Conditional Probability Question?
A committee of 50 senators is formed (from 100). What is the odds that both senators from Alaska are chosen for the committee given that one is chosen
- wiseguyLv 61 decade agoFavorite Answer
Dear John C,
You don't say what process is used in forming the committee. The answer to your question, of course, will be highly dependent on this process. It may very well be that the selection method isn't random, but is determined by a single powerful senator. In this case, the choices would likely depend on that senator's party and the party (or parties) of the Alaskan senators and their relationships. The answer could also hinge on the purpose of the committee, since the senators from Alaska might have expertise or influence in specific areas that would make their selection either more or less likely than just by chance.
However, given that you don't provide other information, I'll guess that this is just a poorly-written problem for which the author is making assumptions about the process without stating them. As such, I'll make the assertion that the process is random so that every senator is equally likely to be selected. Note that this assertion is something which I am adding to the problem, but it is not part of the problem as it is stated.
Now let's define some notation:
A(0) is the event that exactly zero senators from Alaska are chosen,
A(1) is the event that exactly one senator from Alaska is chosen, and
A(2) is the event that exactly two senators from Alaska are chosen.
Based on the assertion about the selection process, we can compute probabilities of the three events just stated. For example, the probability of A(0) can be found by multiplying the number of ways that zero Alaskan senators can be selected from a set of two, by the number of ways that the rest of the committee can be selected (i.e., the number of ways that 50 others can be selected from a set of 98), then dividing by the total number of ways of forming a committee of 50 senators from the set of all 100 senators. These probabilities are described by the hypergeometric distribution.
P(A(0)) = C(2,0) C(98,50) / C(100,50),
P(A(1)) = C(2,1) C(98,49) / C(100,50), and
P(A(2)) = C(2,2) C(98,48) / C(100,50),
where C(n,k) = n! / [k! (n - k)!], which is known as the binomial coefficient, is the number of ways that exactly k items can be chosen from a set of n items. Thus,
P(A(0)) = (50 / 100) (49 / 99) = (1 / 2) (49 / 99) = 49 / 198,
P(A(1)) = 2 (50 / 100) (50 / 99) = 2 (1 / 2) (50 / 99) = 50 / 99, and
P(A(2)) = (50 / 100) (49 / 99) = (1 / 2) (49 / 99) = 49 / 198.
The probability that both senators from Alaska are chosen for the committee of fifty senators, given that at least one is chosen can be written as P(A(2) | [A(1) OR A(2)]).
(Notice here that if we do not infer that "at least" one Alaskan senator is chosen, but instead that "exactly" one is chosen, then we would have P(A(2) | A(1)) = 0, since A(1) and A(2) are mutually exclusive.)
From the formula for conditional probability we can write
P(A(2) | [A(1) OR A(2)])
= P(A(2) & [A(1) OR A(2)]) / P([A(1) OR A(2)])
= P(A(2)) / P([A(1) OR A(2)])
= P(A(2)) / [1 - P(A(0))]
= (49 / 198) / [1 - (49 / 198)]
= 49 / [198 - 49]
= 49 / 149
= 0.32886 (to five decimal places).
- Anonymous1 decade ago