Despite the repeated effort by the government to reform how Wall Street pays its executives,
some of the nation’s biggest banks are continuing to pay out bonuses nearly as large as those in
the best years before the crisis (The Washington Post, January 15, 2010). It is known that 10 out
of 15 members of the board of directors of a company were in favor of a bonus. Suppose three
members were randomly selected by the media.
What is the probability that all of them were in favor of a bonus? (Round your answer to 4
decimal places.)
What is the probability that at least two members were in favor of a bonus? (Round your answer
to 4 decimal places.)
Despite the repeated effort by the government to reform how Wall Street pays its executives,
some of the nation’s biggest banks are continuing to pay out bonuses nearly as large as those in
the best years before the crisis (The Washington Post, January 15, 2010). It is known that 10 out
of 15 members of the board of directors of a company were in favor of a bonus. Suppose three
members were randomly selected by the media.
Solution
a) P(all 3 in favor of bonus) = (10/15) * (9/14) * (8/13) = 24/91 = 0.2637
b) P(at least 2 in favor of the bonus) = P(2 in favor) + P(3 in favor)
The second term of the expression was solved in part (a). For the first term,
P(2 in favor) = (10/15) * (9/14) * (5/13) * 3 = 0.4945
The reason the 3 was included in the expression is to account for the multiple combinations in
which the person who was not in favor could be selected (i.e. first, second, or third order)
Therefore P(at least 2) = P(2) + P(3) = 0.7582.
Despite the repeated effort by the government to reform how Wall Str.pdf
1. Despite the repeated effort by the government to reform how Wall Street pays its executives,
some of the nation’s biggest banks are continuing to pay out bonuses nearly as large as those in
the best years before the crisis (The Washington Post, January 15, 2010). It is known that 10 out
of 15 members of the board of directors of a company were in favor of a bonus. Suppose three
members were randomly selected by the media.
What is the probability that all of them were in favor of a bonus? (Round your answer to 4
decimal places.)
What is the probability that at least two members were in favor of a bonus? (Round your answer
to 4 decimal places.)
Despite the repeated effort by the government to reform how Wall Street pays its executives,
some of the nation’s biggest banks are continuing to pay out bonuses nearly as large as those in
the best years before the crisis (The Washington Post, January 15, 2010). It is known that 10 out
of 15 members of the board of directors of a company were in favor of a bonus. Suppose three
members were randomly selected by the media.
Solution
a) P(all 3 in favor of bonus) = (10/15) * (9/14) * (8/13) = 24/91 = 0.2637
b) P(at least 2 in favor of the bonus) = P(2 in favor) + P(3 in favor)
The second term of the expression was solved in part (a). For the first term,
P(2 in favor) = (10/15) * (9/14) * (5/13) * 3 = 0.4945
The reason the 3 was included in the expression is to account for the multiple combinations in
which the person who was not in favor could be selected (i.e. first, second, or third order)
Therefore P(at least 2) = P(2) + P(3) = 0.7582
