Probability Ex 13.2 Class 12 Maths NCERT Solutions: Independent Events

Understanding the Multiplication Rule and Independent Events

The foundation of this exercise is the Multiplication Theorem on Probability. It states that the probability of the simultaneous occurrence of two events A and B is P(A ∩ B) = P(A) × P(B|A), where P(B|A) is the conditional probability of B given that A has already occurred. This formula is universally true for any two events.

However, Exercise 13.2 focuses on a special, very common scenario: independent events. Two events are considered independent if the occurrence of one has absolutely no impact on the probability of the other. In mathematical terms, this means P(B|A) = P(B) and P(A|B) = P(A). Think about tossing a coin twice; the result of the first toss doesn't change the 50/50 chance for the second toss.

When we substitute this condition of independence into the general multiplication theorem, we get a much simpler formula:

P(A ∩ B) = P(A) × P(B)

This is the key formula you will use throughout this exercise. It tells us that to find the probability of two independent events both happening, you simply multiply their individual probabilities. This principle can also be extended to three or more independent events: P(A ∩ B ∩ C) = P(A) × P(B) × P(C).

Key Definitions for Exercise 13.2

Independent Events

Two events E and F are independent if the probability of occurrence of one of them is not affected by the occurrence of the other. The definitive test for independence is checking if P(E ∩ F) = P(E) × P(F).

Multiplication Rule of Probability

For any two events E and F, the probability of their intersection (both occurring) is given by P(E ∩ F) = P(E) × P(F|E). For independent events, this simplifies to P(E ∩ F) = P(E) × P(F).

Events 'with replacement'

In problems involving drawing items from a collection (like cards from a deck or balls from a bag), 'with replacement' means the item is put back before the next draw. This ensures the sample space remains unchanged, making the successive events independent.

Worked Examples on Independent Events

• Problem 1: A die is thrown. Let E be the event 'the number appearing is a multiple of 3' and F be the event 'the number appearing is even'. Are the events E and F independent? Step 1: Define the Sample Space and Events. Sample Space, S = {1, 2, 3, 4, 5, 6}. Event E = {3, 6}. So, P(E) = 2/6 = 1/3. Event F = {2, 4, 6}. So, P(F) = 3/6 = 1/2. Step 2: Find the Intersection of the Events. The intersection E ∩ F is the event 'the number is a multiple of 3 AND is even'. E ∩ F = {6}. So, P(E ∩ F) = 1/6. Step 3: Check for Independence. We need to check if P(E ∩ F) = P(E) × P(F). L.H.S = P(E ∩ F) = 1/6. R.H.S = P(E) × P(F) = (1/3) × (1/2) = 1/6. Final Answer: Since L.H.S = R.H.S, the events E and F are independent.
• Problem 2: The probability of A solving a problem is 1/2 and B solving it is 1/3. If both try independently, find the probability that the problem is solved. Step 1: Define the Probabilities. P(A) = Probability that A solves it = 1/2. P(B) = Probability that B solves it = 1/3. Since they try independently, the events are independent. Step 2: Define Complementary Probabilities. P(A') = Probability that A does NOT solve it = 1 - P(A) = 1 - 1/2 = 1/2. P(B') = Probability that B does NOT solve it = 1 - P(B) = 1 - 1/3 = 2/3. Step 3: Calculate the Probability that the Problem is NOT Solved. The problem is not solved only if BOTH A and B fail to solve it. Since the events are independent, P(A' ∩ B') = P(A') × P(B') = (1/2) × (2/3) = 1/3. Step 4: Calculate the Probability that the Problem IS Solved. The event 'problem is solved' is the complement of the event 'problem is not solved'. P(Problem is solved) = 1 - P(Problem is not solved) P(Problem is solved) = 1 - P(A' ∩ B') = 1 - 1/3 = 2/3. Final Answer: The probability that the problem is solved is 2/3.

Exam Trap: Independent vs. Mutually Exclusive Events

A very common point of confusion for students is the difference between independent and mutually exclusive events. They are not the same thing; in fact, they are almost opposites!

