Welcome! If you’re looking for the NCERT Solutions for Class 12 Maths Chapter 13, Exercise 13.2, you’re in the right place. Below, you will find detailed, step-by-step solutions for every question in Exercise 13.2, designed to help you clear all your doubts and understand the logic behind each answer.
This exercise has questions based on the concepts of multiplication theorem and independent events, which are important for your board exams.
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:
Step 1: For two independent events A and B we know: P(A \cap B) = P(A) \times P(B)
Step 2: Given probabilities
P(A) = \frac{3}{5}
P(B) = \frac{1}{5}
Step 3: Substitute and solve
P(A \cap B) = \frac{3}{5} \times \frac{1}{5}
P(A \cap B) = \frac{3}{25}
Final Answer: \boxed{ P(A \cap B) = \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:
Step 1: Define the events
Let E be the event that the first card drawn is black.
Let F be the event that the second card drawn is black.
We have to find the probability that both card drawn are black (both events happening), which is P(E \cap F).
Step 2: Use the multiplication rule for probability
Since the cards are drawn without replacement, the events are dependent. We use the multiplication rule:
P(E \cap F) = P(E) \times P(F|E)
Step 3: Calculating the probability of the first event, P(E)
A standard deck has 52 cards, with 26 of them being black.
P(E) = \frac{\text{Number of black cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}
Step 4: Calculating the probability of the second event i.e. the conditional probability, P(F|E)
P(F|E) is the probability of drawing a second black card, given that the first card was black.
After drawing one black card, there are 25 black cards remaining and a total of 51 cards left in the deck.
P(F|E) = \frac{\text{Remaining black cards}}{\text{Remaining total cards}} = \frac{25}{51}
Step 5: Substitute and solve
P(E \cap F) = P(E) \times P(F|E) = \frac{1}{2} \times \frac{25}{51}
P(E \cap F) = \frac{25}{102}
Final Answer: \boxed{ P(E \cap F) = \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:
Step 1: Define the events
The box is approved only if all three oranges drawn are good.
Let G_1 be the event that the first orange is good.
Let G_2 be the event that the second orange is good.
Let G_3 be the event that the third orange is good.
We need to find the probability of all three events happening: P(G_1 \cap G_2 \cap G_3).
Step 2: Use the multiplication rule for probability
Since the oranges are drawn without replacement, the events are dependent. We use the extended multiplication rule:
P(G_1 \cap G_2 \cap G_3) = P(G_1) \times P(G_2|G_1) \times P(G_3|G_1 \cap G_2)
Step 3: Calculating the individual probabilities
P(G₁): There are 12 good oranges out of 15 total.
P(G_1) = \frac{12}{15}
P(G₂|G₁): Given the first was good, there are 11 good oranges left out of 14 total.
P(G_2|G_1) = \frac{11}{14}
P(G₃|G₁ ∩ G₂): Given the first two were good, there are 10 good oranges left out of 13 total.
P(G_3|G_1 \cap G_2) = \frac{10}{13}
Step 4: Substituting
P(\text{Approval}) = P(G_1) \times P(G_2|G_1) \times P(G_3|G_1 \cap G_2)
P(\text{Approval}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}
Let’s simplify the fractions:
P(\text{Approval}) = \left(\frac{4 \times 3}{5 \times 3}\right) \times \left(\frac{11}{14}\right) \times \left(\frac{10}{13}\right)
P(\text{Approval}) = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}
P(\text{Approval}) = \frac{4 \times 11 \times 10}{5 \times 14 \times 13} = \frac{4 \times 11 \times 2}{14 \times 13} = \frac{4 \times 11 \times 1}{7 \times 13}
P(\text{Approval}) = \frac{44}{91}
Final Answer: \boxed{ P(\text{Approval}) = \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:
Step 1: Condition for independence
Two events A and B are independent if and only if P(A \cap B) = P(A) \times P(B).
We must check if this equality holds.
Step 2: Finding P(A) and P(B)
Event A = ‘head appears on the coin’. A fair coin has 2 outcomes {H, T}.
P(A) = \frac{1}{2}
Event B = ‘3 on the die’. An unbiased die has 6 outcomes {1, 2, 3, 4, 5, 6}.
P(B) = \frac{1}{6}
Step 3: Finding P(A) x P(B)
P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}
Step 4: Finding P(A ∩ B)
The event A \cap B is ‘head on the coin AND 3 on the die’.
