a bag contains 2 snicker bars, 8 trix bars, and 5 crunch bars. a candy bar is chosen, replaced, then another…

a bag contains 2 snicker bars, 8 trix bars, and 5 crunch bars. a candy bar is chosen, replaced, then another is chosen. what is the probability of choosing a trix bar then a crunch bar?
Answer
Explanation:
Step1: Calculate total number of candy - bars
The total number of candy - bars is (2 + 8+5=15).
Step2: Calculate probability of choosing a Trix bar first
The probability of choosing a Trix bar first is (P(\text{Trix})=\frac{8}{15}) since there are 8 Trix bars out of 15 total bars.
Step3: Calculate probability of choosing a Crunch bar second (with replacement)
Since the first bar is replaced, the total number of bars remains 15. The probability of choosing a Crunch bar second is (P(\text{Crunch})=\frac{5}{15}) as there are 5 Crunch bars out of 15 total bars.
Step4: Calculate the joint - probability
Since the two events are independent (because of replacement), the probability of choosing a Trix bar then a Crunch bar is (P = P(\text{Trix})\times P(\text{Crunch})). Substitute the values: (P=\frac{8}{15}\times\frac{5}{15}=\frac{8\times5}{15\times15}=\frac{40}{225}=\frac{8}{45}).
Answer:
(\frac{8}{45})