Consider a male with genotype Aa and a female with genotype

Consider a male with genotype Aa and a female with genotype Aa. a) if they have 8 offspring, what is the probability that exactly 5 offspring have genotype AA b) What is the probability that 5 or more of the offspring have a genotype AA

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
p = P(AA) = P(Aa) = 1/2
=0.5
a)
P( X = 5 ) = ( 8 5 ) * ( 0.5^5) * ( 1 - 0.5 )^3
= 0.2188

b)

P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 8 4 ) * 0.5^4 * ( 1- 0.5 ) ^4 + ( 8 3 ) * 0.5^3 * ( 1- 0.5 ) ^5 + ( 8 2 ) * 0.5^2 * ( 1- 0.5 ) ^6 + ( 8 1 ) * 0.5^1 * ( 1- 0.5 ) ^7 + ( 8 0 ) * 0.5^0 * ( 1- 0.5 ) ^8
= 0.6367

P( X > = 5 ) = 1 - P( X < 5) = 0.3633

As per you stated p(Aa) = 0.25, then answer will be

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X = 5 ) = ( 8 5 ) * ( 0.25^5) * ( 1 - 0.25 )^3
= 0.0231

b)

P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0) +   
= ( 8 4 ) * 0.25^4 * ( 1- 0.25 ) ^4 + ( 8 3 ) * 0.25^3 * ( 1- 0.25 ) ^5 + ( 8 2 ) * 0.25^2 * ( 1- 0.25 ) ^6 + ( 8 1 ) * 0.25^1 * ( 1- 0.25 ) ^7 + ( 8 0 ) * 0.25^0 * ( 1- 0.25 ) ^8 +
= 0.9727
P( X > = 5 ) = 1 - P( X < 5) = 0.0273