The total sample space for the combined experiment has 2 \times 6 = 12 outcomes.
S = \{(H,1), (H,2), (H,3), ..., (T,6)\}
The event A \cap B corresponds to exactly one of these outcomes: (H, 3).
P(A \cap B) = \frac{1}{12}
Step 5: Comparing the probabilities
We compare the values from Step 3 and Step 4.
P(A \cap B) = \frac{1}{12}
P(A) \times P(B) = \frac{1}{12}
Since P(A \cap B) = P(A) \times P(B), the events A and B are independent.
Final Answer: \boxed{ \text{A and B are independent events.} }
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:
Step 1: Condition for independence
We must check if P(A \cap B) = P(A) \times P(B)10.
Step 2: Define the sample space and find P(A) and P(B)
The sample space is S = \{1, 2, 3, 4, 5, 6\} .
Event A = ‘the number is even’ = {2, 4, 6}.
P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}
Event B = ‘the number is red’ = {1, 2, 3}.
P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2}
Step 3: Find P(A) x P(B)
P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
Step 4: Find P(A ∩ B)
The event A \cap B is ‘the number is even AND red’.
Looking at the sets: A = \{2, 4, 6\} and B = \{1, 2, 3\}.
The common element is {2}.
So, A \cap B = {2}.
P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{6}
Step 5: Compare the probabilities
P(A \cap B) = \frac{1}{6}
P(A) \times P(B) = \frac{1}{4}
Since \frac{1}{6} \neq \frac{1}{4},
the condition P(A \cap B) = P(A) \times P(B) is not met.
Therefore, A and B are not independent (they are dependent).
Final Answer: \boxed{ \text{A and B are not independent events.} }
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:
Step 1: Conditions for independence
We must check if P(E \cap F) = P(E) \times P(F).
Step 2: Listing the given probabilities
P(E) = \frac{3}{5}
P(F) = \frac{3}{10}
P(E \cap F) = \frac{1}{5}
Step 3: Calculate P(E) x P(F)
P(E) \times P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}
Step 4: Compare the probabilities
We are given P(E \cap F) = \frac{1}{5}.
We calculated P(E) \times P(F) = \frac{9}{50}.
To compare, let’s use a common denominator: P(E \cap F) = \frac{1}{5} = \frac{10}{50}.
Since \frac{10}{50} \neq \frac{9}{50}, the condition P(E \cap F) = P(E) \times P(F) is not met.
Therefore, E and F are not independent.
Final Answer: \boxed{ \text{E and F are not independent events.} }
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 are given P(A) = \frac{1}{2},
P(A \cup B) = \frac{3}{5}, and P(B) = p.
(i) Events are mutually exclusive
Step 1: Recall the formula for mutually exclusive events
If A and B are mutually exclusive, they cannot happen at the same time, so P(A \cap B) = 0.
The addition rule P(A \cup B) = P(A) + P(B) - P(A \cap B) simplifies to:
P(A \cup B) = P(A) + P(B)
Step 2: Substitute and solve for p
\frac{3}{5} = \frac{1}{2} + p
p = \frac{3}{5} - \frac{1}{2}
p = \frac{6}{10} - \frac{5}{10}
p= \frac{1}{10}
(ii) Events are independent
Step 1: Recall the formula for independent events
If A and B are independent, P(A \cap B) = P(A) \times P(B).
Using the addition rule: P(A \cup B) = P(A) + P(B) - P(A \cap B)
Substituting the independence condition, we get:
P(A \cup B) = P(A) + P(B) - (P(A) \times P(B))
Step 2: Substitute and solve for p
\frac{3}{5} = \frac{1}{2} + p - \left(\frac{1}{2} \times p\right)
\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}
\frac{3}{5} = \frac{1}{2} + \frac{p}{2}
\frac{p}{2} = \frac{3}{5} - \frac{1}{2}
\frac{p}{2} = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}
p = 2 \times \left(\frac{1}{10}\right) [.latex][latex]p= \frac{1}{5}
Final Answer: \boxed{ (i) p = \frac{1}{10}, (ii) p = \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)
Solution:
We are given that A and B are independent events, P(A) = 0.3, and P(B) = 0.4.
(i) Find P(A \cap B)
Step 1: Using the definition of independent events.
P(A \cap B) = P(A) \times P(B)
Step 2: Substitute and solve.
P(A \cap B) = 0.3 \times 0.4 = 0.12
(ii) Find P(A \cup B)
Step 1: Use the addition rule for probability.
