Based on a study by Dr. P. Sorita Soni at Indiana University, assume that eye colors in the United States are distributed as follows: 40% brown, 35% blue, 12% green, 7% gray, 6% hazel. If three people are randomly selected, what is the probability that one will have brown eyes, one will have gray eyes, and one will have green eyes (with replacement)

Answer

0.00336

Step-by-step explanation:

Given :

P(Brown), P(Br) = 0.4

P(Blue) = P(b= 0.35

P(green,) = P(g) = 0.12

P(gray), P(Gr) = 0.07

If 3 individuals are selected, the probability that one has brown eye, 1 has Grey and one has green.

Since the probabilities are independent :

P(Br) * P(Gr) * P( g)

0.3 * 0.07 * 0.12

= 0.00336

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

throw the computer

step-by-step explanation: