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Question

(D) Let $A$ be the event that the maximum number on the two chosen tickets is $\le 10$. Let $B$ be the event that the minimum number on the two chosen

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $A$ be the event that the maximum number on the two chosen tickets is $\le 10$. Let $B$ be the event that the minimum number on the two chosen tickets is $5$. We need to find the conditional probability $P(B|A) = \frac{n(A \cap B)}{n(A)}$.First,we determine $n(A)$,the number of ways to choose two distinct tickets such that both are $\le 10$. The number of ways to choose $2$ tickets from $10$ is $\binom{10}{2} = \frac{10 	imes 9}{2} = 45$.Next,we determine $n(A \cap B)$,the number of ways to choose two tickets such that the maximum is $\le 10$ and the minimum is $5$. This means one ticket must be $5$,and the other ticket must be in the set $\{6, 7, 8, 9, 10\}$.There are $5$ such pairs: $(5, 6), (5, 7), (5, 8), (5, 9), (5, 10)$.Thus,$n(A \cap B) = 5$.Therefore,the required probability is $P(B|A) = \frac{5}{45} = \frac{1}{9}$.Since $\frac{1}{9}$ is not among the options,the correct answer is $(D)$ None of these.

$A$ box contains $100$ tickets numbered $1, 2, \dots, 100$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than $10$. What is the probability that the minimum number on them is $5$?

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