P(A \cup B) = P(A) + P(B) - P(A \cap B)
Step 2: Substitute values (using P(A \cap B) from part (i)).
P(A \cup B) = 0.3 + 0.4 - 0.12
P(A \cup B) = 0.7 - 0.12 = 0.58
(iii) Find P(A|B)
Step 1: Use the property of independent events.
For independent events, the occurrence of B does not affect the probability of A.
i.e. P(A|B) = P(A)
Step 2: Substitute the value.
P(A|B) = 0.3
(iv) Find P(B|A)
Step 1: Use the property of independent events.
For independent events, the occurrence of A does not affect the probability of B.
i.e. P(B|A) = P(B)
Step 2: Substitute the value.
P(B|A) = 0.4
Final Answer: \boxed{ \begin{aligned} (i) P(A \cap B) = 0.12 \ \ (ii) P(A \cup B) = 0.58 \ \ (iii) P(A|B) = 0.3 \ \ (iv) P(B|A) = 0.4 \end{aligned} }
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:
We need to find P(not A and not B) , which is written as P(A' \cap B').
Step 1: Use De Morgan's Law
The event A' \cap B' \ (neither A nor B) is the complement of the event A \cup B (at least one of A or B).
P(A' \cap B') = P((A \cup B)')
P(A' \cap B') = 1 - P(A \cup B) ... (1)
Step 2: Find P(A \cup B) using the addition rule
P(A \cup B) = P(A) + P(B) - P(A \cap B)
Step 3: Substitute the given values
P(A) = \frac{1}{4},
P(B) = \frac{1}{2},
P(A \cap B) = \frac{1}{8}
P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}
P(A \cup B) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}
Step 4: Calculate P(A' \cap B') using (1)
P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8}
Final Answer: \boxed{ P(\text{not A and not B}) = \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:
Step 1: State the condition for independence
We must check if P(A \cap B) = P(A) \times P(B).
Step 2: Analyze the given P(not A or not B)
P(not A or not B) is written as P(A' \cup B').
By De Morgan's Law, A' \cup B' is the complement of the event A \cap B.
P(A' \cup B') = P((A \cap B)')
We are given P(A' \cup B') = \frac{1}{4}.
Step 3: Find P(A \cap B)
The probability of an event and its complement sum to 1.
P(A \cap B) + P((A \cap B)') = 1
P(A \cap B) + \frac{1}{4} = 1
P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}
Step 4: Calculate P(A) \times P(B)
P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}
Step 5: Compare the probabilities
We compare the value from Step 3 and Step 4.
P(A \cap B) = \frac{3}{4}
P(A) \times P(B) = \frac{7}{24}
To compare, let's use a common denominator: P(A \cap B) = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}.
Since \frac{18}{24} \neq \frac{7}{24}, the condition P(A \cap B) = P(A) \times P(B) is not met.
Final Answer: \boxed{ \text{A and B are not independent events.} }
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)
Solution: Given:
A and B are independent events,
P(A) = 0.3, and P(B) = 0.6.
(i) P(A and B)
That is P(A \cap B).
Step 1: Using independence formula: P(A \cap B) = P(A) \times P(B)
Step 2: Substitute and solve.
P(A \cap B) = 0.3 \times 0.6 = 0.18
(ii) P(A and not B)
That is P(A \cap B').
Step 1: Since A and B are independent, A and B' are also independent.
P(A \cap B') = P(A) \times P(B')
Step 2: Find P(B').
P(B') = 1 - P(B) = 1 - 0.6 = 0.4
Step 3: Substitute and solve.
P(A \cap B') = 0.3 \times 0.4 = 0.12
(iii) P(A or B)
That is P(A \cup B).
Step 1: Use the addition rule.
P(A \cup B) = P(A) + P(B) - P(A \cap B)
Step 2: Substituting values of P(A \cap B) from part (i)).
P(A \cup B) = 0.3 + 0.6 - 0.18
P(A \cup B) = 0.9 - 0.18 = 0.72
(iv) P(neither A nor B)
This is P(A' \cap B').
Step 1: Since A and B are independent, A' and B' are also independent.
P(A' \cap B') = P(A') \times P(B')
Step 2: Find P(A') and P(B').
