Chapter 13.4, Problem 21E

To determine

To find the probability of removing $13$ cards from a standard deck of cards and have all of them be red.

Answer to Problem 21E

Dependent, $\frac{19}{1160054}$

Explanation of Solution

Given:

A deck of cards.

Calculation:

The events are dependent because outcome of one eventaffects the outcome of another event.

The whole experiment is equivalent to $13$ successive experiments as follows-

Let $A_1$ be the event of selecting the red card and do not put back.

Let $A_2$ be the event of selecting anotherred card and do not put back

Let $A_3$ be the event of selecting another red card and do not put back.

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Let $A_{13}$ be the event of selecting another red card.

Total sample space $=52$

Therefore probability of event $A_1$ is $P(A_2)=\frac{26}{52}$

And probability of event $A_2$ is $P(A_2)=\frac{25}{51}$

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And probability of event $A_{13}$ is $P(A_{13})=\frac{14}{40}$

Hence,

  $\begin{array}{l}P(A_1{\text{ and A}}_2\text{ and}\mathrm{.....}{\text{and A}}_{13})=P(A_1)\times P({\text{A}}_2)\times P({\text{A}}_{13})\\ P(A\text{ and }B)=\frac{26}{52}\times \frac{25}{51}\times \mathrm{.........}\times \frac{14}{40}=\frac{19}{1160054}\end{array}$