The probabiity is 314 that the gestation period of a woman w

The probabiity is .314 that the gestation period of a woman will exceed 9 months. In seven human births, what is the probability that the number in which the gestation period exceeds 9 months is

a) exactly 3

b) exactly 5

c) at least 5

d) between 3 and 5

Solution

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

a)
P( X = 3 ) = ( 7 3 ) * ( 0.314^3) * ( 1 - 0.314 )^4
= 0.24
b)
P( X = 5 ) = ( 7 5 ) * ( 0.314^5) * ( 1 - 0.314 )^2
= 0.0302
c)

P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0) +   
= ( 7 4 ) * 0.314^4 * ( 1- 0.314 ) ^3 + ( 7 3 ) * 0.314^3 * ( 1- 0.314 ) ^4 + ( 7 2 ) * 0.314^2 * ( 1- 0.314 ) ^5 + ( 7 1 ) * 0.314^1 * ( 1- 0.314 ) ^6 + ( 7 0 ) * 0.314^0 * ( 1- 0.314 ) ^7 +
= 0.9649
P( X > = 5 ) = 1 - P( X < 5) = 0.0351

d)
P( X = 4 ) = ( 7 4 ) * ( 0.314^4) * ( 1 - 0.314 )^3
= 0.1098

P( 3 < = X < = 5) = P( X = 3 ) + P( X = 4 )+ P( X = 5 ) = 0.24 + 0.1098 + 0.0302 = 0.38