Mr. And Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 81%. If they have four children, what is the probability that exactly four of their four children will have that trait? Round your answer to the nearest thousandth.

0.430 (nearest thousandth)

Step-by-step explanation:

Using binomial probability relation :

p = 0.81

n = 4

x = 4

1 - p = 1 - 0.81 = 0.19

Recall ;

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x = 4) = 4C4 * 0.81^4 * 0.19^(4-4)

P(x = 4) = 1 * 0.43046721 * 1

P(x = 4) = 0.43046721

P(x = 4) = 0.430 (nearest thousandth)

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.