PEA 2024 PEA Q17 | ISI MSQE

PEAMCQEasy-Moderate

2024 PEA Q17

An urn contains $4$ white, $6$ red, and $5$ black balls. $5$ balls are randomly selected from the urn. Let $X$ and $Y$ denote respectively the number of white and black balls selected. Suppose $Y=2$ , i.e., $2$ of the $5$ selected balls are black. What is the probability that $X$ takes the value $2$ ?

Reveal answer and solution

Answer

Solution

1
Given $Y=2$ , the remaining $3$ balls are drawn from the $4$ white and $6$ red balls (total $10$ ). Conditionally these $3$ balls are uniformly distributed among $\binom{10}{3}=120$ subsets. Hence
2
$P(X=2\mid Y=2)=\frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}}=\frac{6\cdot 6}{120}=\frac{36}{120}=\frac{3}{10}.$

Answer structure / marking notes

After conditioning on $Y=2$ , only the colour distribution of the remaining $3$ balls matters.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2024/ISI_MSQE_PEA_2024_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.