Question

The probability that​ brown-eyed parents, both with the recessive gene for​ blue, will have a child with brown eyes is 0.65. If such parents have 6 ​children, what is the probability that they will have ​(A) All​ blue-eyed children? ​(B) Exactly 5 children with brown​ eyes? ​(C) At least 5 children with brown​ eyes?

Answer 1

Given

p=0.65 -Probablity that colour of eye is brown

q=1-p =0.35 -Probablity that colour of eye is blue

Probablity function for given situation

where x is number of children with brown eyes.

(A) Probablity of having All Blue-eyed children :

Evaluating P(X=0)

=(0.35)6-0

=(0.35)6

=0.00183

(B) Probablity of exactly 5 children with brown eyes :

Evaluating P(X= 5)

$=\frac { 6! }{ 5!\times 1! } \times { (0.65) }^{ 5 }{ (0.35) }^{ 1 }\\ =\quad \frac { 6\times 5! }{ 5!\times 1! } \times { (0.65) }^{ 5 }{ (0.35) }^{ 1 }\\ =6\times { (0.65) }^{ 5 }{ (0.35) }^{ 1 }\\ =6\times 0.116\times 0.35\quad \\ =0.2436$

(C) Evaluating P(X=6)

$=\frac { 6! }{ 6!\times 0! } \times { (0.65) }^{ 6 }{ (0.35) }^{ 0 }\\ =\quad 1\times { (0.65) }^{ 6 }\\ =\quad { (0.65) }^{ 6 }\\ =0.0754$

Probablity that atleast 5 children will have brown eyes :

$P(X\geq 5)=P(5)+P(6)$

=0.2436+0.0754

=0.319