If cards are drawn one by one from a well shuffled pack of $52$ cards without replacement, until an ace appears, find the probability that the fourth card is the first ace to appear.

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Solution

Probability of selecting $3$ non-Ace and $1$ Ace out of $52$ cards is equal to $\frac{^{48} C_{3} \times^{4} C_{1}}{^{52} C_{4}}$
Since we want $4 t h$ card to be first ace, we will also have to consider the arrangement, Now $4$ cards in sample space can be arranged in $4!$ ways and, favorable they can be arranged in $3!$ ways as we want $4 t h$ position to be occupied by ace
Hence required probability $= \frac{^{48} C_{3} \times^{4} C_{1}}{^{52} C_{4}} \times \frac{3!}{4!}$

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If cards are drawn one by one from a well shuffled pack of 52 cards without replacement, until an ace appears, find the probability that the fourth card is the first ace to appear.

If cards are drawn one by one from a well shuffled pack of $52$ cards without replacement, until an ace appears, find the probability that the fourth card is the first ace to appear.