Q. 2.3 - A First Course in Probability

Question

A deck of cards is dealt out. What is the probability that the $14$th card dealt is an ace? What is the probability that the first ace occurs on the $14$th card?

Step-by-Step Solution

Answer

$P\left(14\text{ th card is an ace }\right)=\frac{4·51!}{52!}=\frac{4}{52}=\frac{1}{13}$

$P(14 th card is the first ace$$)\approx 0.03$

Step 1 Given Information.

A deck of cards is dealt out.

Step 2 Explanation.

A deck$\left(52\right)$ of cards is dealt.

Outcome space permutations of cards$\Rightarrow$ each element is equally likely.

A deck has $52$cards so the outcome space has $52 !$elements.

$P\left(14\text{ th card is an ace }\right)=\frac{\text{ \# events in which 14th card is an ace }}{\text{ \# of events in the outcome space }}$

Fix one of the $4$aces in the $14$th place and permute the remaining $51$cards $-4·51$ ! events from the outcome space are such that$14$ th card is an ace

$P(14$th card is th card is the first ace$)=\frac{4·51!}{52!}=\frac{4}{52}=\frac{1}{13}$.

$P(14$th card is the first ace$)=\frac{\text{ \# events in which 14th card is the first ace }}{\#\text{ of events in the outcome space }}$

Fix one of the four aces in the $14$th place and then choose without repetition$13$ cards from the $48$non-ace cards for the first $13$delts. After choosing the first $14$cards, permute the remaining $38$of them.

$4·48·47·\dots 36·38$! possible card dealings where the card $14$is the first ace.

$P\left(14\text{ th card is the first ace }\right)=\frac{4·48·47·\dots 36·38!}{52!}=\frac{4·38·37·36}{52·51·50·49}\approx 0.03$