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Really tough probability question

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by ynatchev » Thu Feb 28, 2013 12:24 am
Hello,

can anyone suggest to me a better way to solve this problem because the book explanation is really hard to understand(although i get it i was just asking if somebody knows a better way to do it) and its long.

Question 7 from the Official gmat study guide(Diagnosis test)

7)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 to be the secretary, one of the remaining 8 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?
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by Anurag@Gurome » Thu Feb 28, 2013 12:31 am
ynatchev 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 to be the secretary, one of the remaining 8 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?
Required probability = 1 - [Probability that Harry is selected for president or Harry is not selected at all]

Probability that Harry is selected for president = 1/10
Probability that Harry is not selected at all = (9/10)*(8/9)*(7/8) = 7/10

Hence, equired probability = 1 - [1/10 + 7/10] = 1 - 8/10 = 2/10 = 1/5
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by Anurag@Gurome » Thu Feb 28, 2013 12:38 am
Another method for solving this problem...

Total number of ways to select the president, secretary and treasurer without any constraint = (10C1)*(9C1)*(8C1) = 10*9*8

Number of ways to select the president, secretary and treasurer such Harry is either Secretary or Treasurer = [(9C1)*(1C1)*(8C1) + (9C1)*(8C1)*(1C1)] = [9*8 + 9*8]

hence, required probability = [9*8 + 9*8]/(10*9*8) = (1 + 1)/10 = 2/10 = 1/5
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by ynatchev » Thu Feb 28, 2013 12:38 am
Thank you so much. This is a way better solution than the textbook one.I'll try remembering it if I run into a question like that. Thanks again :)
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by GMATGuruNY » Thu Feb 28, 2013 4:36 am
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:

https://www.beatthegmat.com/beat-this-pr ... 85719.html
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
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by ynatchev » Thu Feb 28, 2013 5:06 am
Thanks so much Mitch now I feel more confident about probability questions :)
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by Jim@StratusPrep » Thu Feb 28, 2013 1:37 pm
Easiest thing is just to add probabilities of each event.

P Secretary ( 1/10)
P Treasurer (1/10)


Sum = 1/5
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