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Axiomatic Approach

A number deno...

Question

A number denoted by $X$ is chosen at random from the set $1, 2, 3, 4 .$ A second number denoted by $Y$ is chosen at random from the set $1..... X,$ the probability that $Y = 1$ is

\frac{1}{4}

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\frac{1}{2}

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\frac{25}{48}

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\frac{23}{48}

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Solution

The correct option is A $\frac{25}{48}$
Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the events of choosing $1, 2, 3, 4$ respectively from the first draw.
$P (A_{1}) = P (A_{2}) = P (A_{3}) = P (A_{4}) = \frac{1}{4}$

Let $B$ be the event of getting $1$ in the second draw.
$A_{1}, A_{2}, A_{3}, A_{4}$ are mutually exclusive and exhaustive. That is, one of them necessarily occurs.

Therefore $B = B A_{1} \cup B A_{2} \cup B A_{3} \cup B A_{4}$

Hence $P (B) = P (B A_{1}) + P (B A_{2}) + P (B A_{3}) + P (B A_{4})$

$= P (A_{1}) P (\frac{B}{A_{1}}) + P (A_{2}) P (\frac{B}{A_{2}}) + P (A_{3}) P (\frac{B}{A_{3}}) + P (A_{4}) P (\frac{B}{A_{4}})$

$= \frac{1}{4} \times 1 + \frac{1}{4} \times \frac{1}{2} + \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{4} = \frac{25}{48}$

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