii) one is black and the other red
Show solution
- Black then red, or red then black: \(2 \times \tfrac{20}{81}\).
Answer: \(\tfrac{40}{81}\)
Exercise 13.2, Question 14
A and B solve a problem independently with probabilities \(\tfrac12\) and \(\tfrac13\). Find the probability that:
(i) the problem is solved
Show solution
- \(1 - \tfrac12 \times \tfrac23\).
Answer: \(\tfrac23\)
(ii) exactly one of them solves it
Show solution
- \[\tfrac12 \times \tfrac23 + \tfrac12 \times \tfrac13\]
Answer: \(\tfrac12\)
Exercise 13.2, Question 15
One card is drawn from a pack of 52. In which cases are E and F independent?
(i) E: spade, F: ace (ii) E: black, F: king (iii) E: king or queen, F: queen or jack
Show solution
- (i) \[\begin{aligned}\tfrac14 \cdot \tfrac{1}{13} &= \tfrac{1}{52} \\ &= P(\text{ace of spades})\end{aligned}\]: independent.
- (ii) \[\begin{aligned}\tfrac12 \cdot \tfrac{1}{13} &= \tfrac{1}{26} \\ &= P(\text{black king}) \\ &= \tfrac{2}{52}\end{aligned}\]: independent.
- (iii) \[\begin{aligned}\tfrac{2}{13} \cdot \tfrac{2}{13} &= \tfrac{4}{169} \ne \tfrac{1}{13} \\ &= P(\text{queen})\end{aligned}\]: not independent.
Answer: (i) and (ii)
Exercise 13.2, Question 16
In a hostel 60% read a Hindi paper, 40% an English paper and 20% both. A student is chosen at random.
(a) Find the probability that she reads neither.
Show solution
- \(1 - (0.6 + 0.4 - 0.2)\).
Answer: \(0.2\) (i.e. \(\tfrac15\))
(b) If she reads Hindi, find the probability she reads English.
Show solution
- \(\dfrac{0.2}{0.6}\).
Answer: \(\tfrac13\)
(c) If she reads English, find the probability she reads Hindi.
Show solution
- \(\dfrac{0.2}{0.4}\).
Answer: \(\tfrac12\)
Exercise 13.2, Question 17
The probability of an even prime on each die when a pair of dice is rolled is: (A) 0 (B) \(\tfrac13\) (C) \(\tfrac{1}{12}\) (D) \(\tfrac{1}{36}\)
Show solution
- The only even prime is 2: \(\tfrac16 \times \tfrac16\).
Answer: (D)
Exercise 13.2, Question 18
A and B are 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) \(P(A) + P(B) = 1\)
Show solution
- \[P(A' \cap B') = 1 - P(A) - P(B) + P(A \cap B)\], which equals \((1 - P(A))(1 - P(B))\) exactly when \(P(A \cap B) = P(A)P(B)\).
Answer: (B)
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