Chapter 0, Problem 20PR

To determine

To calculate: the probability in the card.

Answer to Problem 20PR

The probability of getting two aces without replacement is $\frac{1}{221}.$

Explanation of Solution

Given information:

Total number of cards = 52

Total number of aces=4

Used formula:

  $P(B|A)=\frac{P(A\cap B)}{P(A).}$

Calculation:

According to question,

There are 52 cards in a pack and each pack contains 4 aces

So, probability of drawing first card aces $=\frac{4}{52}=\frac{1}{13}.$

Now, there are total 51 cards and out of those, 3 are aces

  $P(B|A)=$ probability of drawing second card $=\frac{3}{51}.$

what we want to calculate is $P(A\cap B).$

Now, by bayes rule-

  $P(B|A)=\frac{P(A\cap B)}{P(A).}$

Hence,

  $\begin{array}{l}P(A\cap B)=P(B|A).P(A)\\ =\frac{1}{13}\times \frac{3}{51}\\ =\frac{3}{663}=\frac{1}{221}.\end{array}$

So, the probability of getting two aces without replacement is $\frac{1}{221}.$