Q.3.7

Question

A friend randomly chooses two cards, without replacement, from an ordinary deck of $52$ playing cards. In each of the following situations, determine the conditional probability that both cards are aces. $110$ Chapter $3$ Conditional Probability and Independence.

(a) You ask your friend if one of the cards is the ace of spades, and your friend answers in the affirmative.

(b) You ask your friend if the first card selected is an ace, and your friend answers in the affirmative.

(d) You ask your friend if either of the cards selected is an ace, and your friend answers in the affirmative.

Step-by-Step Solution

Verified

Answer

The required probabilities are,

a) $P (B ∣ S) = \frac{1}{17}$

b) $P (B ∣ A_{1}) = \frac{1}{17}$

c) $P (B ∣ A_{2}) = \frac{1}{17}$

d) $P (B ∣ A) = \frac{1}{33}$

1Step 1: Given Information (Part a)

Two cards are selected, in an order from a deck of 52 cards

Events:

$B$ - both cards are aces.

$S$ - one card is the ace of spades.

2Step 2: Explanation (Part a)

Every ordered pair of different cards is equally likely to be the drawn pair: $52 \cdot 51$ possibilities.

If one card is an ace of spades, the other can be any of the $51$ other cards, taking into consideration the order -

$2 \cdot 51$ equally likely possibilities in the reduced outcome space

If additionally both cards are aces, there are $3$ possibilities for the other card and $2$ for the order $- 6$ possibilities.

Therefore:

P (B ∣ S) = \frac{# of events from the reduced space that are in B}{# of events from the reduced space} = \frac{6}{2 \cdot 51} = \frac{1}{17}

3Step 3: Final Answer (Part a)

The required probability is $P (B | S) = \frac{1}{17}$ .

4Step 4: Given Information (Part b)

$B -$ both cards are aces.

$A_{1} -$ first card in the ace.

5Step 5: Explanation (Part b)

The number of events in the reduced outcome space:

The first card can be any of the $4$ aces, and the second one any of the remaining $51 - 4 \cdot 51$

If both cars are aces:

First can be any of the $4$ aces, and the second any of the remaining $3 - 4 \cdot 3$

P (B ∣ A_{1}) = \frac{# of events from the reduced space that are in B}{# of events from the reduced space} = \frac{4 \cdot 3}{4 \cdot 51} = \frac{1}{17}

6Step 6: Final Answer (Part b)

The required probability is $P (B | A_{1}) = \frac{1}{17}$ .

7Step 7: Given Information (Part c)

$B -$ both cards are aces.

$A_{2} -$ second card is an ace.

8Step 8: Explanation (Part c)

This is identical as b), because two cards are drawn randomly

P (B ∣ A_{2}) = \frac{of events from the reduced space that are in B}{# of events from the reduced space} = \frac{4 \cdot 3}{4 \cdot 51} = \frac{1}{17}

9Step 9: Final Answer (Part c)

The required probability is $P (B | A_{2}) = \frac{1}{17}$ .

10Step 10: Given Information (Part d)

$B -$ both cards are aces.

$A = A_{1} \cup A_{2}$ - one of the cards is an ace.

11Step 11: Explanation (Part d)

According to the formula of inclusion and exclusion for the number of elements in set:

n (A) = n (A_{1} \cup A_{2}) = n (A_{1}) + n (A_{2}) - n (A_{1} A_{2})

From sections b) and c) the number of events in $A_{1} - n (A_{1})$ and $A_{2}- $A_{2} - n (A_{2})$ are $4 \cdot 51$ . And there are $4 \cdot 3$ draws in which both cards are aces $A_{1} A_{2}$ .