Exercise 5 Page 678

Since we do not replace each card before we draw the next one, the occurrence of each event affects the occurrence of the others. Therefore, they are dependent.

Probability of Dependent Events

If the events $A,$ $B,$ and $C$ are dependent, then the probability that $A,$ $B,$ and $C$ will occur is $P(A\text{ and }B\text{ and }C)=P(A)\cdot P(B|A)\cdot P(C|A\text{ and }B).$

From Part A, we already know that $P(A)$ is $\frac{3}{4}.$ The difference this time is that we need to calculate $P(B|A)$ and $P(C|A\text{ and }B).$ Let's do it one at a time. The number of $\text{possible}$ $\text{outcomes}$ in the second drawing is different than in the first one, because we do not replace the first card. Because we selected a card that is not hearts in the first drawing, there remains only $39-1=38$ cards that are not hearts. This means that we have enough information to calculate $P(B|A).$ The number of $\text{possible}$ $\text{outcomes}$ in the third drawing also changes, because we again do not replace the cards. Moreover, there are only $38-1=37$ cards that are not hearts left to select from. We are now able to calculate $P(C|A\text{ and }B).$ According to the formula, to calculate $P(A\text{ and }B\text{ and }C)$ we have to multiply the obatined probabilities $P(A),$ $P(B|A),$ and $P(C|A\text{ and }B).$ Next, we want to compare the probabilities from Part A and Part B. In order to compare the situations in Part A and Part B, we will divide the probability obtained with replacing by the probability obtained without replacing. Therefore, we are $1.02$ times more likely to select $3$ cards of which none are hearts when we replace each card before selecting the next card.