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Question

A dice is cast twice, and the sum of the appearing numbers is 10. Then the probability that the number 5 has appeared at least once is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{1}{3}$

Understanding the Conditional Probability Problem

We need to find the conditional probability. The condition is that the sum of the numbers appearing on two dice casts is 10. We want to find the probability that the number 5 appeared at least once, given this condition.

Determining the Conditional Sample Space

Let A be the event that the sum of the numbers is 10. The possible outcomes $(x, y)$ where $x$ is the result of the first cast and $y$ is the result of the second cast are:

  • (4, 6)
  • (5, 5)
  • (6, 4)

The total number of outcomes where the sum is 10 is $n(A) = 3$. This is our reduced sample space.

Identifying Favorable Outcomes

Let B be the event that the number 5 appears at least once.

We need to find the outcomes in event A (sum is 10) where the number 5 appears at least once. Looking at the outcomes in A:

  • (4, 6) - Does not contain 5.
  • (5, 5) - Contains 5 at least once.
  • (6, 4) - Does not contain 5.

The only outcome where the sum is 10 AND the number 5 appears at least once is (5, 5). So, the number of favorable outcomes is $n(A \cap B) = 1$.

Calculating the Conditional Probability

The conditional probability $P(B|A)$ is calculated as:

$ P(B|A) = \frac{n(A \cap B)}{n(A)} $

Substituting the values we found:

$ P(B|A) = \frac{1}{3} $

Therefore, the probability that the number 5 has appeared at least once, given that the sum is 10, is $\frac{1}{3}$.

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