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.

x = -2

−12x−2(x+9)=5(x+4)

Use the distributive property to multiply −2 by x+9.

−12x−2x−18=5(x+4)

Combine −12x and −2x to get −14x.

−14x−18=5(x+4)

Use the distributive property to multiply 5 by x+4.

−14x−18=5x+20

Subtract 5x from both sides.

−14x−18−5x=20

Combine −14x and −5x to get −19x.

−19x−18=20

Add 18 to both sides.

−19x=20+18

Add 20 and 18 to get 38.

−19x=38

Divide both sides by −19.

x=  

−19

38

​

Divide 38 by −19 to get −2.

x=−2