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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by GmatKiss » Sat May 05, 2012 12:09 pm
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
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by aneesh.kg » Sun May 06, 2012 12:03 am
Probability of Harry being Secretary or Treasurer = (Total number of ways in Harry can be made treasurer or secretary)/(Total number of ways in which Selections can be made)

Total number of ways of selecting Harry as Secretary can be found by selecting one among anyone except Harry for the president (9C1) because if Harry is selected as the president he cant become Treasurer or Secretary, then select Harry (1C1) for Secretary and then select one among the remaining (8C1) for Treasurer.

Total number of ways of selecting Harry as Treasurer can be found by selecting one among anyone except Harry for the president (9C1), then select any of remaining people (8C1) except Harry for Secretary then Harry (1C1) for Treasurer.

Total number of ways in Harry can be made treasurer or secretary =
9C1 * 1C1 * 8C1 + 9C1 * 8C1 * 1C1
Total number of ways in which Selections can be made (as specified in the problem) = 10C1 * 9C1 * 8C1

therefore,
The required probability =
(9*1*8) + (9*8*1)/(10*9*8)
=
(2*9*8/10*9*8)
=
[spoiler]1/5[/spoiler]

[spoiler](E)[/spoiler] is the answer.

Alternatively:

Required Probability = 1 - Probability of (Harry being the president OR Harry being neither Secretary nor the Treasurer)
= 1 - (1C1*9C1*8C1 + 9C1*8C1*7C1)/(10C1*9C1*8C1)
= 1 - (9*8*8)/(10*9*8)
= 1 - (4/5)
= [spoiler]1/5[/spoiler]

Same answer.
Aneesh Bangia
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by Brent@GMATPrepNow » Sun May 06, 2012 7:59 am
GmatKiss 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
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
[spoiler]= E[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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