binomial example

The Social Science Coffee bar has two early morning servers, Ben and Sean. The manager estimates that each has a 2 in 7 chance of being early or on time to work. They act independently from each other and travel to work by different means.

What are the chances that

1) they will both be late

2) one and only one will be on time and

3) they will both be on time?

Solution method one:

1) The probability of Ben being late is 5/7, p(B)

The probability of Sean being late is also 5/7, p(S)

Therefore the probability of both being late is an application of the multiplication rule so p(B) x p(S) =

5/7 x 5/7 = 25/49 =.5102

2) The probability of Ben being late but Sean being on time is 5/7 x 2/7 = 10/49; which is the same for Sean being late but Ben being on time.

So the probability of one and only one of them being on time is an application of the addition rule. So the probability of one and only one being on time is

10/49 + 10/49 = 20/49 = .4082

3) Finally the probability of both being on time is

2/7 x 2/7 = .0816.

Using the binomial formula

The arrival of a staff member can be viewed as a trial. In our case the trial is a success if the staff member arrives early or on time. Otherwise it’s a failure. In our case we have two trials so n = 2 and p = 2/7 and q = 1-p= 5/7

so B^x = 2/7, B^n-x= 5/7

formula is:

1) Probability of both being late (0 successes) is

2) Probability of only one being early or on time (1 success) is

3) Probability of both being on or before time (2 successes)

.5102 + .4082 + 0816 = 1