Question

julio has two bags of gummy candies. one at a time, he randomly pulled 8 pineapple lime, and 7 strawberry candies out of the first bag and put them back each time. he also randomly pulled 11 mango, 3 tamarind, and 6 watermelon candies out of the second bag and put them back each time. based on this information, what is the probability that julio will select a strawberry candy from the first bag and a tamarind candy from the second bag the next time he takes a candy from each bag? type a number in each box.

Explanation:

Step1: Calculate probability of strawberry from first bag

The first bag has \(8 + 7=15\) candies in total. The probability of selecting a strawberry candy from the first bag, \(P(S_1)\), is the number of strawberry candies divided by the total number of candies in the first bag. So \(P(S_1)=\frac{7}{15}\).

Step2: Calculate probability of tamarind from second bag

The second bag has \(11 + 3+6 = 20\) candies in total. The probability of selecting a tamarind candy from the second bag, \(P(T_2)\), is the number of tamarind candies divided by the total number of candies in the second bag. So \(P(T_2)=\frac{3}{20}\).

Step3: Calculate joint - probability

Since the events of selecting a candy from the first bag and the second bag are independent, the probability of both events occurring is the product of their individual probabilities. So \(P = P(S_1)\times P(T_2)=\frac{7}{15}\times\frac{3}{20}\).
\[

$$\begin{align*} \frac{7}{15}\times\frac{3}{20}&=\frac{7\times3}{15\times20}\\ &=\frac{21}{300}\\ &=\frac{7}{100} = 0.07 \end{align*}$$

\]

Answer:

0.07