Question

A bag of Jolly Ranchers has 35 blue raspberry, 22 watermelon, 18 cherry, and 15 green apple. If Addison takes out one, does not replace it, and takes out a second, what is the probability that both will be watermelon?

Answer 1

Answer:

The probability that both Jolly Ranchers will be watermelon is $\displaystyle\bf \frac{77}{1335}$.

Step-by-step explanation:

To find the probability of taking 2 watermelons, we create 2 events where:

1. A = 1st Jolly Rancher is watermelon
2. B = 2nd Jolly Rancher is watermelon

Hence, the event of taking 2 watermelons on both 1st and 2nd is P(A and B) or P(A∩B).

Since the outcome of event B depends on event A, then these events are dependent events (conditional probability), where:

$\boxed{P(A\cap B)=P(A)\times P(B|A)}$

Event A:

• Total number of Jolly Ranchers $(n(S)_A)$ = 35 + 22 + 18 + 15 = 90
• Total number of watermelon $(n(A))$ = 22

$\displaystyle P(A)=\frac{n(A)}{n(S)_A}$

        $\displaystyle =\frac{22}{90}$

Event B:

• Total number of marbles Jolly Ranchers $(n(S)_B)$ = 90 - 1 = 89 (1 Jolly Rancher was taken at event A)
• Total number of watermelon $(n(B))$ = 22 - 1 = 21 (1 watermelon was taken at event A)

$\displaystyle P(B)=\frac{n(B)}{n(S)_B}$

        $\displaystyle =\frac{21}{89}$

Therefore:

$P(A\cap B)=P(A)\times P(B|A)$

               $\displaystyle=\frac{22}{90} \times\frac{21}{89}$

               $\displaystyle=\bf \frac{77}{1335}$