• Independent Events: The occurrence of one event does not affect the probability of the other. Their probabilities multiply: P(A ∩ B) = P(A) × P(B). They can happen at the same time.

Example: Getting heads on a coin toss (Event A) and rolling a 6 on a die (Event B). P(A ∩ B) = (1/2) × (1/6) = 1/12.

• Mutually Exclusive (or Disjoint) Events: The events cannot happen at the same time. The occurrence of one prevents the other. Their intersection is the empty set. Therefore, P(A ∩ B) = 0.

Example: On a single die roll, getting a 1 (Event A) and getting a 6 (Event B). You cannot get both simultaneously.

Key Takeaway: If two events A and B have non-zero probabilities (P(A) > 0 and P(B) > 0), they cannot be both independent and mutually exclusive. If they are independent, P(A ∩ B) > 0. If they are mutually exclusive, P(A ∩ B) = 0.

Practice Questions with Solutions

• Q: Two cards are drawn from a well-shuffled deck of 52 cards with replacement. What is the probability that the first card is a king and the second card is a heart? A: Step 1: Define the events. Let K be the event 'the first card is a king' and H be the event 'the second card is a heart'. Since the cards are drawn with replacement, the events are independent. Step 2: Calculate the probability of each event. There are 4 kings in a deck of 52 cards. So, P(K) = 4/52 = 1/13. There are 13 hearts in a deck of 52 cards. So, P(H) = 13/52 = 1/4. Step 3: Use the multiplication rule for independent events. We need to find P(K ∩ H). Since the events are independent, P(K ∩ H) = P(K) × P(H). P(K ∩ H) = (1/13) × (1/4) = 1/52. Final answer: The probability is 1/52.
• Q: Given P(A) = 0.3, P(B) = 0.6 and P(A ∩ B) = 0.18. Are events A and B independent? A: Step 1: State the condition for independence. Two events A and B are independent if P(A ∩ B) = P(A) × P(B). Step 2: Calculate P(A) × P(B). Given P(A) = 0.3 and P(B) = 0.6. P(A) × P(B) = 0.3 × 0.6 = 0.18. Step 3: Compare the calculated value with the given P(A ∩ B). We calculated P(A) × P(B) = 0.18. We are given P(A ∩ B) = 0.18. Step 4: Conclude based on the comparison. Since P(A ∩ B) = P(A) × P(B), the events are independent. Final answer: Yes, events A and B are independent.
• Q: A bag contains 4 red and 6 black balls. Three balls are drawn one by one with replacement. What is the probability that all three balls are black? A: Step 1: Define the events and their probabilities. Let B1, B2, and B3 be the events of drawing a black ball on the first, second, and third draw, respectively. Total balls = 4 + 6 = 10. Number of black balls = 6. The probability of drawing a black ball in any single draw is P(B) = 6/10 = 3/5. Step 2: Recognize that the events are independent. Since the balls are drawn with replacement, the outcome of each draw does not affect the next. Therefore, B1, B2, and B3 are independent events. Step 3: Apply the multiplication rule for three independent events. We want to find the probability of getting a black ball on all three draws, which is P(B1 ∩ B2 ∩ B3). P(B1 ∩ B2 ∩ B3) = P(B1) × P(B2) × P(B3) = (3/5) × (3/5) × (3/5) = 27/125. Final answer: The probability that all three balls are black is 27/125.
• Q: Let A and B be independent events with P(A) = 1/4 and P(B) = 1/2. Find P(A' ∩ B'). A: Step 1: Use the property that if A and B are independent, then their complements A' and B' are also independent. This means P(A' ∩ B') = P(A') × P(B'). Step 2: Calculate the probabilities of the complementary events. P(A') = 1 - P(A) = 1 - 1/4 = 3/4. P(B') = 1 - P(B) = 1 - 1/2 = 1/2. Step 3: Multiply the probabilities of the complements. P(A' ∩ B') = P(A') × P(B') = (3/4) × (1/2) = 3/8. Final answer: The probability of P(A' ∩ B') is 3/8.

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