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Question

Suppose a number is chosen at random from the set {1,2,3,4}. For events A: {1,3}. B: {1,2} and C: {1,4}, which of the following does NOT hold?

This question was previously asked in
SSC CGL 2022 Tier-II (Paper 2 JSO) Previous Year Paper (04-Mar-2023)
The correct answer is

A, B and C are independent  

Understanding Events and Independence in Probability

This solution explains the concept of event independence in probability using a specific example. We are given a set of numbers (sample space) and three events A, B, and C defined on this set. The goal is to determine which statement about the independence of these events is incorrect.

Defining the Sample Space and Events

The problem provides the following information:

  • Sample Space (S): The set of all possible outcomes is S = {1, 2, 3, 4}. The total number of possible outcomes is N = 4.
  • Event A: A = {1, 3}. The number of outcomes in A is n(A) = 2.
  • Event B: B = {1, 2}. The number of outcomes in B is n(B) = 2.
  • Event C: C = {1, 4}. The number of outcomes in C is n(C) = 2.

Calculating Probabilities of Individual Events

The probability of an event is calculated as the number of favourable outcomes divided by the total number of outcomes.

  • Probability of event A, P(A) = $\frac{n(A)}{N} = \frac{2}{4} = \frac{1}{2}$
  • Probability of event B, P(B) = $\frac{n(B)}{N} = \frac{2}{4} = \frac{1}{2}$
  • Probability of event C, P(C) = $\frac{n(C)}{N} = \frac{2}{4} = \frac{1}{2}$

Assessing Pairwise Independence of Events

Two events, say X and Y, are considered independent if the probability of both occurring together is equal to the product of their individual probabilities, i.e., P(X ∩ Y) = P(X) * P(Y). Let's check this for the pairs (A, B), (A, C), and (B, C).

Checking Independence of A and B

First, find the intersection of A and B:

  • A ∩ B = {1, 3} ∩ {1, 2} = {1}
  • The number of outcomes in A ∩ B is n(A ∩ B) = 1.
  • The probability of A and B occurring together is P(A ∩ B) = $\frac{n(A \cap B)}{N} = \frac{1}{4}$.

Now, calculate the product of their individual probabilities:

  • P(A) * P(B) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since P(A ∩ B) = P(A) * P(B) ($\frac{1}{4} = \frac{1}{4}$), events A and B are independent.

Checking Independence of A and C

First, find the intersection of A and C:

  • A ∩ C = {1, 3} ∩ {1, 4} = {1}
  • The number of outcomes in A ∩ C is n(A ∩ C) = 1.
  • The probability of A and C occurring together is P(A ∩ C) = $\frac{n(A \cap C)}{N} = \frac{1}{4}$.

Now, calculate the product of their individual probabilities:

  • P(A) * P(C) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since P(A ∩ C) = P(A) * P(C) ($\frac{1}{4} = \frac{1}{4}$), events A and C are independent.

Checking Independence of B and C

First, find the intersection of B and C:

  • B ∩ C = {1, 2} ∩ {1, 4} = {1}
  • The number of outcomes in B ∩ C is n(B ∩ C) = 1.
  • The probability of B and C occurring together is P(B ∩ C) = $\frac{n(B \cap C)}{N} = \frac{1}{4}$.

Now, calculate the product of their individual probabilities:

  • P(B) * P(C) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since P(B ∩ C) = P(B) * P(C) ($\frac{1}{4} = \frac{1}{4}$), events B and C are independent.

Assessing Mutual Independence of A, B, and C

For three events A, B, and C to be considered mutually independent, all pairwise independence conditions must be met, and additionally, the probability of all three occurring together must equal the product of their individual probabilities.

The conditions are:

  • P(A ∩ B) = P(A)P(B)
  • P(A ∩ C) = P(A)P(C)
  • P(B ∩ C) = P(B)P(C)
  • P(A ∩ B ∩ C) = P(A)P(B)P(C)

We have already confirmed that A, B, and C satisfy the first three conditions (pairwise independence).

Checking Independence of the Triple Intersection

First, find the intersection of A, B, and C:

  • A ∩ B ∩ C = {1, 3} ∩ {1, 2} ∩ {1, 4} = {1}
  • The number of outcomes in A ∩ B ∩ C is n(A ∩ B ∩ C) = 1.
  • The probability of A, B, and C occurring together is P(A ∩ B ∩ C) = $\frac{n(A \cap B \cap C)}{N} = \frac{1}{4}$.

Now, calculate the product of their individual probabilities:

  • P(A) * P(B) * P(C) = $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

Compare the results:

  • P(A ∩ B ∩ C) = $\frac{1}{4}$
  • P(A) * P(B) * P(C) = $\frac{1}{8}$

Since P(A ∩ B ∩ C) ≠ P(A) * P(B) * P(C) ($\frac{1}{4} \neq \frac{1}{8}$), the events A, B, and C are NOT mutually independent.

Identifying the Statement That Does NOT Hold

Let's review the options based on our calculations:

  • Option 1: B and C are independent. This statement holds true because P(B ∩ C) = P(B)P(C).
  • Option 2: A and C are independent. This statement holds true because P(A ∩ C) = P(A)P(C).
  • Option 3: A and B are independent. This statement holds true because P(A ∩ B) = P(A)P(B).
  • Option 4: A, B and C are independent. This statement does NOT hold true because P(A ∩ B ∩ C) ≠ P(A)P(B)P(C).

The question asks to identify the statement that does NOT hold. Therefore, the statement "A, B and C are independent" is the correct choice.

Final Answer Summary

The analysis demonstrates that while the events A, B, and C are independent in pairs (meaning each pair satisfies the independence condition P(X ∩ Y) = P(X)P(Y)), they are not mutually independent. This is because the condition for mutual independence, specifically P(A ∩ B ∩ C) = P(A)P(B)P(C), is not satisfied.

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