A certain club has 10 members, including Harry. One of the 1

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A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/80
(C) 1/10
(D) 1/9
(E) 1/5

OA E

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by nonplus2 » Fri Oct 12, 2018 11:50 pm
Each member out of 10, including Harry, has equal chances to be selected for any of the positions (the sequence of the selection is given just to confuse us). The probability that Harry will be selected to be the secretary is 1/10 and the probability that Harry will be selected to be the treasurer is also 1/10. So, the probability that Harry will be selected to be either the secretary or the the treasurer is 1/10+1/10=2/10.

Correct Answer E

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BTGmoderatorDC wrote:A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the 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.

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by Brent@GMATPrepNow » Sat Oct 13, 2018 5:03 am
A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

A. 1/720
B. 1/80
C. 1/10
D. 1/9
E. 1/5
As with many probability questions, we can solve this using counting techniques or probability rules. So, before you embark on one approach, try to determine which one is faster.

Here's the probability approach:

P(Harry selected Secretary or Treasurer) = 1 - P(Harry selected president OR Harry selected for no positions)
P(Harry selected president) = 1/10
P(Harry selected for no positions)= (9/10)(8/9)(7/8) = 7/10

Aside: For the last probability, the probability is 9/10 that Harry is not selected president. Once a president is selected, there are 9 members remaining so the probability is 8/9 that Harry is not selected secretary. Once a secretary is selected, there are 8 members remaining so the probability is 7/8 that Harry is not selected treasurer.

Okay, so P(Harry selected Secretary or Treasurer) = 1 - [1/10 + 7/10]
= 1 - [8/10]
= 2/10
= 1/5

Answer: E

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by Scott@TargetTestPrep » Wed Oct 17, 2018 6:09 pm
BTGmoderatorDC wrote:A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/80
(C) 1/10
(D) 1/9
(E) 1/5
We are given that a club has 10 members, including Harry. When selecting a president, secretary, and treasurer from the 10 members, we must determine the probability that Harry will either be chosen secretary or treasurer.

Since we have 10 total people the probability that Harry is chosen to be the secretary is 1/10 and the probability that he is chosen to be the treasurer is 1/10.

Thus, the probability that he is chosen to be the secretary or treasurer is 1/10 +1/10 = 1/5.

Alternate Solution:

There are 10P3 = 10!/(10 - 3)! = 10!/7! = 10*9*8 = 720 different ways of assigning a president, a secretary and a treasurer to the club.

If we assume that Harry is chosen to be the secretary, then from the remaining 9 people, there are 9P2 = 9!/7! = 72 ways to assign the president and the treasurer positions.

Similarly, if we assume that Harry is chosen to be the treasurer, then there are 9P2 = 72 ways to assign the president and the secretary positions.

Thus, the probability of Harry being either the treasurer or the secretary is 72/720 + 72/720 = 1/10 + 1/10 = 1/5.

Answer: E

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