P(A') = 1 - P(A) = 1 - 0.3 = 0.7
P(B') = 1 - P(B) = 1 - 0.6 = 0.4
Step 3: Substituting
P(A' \cap B') = 0.7 \times 0.4 = 0.28
(Alternative check: P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.72 = 0.28 )
Final Answer:
\boxed{ \begin{aligned} (i) P(A \text{ and } B) = 0.18 \ \ (ii) P(A \text{ and not } B) = 0.12 \ \ (iii) P(A \text{ or } B) = 0.72 \ \ (iv) P(\text{neither } A \text{ nor } B) = 0.28 \end{aligned} }
Q12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
Step 1: Defining the events
Let A be the event 'getting an odd number at least once'.
A' is the event 'not getting an odd number at all', which is 'getting an even number on all three tosses'.
Using the formula P(A) = 1 - P(A').
Step 2: Find the probability of getting an even number in one toss
A die has 6 faces: S = \{1, 2, 3, 4, 5, 6\}.
The odd numbers are \{1, 3, 5\}.
The even numbers are \{2, 4, 6\}.
The probability of getting an even number in a single toss is P(\text{Even}) = \frac{3}{6} = \frac{1}{2}.
Step 3: Find the probability of the complement event, A'
Let E_1, E_2, E_3 be the events of getting an even number on the first, second, and third tosses, respectively.
Since the tosses are independent events:
P(A') = P(E_1 \cap E_2 \cap E_3)
P(A') = P(E_1) \times P(E_2) \times P(E_3)
P(A') = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}
Step 4: Calculate the final probability, P(A)
P(A) = 1 - P(A')
P(A) = 1 - \frac{1}{8} = \frac{7}{8}
Final Answer: \boxed{ P(\text{getting an odd number at least once}) = \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.
Solution:
The draw is with replacement, so the outcome of the first draw does not affect the second. The events are independent.
Total balls = 10 Black + 8 Red = 18 balls.
Probability of drawing a Red ball: P(R) = \frac{8}{18} = \frac{4}{9}
Probability of drawing a Black ball: P(B) = \frac{10}{18} = \frac{5}{9}
(i) both balls are red
This is P(R \text{ on first} \cap R \text{ on second}).
Since the events are independent:
P(R \cap R) = P(R) \times P(R) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}
(ii) first ball is black and second is red
This is P(B \text{ on first} \cap R \text{ on second}).
Since the events are independent:
P(B \cap R) = P(B) \times P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}
(iii) one of them is black and other is red
This can happen in two mutually exclusive ways:
1. First is Black AND second is Red (calculated in part ii)
2. First is Red AND second is Black
We need to find the probability for the second case:
P(R \text{ on first} \cap B \text{ on second}) = P(R) \times P(B) = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}
Now, we add the probabilities of these two cases:
P(\text{one B, one R}) = P(B \cap R) + P(R \cap B) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}
Final Answer: \boxed{ (i) \frac{16}{81}, (ii) \frac{20}{81}, (iii) \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.
Solution:
Let A be the event 'problem is solved by A' and B be the event 'problem is solved by B'.
We are given:
P(A) = \frac{1}{2}
P(B) = \frac{1}{3}
We also need the complements (probabilities of not solving):
P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}
P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}
(i) the problem is solved
The problem is solved if at least one of them solves it. This is P(A \cup B).
It's easier to find the complement: P(\text{problem is not solved}) = P(A' \cap B').
Since A and B are independent, A' and B' are also independent.
P(A' \cap B') = P(A') \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}
The probability that the problem is solved is:
P(A \cup B) = 1 - P(A' \cap B') = 1 - \frac{1}{3} = \frac{2}{3}
(ii) exactly one of them solves the problem
This can happen in two mutually exclusive ways:
A solves AND B does not solve (A \cap B' )
A does not solve AND B solves (A' \cap B )
We add the probabilities of these two independent events:
P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)
P(\text{exactly one}) = (P(A) \times P(B')) + (P(A') \times P(B))
P(\text{exactly one}) = \left(\frac{1}{2} \times \frac{2}{3}\right) + \left(\frac{1}{2} \times \frac{1}{3}\right)
P(\text{exactly one}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
Final Answer: \boxed{ (i) \frac{2}{3}, (ii) \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'. 3</h4>
Solution:
We must check the condition P(E \cap F) = P(E) \times P(F) for each case.
(i) E: spade, F: ace
P(E) = \frac{13 \text{ spades}}{52} = \frac{1}{4}
P(F) = \frac{4 \text{ aces}}{52} = \frac{1}{13}
P(E) \times P(F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}
E \cap F is 'the Ace of Spades'.
