According to the Humane Society of the United States there a

According to the Humane Society of the United States, there are approximately 77.5 million owned dogs in the United States, and approximately 40% of all U.S. households own at least one dog.† Suppose that the 40% figure is correct and that 20 households are randomly selected for a pet ownership survey

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) What is the probability that exactly eight of the householdshave at least one dog?

b) What is the probability that at most four of the householdshave at least one dog?

c) What is the probability that more than ten households have atleast one dog?

a)
P( X = 8 ) = ( 20 8 ) * ( 0.4^8) * ( 1 - 0.4 )^12
= 0.1797

b)
P( X < = 4) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 20 4 ) * 0.4^4 * ( 1- 0.4 ) ^16 + ( 20 3 ) * 0.4^3 * ( 1- 0.4 ) ^17 + ( 20 2 ) * 0.4^2 * ( 1- 0.4 ) ^18 + ( 20 1 ) * 0.4^1 * ( 1- 0.4 ) ^19 + ( 20 0 ) * 0.4^0 * ( 1- 0.4 ) ^20   
= 0.051

c)
P( X < = 10) = P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 20 10 ) * 0.4^10 * ( 1- 0.4 ) ^10 + ( 20 9 ) * 0.4^9 * ( 1- 0.4 ) ^11 + ( 20 8 ) * 0.4^8 * ( 1- 0.4 ) ^12 + ( 20 7 ) * 0.4^7 * ( 1- 0.4 ) ^13 + ( 20 6 ) * 0.4^6 * ( 1- 0.4 ) ^14 + ( 20 5 ) * 0.4^5 * ( 1- 0.4 ) ^15 + ( 20 4 ) * 0.4^4 * ( 1- 0.4 ) ^16 + ( 20 3 ) * 0.4^3 * ( 1- 0.4 ) ^17 + ( 20 2 ) * 0.4^2 * ( 1- 0.4 ) ^18 + ( 20 1 ) * 0.4^1 * ( 1- 0.4 ) ^19 + ( 20 0 ) * 0.4^0 * ( 1- 0.4 ) ^20
= 0.8725
P( X > 10) = 1 - P ( X <=10) = 1 -0.8725 = 0.1275