Questions asked at end of WIR:
Five people are asked to write down an integer between 1 and 19 inclusive. What is the
probability that exactly 3 of the people pick an even number? (Round your answer to 4 decimal
digits.)
The total ways (denominator) for five people to pick an integer between 1 and 19 inclusive is
195. The number of ways exactly 3 pick an even number is 9*9*9*10*10*C(5,3) because there
are 9 even and 10 odd numbers and then we multiply by the ways that can be done.
So [93*102*C(5,3)]/195 ≈ 0.2944
Suppose that there are eight corporations competing for nine different contracts. If the contracts
are awarded randomly, what is the probability that each corporation will get at least one
contract?
The total ways (denominator) the contracts can be awarded is 89 because each of the 9 contracts
has 8 corporations that can be chosen. The number of ways they can be awarded so that each gets
at least one:
First, choose 8 of the 9 contracts to award to the 8 corporations, P(9,8). Then, award the last
contract to any one of the 8 corporations. So you have 8*P(9, 8), but there are duplicates of each,
so divide by 2.
One challenge to this problem is that you are not asked to round, and the calculator does not
easily give you a simplified fraction. So, dividing by 8 as many times as possible, and then
8 P (9,8) P (9,8) 362880 45360 5670
2835
=
=
=
=
=
dividing by 2, we have
9
8
8
7
6
2*8
2 *8
2 *8
2 *8
2*8
262144
An oil company estimates that only 1 oil well in 11 will yield commercial quantities of oil.
Assume that successful drilled wells represent independent events. If 8 wells are drilled, find the
probability of obtaining commercially successful wells only in the first two wells. (Round your
answer to five decimal places.)
The problem with using binompdf on this one is that we are not just looking for 2 successes, but
specifically the first 2 wells. So we have (1/11)2*(10/11)6 ≈.00467