If 60 of Americans are right handed and 40 are left handed a

If 60% of Americans are right handed and 40% are left handed.

a) If we randomly pick 15 Americans what is the probablity that all 15 are right handed?

b) What is the probability that exactly 10 are right handed

c) what is the probability that fewer than 5 are right handed

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( X = 15 ) = ( 15 15 ) * ( 0.6^15) * ( 1 - 0.6 )^0
= 0.00047

b)
P( X = 10 ) = ( 15 10 ) * ( 0.6^10) * ( 1 - 0.6 )^5
= 0.185938

c)
P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 4 ) * 0.6^4 * ( 1- 0.6 ) ^11 + ( 15 3 ) * 0.6^3 * ( 1- 0.6 ) ^12 + ( 15 2 ) * 0.6^2 * ( 1- 0.6 ) ^13 + ( 15 1 ) * 0.6^1 * ( 1- 0.6 ) ^14 + ( 15 0 ) * 0.6^0 * ( 1- 0.6 ) ^15
= 0.009348