ynatchev wrote:A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be president, one of the remaining 9 members is to be chosen at random to be secretary, and one of the remaining 8 members is to be chosen at random to be treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be treasurer?
a. 1/720
b. 1/80
c. 1/10
d. 1/9
e. 1/5
Each of the 10 members has an EQUAL chance of being elected secretary.
P(Harry is elected secretary) = 1/10.
Each of the 10 members has an EQUAL chance of being elected treasurer.
P(Harry is elected treasurer) = 1/10.
Since EITHER outcome is favorable, we ADD the fractions:
1/10 + 1/10 = 2/10 = 1/5.
The correct answer is
E.
An alternate approach:
For Harry to be secretary, he must be the SECOND person selected.
For Harry to be treasurer, he must be the THIRD person selected.
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
P(Harry is chosen second) = P(Harry is chosen first) = 1/10.
P(Harry is chosen third) = P(Harry is chosen first) = 1/10.
Since EITHER outcome is favorable, we ADD the fractions:
1/10 + 1/10 = 2/10 = 1/5.
The answer would be the same even if the problem were as follows:
A certain club has 10 members, including Harry. Seven of the 10 members are to be chosen at random to be co-presidents, one of the remaining 3 members is to be chosen at random to be secretary, and one of the remaining 2 members is to be chosen at random to be treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be treasurer?
P(Harry is chosen eighth) = P(Harry is chosen first) = 1/10.
P(Harry is chosen ninth) = P{Harry is chosen first) = 1/10.
Adding the fractions, we get:
1/10 + 1/10 = 2/10 = 1/5.
Other problems that test this concept:
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https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
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