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A certain club has 10 members-including Harry.

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by Mathsbuddy » Tue Nov 18, 2014 7:47 am
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Difficulty–

right wrong

Pick an answer to see how members did

We are looking for the probability of:
A{Not President then secretary} OR B{Not president then not secretary then treasurer}
p(not president) = 9/10
p(A) = p(not president then secretary) = 9/10 x 1/9 = 1/10
p(B) = p(not president then not secretary then treasurer) = 9/10 x 8/9 x 1/8 = 1/10
p(A or B) = p(A) + p(B) = 1/10 + 1/10 = 1/5
ANSWER = E
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by kaushalik » Thu Feb 09, 2017 2:15 am
Some please help 10C3 method, i am not getting the answer using that method
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by Brent@GMATPrepNow » Thu Feb 09, 2017 9:15 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
[spoiler]= E[/spoiler]

Cheers,
Brent
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by Matt@VeritasPrep » Fri Feb 17, 2017 2:19 am
kaushalik wrote:Some please help 10C3 method, i am not getting the answer using that method
This isn't the best way to solve. We aren't looking for the number of ways we could choose 3 people from the group of 45, and trying to fit that concept to the problem will probably do more harm than good! Look for the simplest, most direct approach you can - don't necessarily stick with your first instinct if it would force you to do a lot of extra work.
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by Matt@VeritasPrep » Fri Feb 17, 2017 2:23 am
KristenH88 wrote: The question says: 10 members
1/10 chance of P
1/9 chance of S
1/8 chance of T

and everyone is using 1/10 + 1/10 instead of what they gave. How can I use what they gave to find the Probability of (S+T)? And please don't say we know its a 1/10 chance for each bc blah blah.
But that isn't what they gave! :)

Your chance of being selected treasurer IS NOT SIMPLY 1/8 because TWO OTHER THINGS HAVE TO HAPPEN FIRST. For you to be the treasurer, you must:

1) Not be selected president, then
2) Not be selected vice president

So you need to run the following gauntlet:

1) Avoid being selected president. p = 9/10
2) Avoid being selected vice president. p = 8/9, since there are only 9 people left at this point.
3) Be selected treasurer. After going through the first two steps (and ONLY after going through those steps), your p = 1/8.

Since you need ALL THREE OF THESE, your p is (9/10) * (8/9) * (1/8), or 1/10.
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by Matt@VeritasPrep » Fri Feb 17, 2017 2:25 am
KristenH88 wrote: My approach was wrong but none of these look simple. It's not clicking.
After going through this thread, though, it seems to me that everyone has not posted the most intuitive explanation.

If you're in this group, one person (of 10) will be president, one person (of 10) will be vice president, and one person (of 10) will be treasurer.

There isn't anything that makes anyone any more likely to win any of these three positions, so each position has p = 1/10.
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by toby001 » Tue May 09, 2017 10:33 am
Brent@GMATPrepNow 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
Hi Brent,

If the president had to be selected first, this would be different right? Would it then be 17/80?

Thanks!
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by Brent@GMATPrepNow » Tue May 09, 2017 2:59 pm
toby001 wrote: Hi Brent,

If the president had to be selected first, this would be different right? Would it then be 17/80?

Thanks!
I'm not sure what you're asking. The president IS selected first.

That said, the answer will always be 1/5, regardless of the order in which the three positions are filled.

Cheers,
Brent
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by toby001 » Wed May 10, 2017 6:24 am
Brent@GMATPrepNow wrote:
toby001 wrote: Hi Brent,

If the president had to be selected first, this would be different right? Would it then be 17/80?

Thanks!
I'm not sure what you're asking. The president IS selected first.

That said, the answer will always be 1/5, regardless of the order in which the three positions are filled.

Cheers,
Brent
Thanks for responding Brent. For some reason, I interpreted it as (9/10)*[(1/9)+(1/8)]. That is, the the probability that he does not get selected as president and the probability that he gets selected as the treasurer or the secretary. Can you help me understand why that is wrong? Thanks!
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by Matt@VeritasPrep » Thu May 11, 2017 8:42 pm
toby001 wrote:
Brent@GMATPrepNow wrote:
toby001 wrote: Hi Brent,

If the president had to be selected first, this would be different right? Would it then be 17/80?

Thanks!
I'm not sure what you're asking. The president IS selected first.

That said, the answer will always be 1/5, regardless of the order in which the three positions are filled.

Cheers,
Brent
Thanks for responding Brent. For some reason, I interpreted it as (9/10)*[(1/9)+(1/8)]. That is, the the probability that he does not get selected as president and the probability that he gets selected as the treasurer or the secretary. Can you help me understand why that is wrong? Thanks!
Because to be selected secretary, he must NOT be selected treasurer. So just as you did with the first step, you need to have (8/9)*(1/8) for the last piece.

Basic idea:

President = 1/10

Treasurer = Not President * Yes Treasurer = (9/10)*(1/9)

Secretary = Not President * Not Treasurer * Yes Secretary = (9/10)*(8/9)*(1/8)
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by Scott@TargetTestPrep » Tue Dec 05, 2017 5:29 pm
mitzwillrockgmat 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
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.

Answer: E

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by [email protected] » Mon Feb 12, 2018 11:39 am
Hi All,

We're told that 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. We're asked for the probability that Harry will be EITHER the member chosen to be the secretary or the member chosen to be treasurer. There are a couple of different ways to do this type of math; here's how you can break the calculation down into two 'pieces':

The probability that Harry is chosen to be secretary is:
(Not Harry for President)(Harry for secretary)(Not Harry for treasurer) = (9/10)(1/9)(8/8) = 72/720 = 1/10

The probability that Harry is chosen to be treasurer is:
(Not Harry for President)(Not Harry for secretary)(Harry for treasurer) = (9/10)(8/9)(1/8) = 72/720 = 1/10

Thus, the total probability of either event occurring is 1/10 + 1/10 = 2/10 = 1/5

Final Answer: E

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