Question

Two candy bowls contain red and green candies. The first bowl contains an even mix of red and green​ candies, while the second bowl contains 80​% red and 20​% green candies. A bowl is​ chosen, and a candy from that bowl is randomly selected. Suppose that whenever a red candy is​ chosen, it came from the first bowl 50.0​% of the time. What is the probability that the first bowl is​ chosen?

Answer 1

Answer: The probability that the first bowl is chosen is 0.38 = 38%.

Step-by-step explanation:

Number of bowls = 2

Probability of getting first bowl P(A)= 50%

Probability of getting second bowl P(B) = 100% - 50% = 50%

Percentage of red candies in first bowl P(R|A)= 50%

Percentage of green candies in first bowl P(G|A) = 50%

Percentage of red candies in second bowl P(R|B)= 80%

Percentage of green candies in second bowl P(G|B)= 20%

We have given that a red candy is chosen, so the probability that the first bowl is chosen.

$P(A|R)=\dfrac{P(A).P(R|A)}{P(A).P(R|A)+P(B).P(R|B)}\\\\P(A|R)=\dfrac{0.5\times 0.5}{0.5\times 0.5+0.5\times 0.8}\\\\P(A|R)=\dfrac{0.25}{0.65}\\\\P(A|R)=0.38$

Hence, the probability that the first bowl is chosen is 0.38 = 38%.