A club has 30 members.
1. How many different 5-member committees can be formed?
(30*29*28*27*26) / (5*4*3*2*1) = 142,506
2. What is the probability that Tom is a member of the committee assuming that there is only one Tom in the club?
The number of favorable outcomes is the number of 4-member committees that do not include Tom:
(29*28*27*26) / (4*3*2*1) = 23,751
P(Tom is on committee) = 23,751 / 142,506 = 1/6 ≈ 0.1667
3. What is the probability that the committee includes a member named Tom assuming that there are two Toms in the club?
Solution 1
There are two possibilities: just one of the two Toms is on the committee or both Toms are on the committee.
a) To find the number of 5-member committees with just one Tom find the number of 4-member committees with no Tom and then add either of the two Toms to make a 5-member committee:
2 * [(28*27*26*25) / (4*3*2*1)] = 40,950
b) To find the number of committees with both Toms, find the number of 3-member committees with no Toms and add both Toms to make a 5-member committee:
(28*27*26) / (3*2*1) = 3,276
The total number of committees that include at least one Tom is 40,959 + 3,276 = 44,226
P(Tom is on the committee) = 44,226 / 142,506 ≈ .3103
Solution 2
A simpler approach is to calculate the number of 5-member committees that do not include a member named Tom:
(28*27*26*25*24) / (5*4*3*2*1) = 98280
P(Tom is not on the committee) = 98280 / 142506
P(Tom is on the committee) = 1 - P(Tom is not on the committee) = 1 - 98280/142506 ≈ .3103