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?

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.

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

let's fu(!

step-by-step explanation:

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake