Q7.24 - A First Course in Probability

Question

Three cards are randomly chosen without replacement from an ordinary deck of $52$ cards. Let $X$ denote the number of aces chosen.
($a$) Find $E$[$X\mid$ the ace of spades is chosen].
($b$) Find $E$[$X\mid$ at least one ace is chosen ].

Step-by-Step Solution

Answer

(a)Therefore, the required value is $E[X\mid D]=1.1176$.

(b)Therefore, the required value is $E[X\mid F]=1.0616$.

Step 1: Concept Introductions(part a)

Three cards are randomly chosen without replacement from an ordinary deck of $52$ cards. The number of aces chosen is denoted by $X$.

Step 2 : Explanation(part a)

Three cards are randomly chosen without replacement from an ordinary deck of $52$ cards. The number of aces chosen is denoted by $X$.
Moreover, let's define $X_i$ as follows:
$X_i=\left\{\begin{array}{c}1\text{ if the }{\mathrm{i}}^{th}\text{ card drawn is an ace }\\ 0\text{ Otherwise }\end{array}\right.$

Step 3 :Explanation(part a)

From the definition:
$X=X_1+X_2+X_3$

$E\left[X\right]=E\left[X_1\right]+E\left[X_2\right]+E\left[X_3\right]$

Step 4 :Explanation(part a)

There are four aces in a deck of cards. Hence:
$E\left[X_i\right]=\frac{1}{13}$

$=\frac{1}{13}+\frac{1}{13}+\frac{1}{13}$

$=\frac{3}{13}$

Step 5 :Explanation(part a)

$D^c$Now, let $D$ be the event that the ace of spades is chosen. Then conditioning on the event $D$, the expectation value of $X$ is given by,
$E\left[X\right]=E[X\mid D]P\left(D\right)+E\left[X\mid D^C\right]P\left(D^c\right)$

$=E[X\mid D]\frac{3}{52}+E\left[X\mid D^c\right]\frac{49}{52}$

The event  is that no ace of spades is chosen.

Step 6 :Explanation(part a)

Hence: 

$E\left[X\mid D^c\right]=E\left[\sum _{i=1}^3{X}_i\mid D^c\right]$

$=\sum _{i=1}^{3}E\left[X_i\mid D^c\right]$$=\sum _{i=1}^{3}E\left[X_i\mid D^C\right]$

$=\frac{3}{51}+\frac{3}{51}+\frac{3}{51}$

$=\frac{3}{51}+\frac{3}{51}+\frac{3}{51}$

$=\frac{9}{51}$

Step 7 :Explanation(part a)

From the above calculations,
$E\left[X\right]=E[X\mid D]P\left(D\right)+E\left[X\mid D^c\right]P\left(D^c\right)$

$=E[X\mid D]\frac{3}{52}+E\left[X\mid D^c\right]\frac{49}{52}$

$=E[X\mid D]\frac{3}{52}+\frac{49}{52}\frac{9}{51}$

Step 7 :Explanation(part a)

Using the above result for the expectation value of $X$, one has that:
$\frac{3}{13}=E[X\mid D]\frac{3}{52}+\frac{49}{52}\frac{9}{51}$

$E[X\mid D]=4-3\frac{49}{51}$

$=1.1176$

Step9:Final Answer(part a)

Therefore, the required value is $E[X\mid D]=1.1176$.

Step 10 :Concept Introduction(part b)

Now, let $F$ be the event that at least one ace is chosen. Then, by conditioning on the event $F$

Step 11 :Explanation(part b)

Now, let $F$ be the event that at least one ace is chosen. Then, by conditioning on the event $F$, one has that:
$E\left[X\right]=E[X\mid F]P\left(F\right)+E\left[X\mid F^c\right]P\left(F^c\right)$

The event that no ace chosen is denoted by $F^c$,

Step 12 :Explanation(part b)

which has $0$ probability of occurring: hence, $E\left[X\right]=E[X\mid F]P\left(F\right)+E\left[X\mid F^c\right]P\left(F^c\right)$

$=E[X\mid F]P\left(F\right)$

Step 13 :Explanation(part b)

The probability for event $F$ is given by:

$P\left(F\right)=1-\frac{48}{52}\frac{47}{51}\frac{46}{50}$

$=\frac{52·51·50-48·47·46}{52·51·50}$

Step 14 :Explanation(part b)

Now, putting it all together and solving for $E[X\mid F]$, one has that:
$E\left[X\right]=E[X\mid F]P\left(F\right)$

$\Rightarrow E[X\mid F]=\frac{E\left[X\right]}{P\left(F\right)}$

$=\frac{3/13}{\frac{52·51·50-48·47·46}{52·51·50}}$

$\simeq 1.0616$

Step15:Final Answer(part b)

Therefore, the required value is $E[X\mid F]=1.0616$