Question

The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women (50% chance of choosing either gender). If participants are picked randomly, what is the binomial probability of getting at least 7 women when 10 people are picked?

There is a 75% chance a Starbuck’s customer will want room for milk in a coffee order. Find the probability that less than 8 out of the next 12 customers will want room for milk?

Answer 1

(a) Let X be the number of female participants selected, then it is given that X~Bin (n=10,p=0.5) , we have to calculate the probability of getting atleast 7 women i.e P(X>=7) . We will use the result P(X=k) = ⁿC_k * p^k * (1-p)^n-k where n= 10, p=0.5, k=7,8,9,10 and add these probabilities

Now P(X>=7) = P(X=7)+ P(X=8)+P(X=9) + P(X=10) = 0.1172 + 0.0439 + 0.00976 + 0.000976 = 0.1718

(b) Let X be the number of Starbuck's customer who will want room for milk, then it is given that X~Bin (n=12,p=0.75) , we have to calculate the probability that less than 8 out of the next 12 customers will want room for milk i.e P(X<8) . We will use the result P(X=k) = ⁿC_k * p^k * (1-p)^n-k where n= 12, p=0.25, k=1,2,3,4,5,6,7 and add these probabilities.

Thus P(X<8) = P(X=1) +P(X=2) +P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 0.1576

This result can also be found out by using probabilities for the binomial distribution. by using n=12, X=7 and p=0.75.

This would give the result P(X<=7) = P(X < 8 ) = 0.1576 which is the same as above.