Answer to Question #342783 in Statistics and Probability for Ahmad

Question

In a restaurant an average of two out of every five customers asks for water with their meal. A family

of ten members arrives in the restaurant. What is the probability that:

a. Exactly seven members ask for water?

b. Less than nine members ask for water?

Answer 1

Here we have a binomial distribution:

$P(x)=\frac {n!} {(n-x)!x!}p^xq^{n-x}$

where $n=10, p=2/5,q=1-p=3/5$

So,

a. $P(7)=\frac {10!} {(10-7)!7!}(\frac 2 5)^7(\frac 3 5 )^{10-7}=0.197*0.216=0.0425=4.25\%$

b. $P(<9)=1-P(=9)-P(=10)$

$=1-\frac {10!} {(10-9)!9!}(\frac 2 5)^9(\frac 3 5 )^{10-9}-(\frac 2 5)^{10}=1-0.00157-0.0001=0.9983=99.83\%$