You receive calls from either your family or friends Suppose

You receive calls from either your family or friends. Suppose that the probability to get calls from family is 1/4. Compute the probability that in a sample of 25 (independent) calls, you get exactly 6 calls from friends? Compute the probability that in a sample of 15 (independent) calls, you get exactly 5 calls from friends? Compute the probability that in a sample of 10 (independent) calls, you get exactly 3 calls from friends?

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

a)
P = 3/4 calls from friends = 0.75
P( X = 6 ) = ( 25 6 ) * ( 0.75^6) * ( 1 - 0.75 )^19
= 0.0000001147

b)
P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 5 ) * 0.75^5 * ( 1- 0.75 ) ^10 + ( 15 4 ) * 0.75^4 * ( 1- 0.75 ) ^11 + ( 15 3 ) * 0.75^3 * ( 1- 0.75 ) ^12 + ( 15 2 ) * 0.75^2 * ( 1- 0.75 ) ^13 + ( 15 1 ) * 0.75^1 * ( 1- 0.75 ) ^14 + ( 15 0 ) * 0.75^0 * ( 1- 0.75 ) ^15   
= 0.000795

c)
P = Call from family = 1/4 = 0.25

P( X < 3) = P(X=2) + P(X=1) + P(X=0)
= ( 10 2 ) * 0.25^2 * ( 1- 0.25 ) ^8 + ( 10 1 ) * 0.25^1 * ( 1- 0.25 ) ^9 + ( 10 0 ) * 0.25^0 * ( 1- 0.25 ) ^10 + = 0.525593
P( X > = 3 ) = 1 - P( X < 3) = 0.474407

d)

P( X < 1) = P(X=0)
= ( 25 0 ) * 0.75^0 * ( 1- 0.75 ) ^25
= 0
P( X > = 1 ) = 1 - P( X < 1) = 1