There is 1 such card.P(E \cap F) = \frac{1}{52}
Since P(E \cap F) = P(E) \times P(F), the events are independent.
(ii) E: black, F: king
P(E) = \frac{26 \text{ black cards}}{52} = \frac{1}{2}
P(F) = \frac{4 \text{ kings}}{52} = \frac{1}{13}
P(E) \times P(F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}
E \cap F is 'a black king'.
There are 2 such cards (King of Spades, King of Clubs).
P(E \cap F) = \frac{2}{52} = \frac{1}{26}
Since P(E \cap F) = P(E) \times P(F), the events are independent.
(iii) E: king or queen, F: queen or jack
E = {King, Queen}.
n(E) = 4 + 4 = 8.
P(E) = \frac{8}{52} = \frac{2}{13}
F = {Queen, Jack}.
n(F) = 4 + 4 = 8.
P(F) = \frac{8}{52} = \frac{2}{13}
P(E) \times P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169}
E \cap F is 'Queen'. There are 4 Queens.
P(E \cap F) = \frac{4}{52} = \frac{1}{13}
Since \frac{1}{13} \neq \frac{4}{169}, the events are not independent.
Final Answer: \boxed{ \text{Cases (i) and (ii) are independent. Case (iii) is 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. 4</h4>
Solution:
Let H be the event 'reads Hindi newspaper' and E be the event 'reads English newspaper'.
We are given:
P(H) = 60\% = 0.6
P(E) = 40\% = 0.4
P(H \cap E) = 20\% = 0.2
(a) Find P(neither Hindi nor English)
This is P(H' \cap E').
Step 1: Use De Morgan's Law: P(H' \cap E') = 1 - P(H \cup E).
Step 2: Finding P(H \cup E) using the addition rule.
P(H \cup E) = P(H) + P(E) - P(H \cap E)
P(H \cup E) = 0.6 + 0.4 - 0.2 = 0.8
Step 3: Calculate the final probability.
P(H' \cap E') = 1 - 0.8 = 0.2
(b) If she reads Hindi, find P(she reads English)
This is conditional probability P(E|H).
Step 1: Using conditional probability formula.
P(E|H) = \frac{P(E \cap H)}{P(H)}
Step 2: Substitute and solve.
P(E|H) = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3}
(c) If she reads English, find P(she reads Hindi)
This is conditional probability P(H|E).
Step 1: Use the conditional probability formula.
P(H|E) = \frac{P(H \cap E)}{P(E)}
Step 2: Substitute and solve.
P(H|E) = \frac{0.2}{0.4} = \frac{2}{4} = \frac{1}{2}
Final Answer: \boxed{ (a) 0.2, (b) \frac{1}{3}, (c) \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}
Solution:
Step 1: Identify the event
The only even prime number on a die {1, 2, 3, 4, 5, 6} is the number 2.
We have to calculate probability of getting 2 on the first die AND 2 on the second die.
Let A = '2 on the first die' and B = '2 on the second die'.
We need to find P(A \cap B).
Step 2: Finding the individual probabilities
The two die rolls are independent events.
P(A) = \frac{1}{6}
P(B) = \frac{1}{6}
Step 3: Use the multiplication rule for independent events
P(A \cap B) = P(A) \times P(B)
P(A \cap B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}
\boxed{P(A \cap B) = \frac{1}{36} }
The correct option is (D).
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
Solution:
Two events A and B are independent if P(A \cap B) = P(A) \times P(B).
Let's analyze the given options based on this and related properties.
(A) If A and B are mutually exclusive (and P(A) \neq 0, P(B) \neq 0),
P(A \cap B) = 0.
But P(A) \times P(B) \neq 0.
So, mutually exclusive events are (generally) dependent, not independent. This is incorrect.
(C) P(A) = P(B) is not a condition for independence.
(e.g., rolling two 6s on dice: P(A)=1/6, P(B)=1/6, but P(A \cap B) \neq P(A)). This is incorrect.
(D) P(A) + P(B) = 1 is not a condition for independence. This is incorrect.
(B) This option states P(A'B') = [1 - P(A)][1 - P(B)]
P(A'B') is shorthand for P(A' \cap B') (A' and B').
[1 - P(A)] is the definition of P(A').
[1 - P(B)] is the definition of P(B').
So, the equation is P(A' \cap B') = P(A') \times P(B').
If A' and B' are independent, then their complements A and B are also independent.
Therefore, this statement is a correct.
The correct option is (B).