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 30% red and 70% 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 75.0% of the time. what is the probability that the first bowl is chosen? g

Our required probability is 0.642.

Step-by-step explanation:

Since we have given that

Probability of red candies from first bowl = 50% = 0.5

Probability of green candies from second bowl = 50% = 0.5

Let the probability of chosing first bowl be 'x'.

Let the probability of chosing second bowl be '1-x'

Probability of red candies from second bowl = 30% = 0.3

Probability of green candies from second bowl = 70% =0.7

Probability of red that comes from the first bowl = 75% =

According to question,

Hence, our required probability is 0.642.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: