Chapter 13.10, Problem 9PPS

To determine

The probability of each card that is not replaced and second card is drawn.

  $P\left(\text{B}\text{and}\text{F}\right)$

Answer to Problem 9PPS

The probability of each card that is not replaced and second card is drawn is $P\left(\text{B}\text{and}\text{F}\right)=0.0179$

Explanation of Solution

Given:

Letters: $A,B,C,D,E,F,G,H.$

First consider a deck $M$ of eight cards with letters $A,B,C,D,E,F,G,H.$

So, think that a card is drawn at random from $M$ such that it is not replaced and a second one is drawn.

So as the first card is not replaced the first event affects the second event so these are the dependent events.

Besides the overall probability of two dependent events is calculated by multiplying the probability with the first event by the second event after the occurrences of first event.

The probability of $P\left(\text{B}\text{and}\text{F}\right)$ as follows:

  $P\left(B\right)=\frac{1}{8}$

As there is a card with the letter $B$ is drawn, there are $7$ cards left in the deck.

  $P\left(F\right)=\frac{1}{7}$

So, the probability of $P\left(\text{B}\text{and}\text{F}\right)$ is

  $\begin{array}{l}P\left(\text{B}\text{and}\text{F}\right)=P\left(B\right)P\left(F\right)\\ P\left(\text{B}\text{and}\text{F}\right)=\frac{1}{8}\times \frac{1}{7}\\ P\left(\text{B}\text{and}\text{F}\right)=\frac{1}{56}=0.0179\end{array}$

Conclusion:

Hence, $P\left(\text{B}\text{and}\text{F}\right)=0.0179$