Ahuman gene carries a certain disease from the mother to the child with a probability rate of 3939%. that is, there is a 3939% chance that the child becomes infected with the disease. suppose a female carrier of the gene has threethree children. assume that the infections of the threethree children are independent of one another. find the probability that at least one of the children get the disease from their mother. round to the nearest thousandth.

The probability that human gene carries a certain disease from the mother to the children is 0.39. A female carrier of gene has three children and infection of the three children are independent of one another.

P(getting disease from mother) = 0.39

P(not getting disease ) = 1- 0.39 = 0.61

Probability that at least one of the children get the disease from their mother

P(at least one out of 3 get disease) = 1 - P( None out of 3 get disease)

= 1 - P( all three do not get disease)

= 1 - (0.61)^3

= 1 - 0.2269

P(at least one out of 3 get disease) = 0.7731

The probability that at least one out of three get disease is 0.773

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: