Question 1011016
 
Question:
the probability that a double glazing salesperson will make a sale on a particular day is 0.05. what is the probability that over a three day period the sale person will make: 3 sales, exactly 1 sale, at least 1 sale?   
 
Solution:
We note that the questions satisfies the requirements of the binomial distribution, namely, known number of trials (n=3), constant probability (p=0.05), trials (days) are independent.
We need to assume that the salesman can only make a maximum of one sale a day.
The probability of making x sales in three days is given by the binomial distribution:
{{{P(X=x)=C(n,x)*p^(x)*q^(n-x)}}}
Where C(n,x) is the combination function where C(n,x)=n!/(x!(n-x)!)
 
A. Probability of making 3 sales in three days, P(X=3)
{{{P(X=3)=C(3,3)*0.05^3*(0.95)^0=0.000125}}}
 
B. Probibility of making exactly one sale
{{{P(X=1)=C(3,1)*0.05^1*(0.95)^2=0.13538}}}
 
C. Probability of making at least one sale
{{{P(X>=1) = 1-P(X=0) = 1-C(3,0)*0.05^0*0.95^(3) = 0.14263}}}