Chapter 4, Problem 29CQ

a.

To determine

To find: The probability that all 5 cards are red.

Answer to Problem 29CQ

The probability that all 5 cards are red is $\frac{253}{9,996}$.

Explanation of Solution

The number of red cards in a desk is 26.

Calculation:

Here, the selections are made without replacement. Thus, the selections are dependent.

Multiplication rule for dependent events:

If the events A and B are dependent, the probability of occurring of event B is affected by event A, that is $P\left(A\text{ and }B\right)$ is $P\left(A\right)\times P\left(B|A\right)$.

Then the probability that all 5 cards are red is,

$P\left(\begin{array}{l}\text{All 5 cards}\\ \text{are Red cards}\end{array}\right)=P\left(\begin{array}{l}\left(\text{First card is Red}\right)\text{ and }\\ \left(\begin{array}{l}\text{Second card is Red given that}\\ \text{first card is Red}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Third card is Red given that}\\ \text{both first and second cards are Red}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fourth card is Red given that}\\ \text{first, second and third cards are Red}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fifth card is Red given that}\\ \text{first, second, third and fourth cards are Red}\end{array}\right)\end{array}\right)$

                                                                                                        $\begin{array}{c}=\left[\begin{array}{l}P\left(\text{First card is Red}\right)\times \\ P\left(\text{Second card was Red | First card is Red }\right)\times \\ P\left(\text{Third card was Red | First and second cards are Red}\right)\times \\ P\left(\text{Fourth card was Red | First, second and third cards are Red}\right)\times \\ P\left(\text{Fifth card was Red | First,second,third and fourth cards are Red}\right)\end{array}\right]\\ =\frac{26}{52}\times \frac{25}{51}\times \frac{24}{50}\times \frac{23}{49}\times \frac{22}{48}\\ =\frac{253}{9,996}\end{array}$

Thus, theprobability that all 5 cards are red is $\frac{253}{9,996}$.

b.

To determine

To find: The probability that all 5 cards are diamonds.

Answer to Problem 29CQ

The probability that all 5 cards are diamonds is $\frac{33}{66,640}$.

Explanation of Solution

The number of diamonds in a desk is 13.

Calculation:

Here, the selections are made without replacement. Thus, the selections are dependent.

Then the probability that all 5 cards are diamonds is,

$P\left(\begin{array}{l}\text{All 5 cards}\\ \text{are diamonds}\end{array}\right)=P\left(\begin{array}{l}\left(\text{First card is diamond }\right)\text{and }\\ \left(\begin{array}{l}\text{Second card is diamond given that}\\ \text{first card is diamond }\end{array}\right)\text{and}\\ \left(\begin{array}{l}\text{Third card is diamond given that}\\ \text{both first and second cards are diamonds}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fourth card is diamond given that}\\ \text{first, second and third cards are diamonds}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fifth card is diamond given that}\\ \text{first, second, third and fourth cards are diamonds}\end{array}\right)\end{array}\right)$

                          $\begin{array}{c}=\left[\begin{array}{l}P\left(\text{First card is diamond}\right)\times \\ P\left(\begin{array}{l}\text{Second card was diamond | }\\ \text{First card is diamond }\end{array}\right)\times \\ P\left(\begin{array}{l}\text{Third card was diamond | }\\ \text{First and second cards are diamonds}\end{array}\right)\times \\ P\left(\begin{array}{l}\text{Fourth card was diamond | }\\ \text{First, second and third cards are diamonds}\end{array}\right)\times \\ P\left(\begin{array}{l}\text{Fifth card was diamond | }\\ \text{First,second,third and fourth cards are diamond}s\end{array}\right)\end{array}\right]\\ =\frac{13}{52}\times \frac{12}{51}\times \frac{11}{50}\times \frac{10}{49}\times \frac{9}{48}\\ =\frac{33}{66,640}\end{array}$

Thus, the probability that all 5 cards are diamonds is $\frac{33}{66,640}$.

c.

To determine

To find: The probability that all 5 cards are aces.

Answer to Problem 29CQ

The probability that all 5 cards are aces is 0.

Explanation of Solution

The given number of aces in a desk is 4.

Calculation:

Here, the selections are made without replacement. Thus, the selections are dependent.

Then the probability that all 5 cards are aces is,

$P\left(\begin{array}{l}\text{All 5 cards}\\ \text{are aces}\end{array}\right)=P\left(\begin{array}{l}\left(\text{First card is ace}\right)\text{and }\\ \left(\begin{array}{l}\text{Second card is ace given that}\\ \text{first card is ace}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Third card is ace given that}\\ \text{both first and second cards are aces}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fourth card is ace given that}\\ \text{first, second and third cards are aces}\end{array}\right)\text{ and}\\ \left(\begin{array}{l}\text{Fifth card is ace given that}\\ \text{first, second, third and fourth cards are aces}\end{array}\right)\end{array}\right)$

                 $\begin{array}{c}=\left[\begin{array}{l}P\left(\text{First card is ace}\right)\times \\ P\left(\text{Second card was ace | First card is ace}\right)\times \\ P\left(\text{Third card was ace | First and second cards are ace}\right)\times \\ P\left(\text{Fourth card was ace | First, second and third cards are ace}\right)\times \\ P\left(\text{Fifth card was ace | First,second,third and fourth cards are ace}\right)\end{array}\right]\\ =\frac{4}{52}\times \frac{3}{51}\times \frac{2}{50}\times \frac{1}{49}\times \frac{0}{48}\\ =0\end{array}$

Thus, the probability that all 5 cards are aces is 0.