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## A certain club has 10 members-including Harry.

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mitzwillrockgmat Master | Next Rank: 500 Posts
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#### A certain club has 10 members-including Harry.

Fri Jun 18, 2010 2:19 am
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

Hi! Can someone help me out with this question? I eventually found a way to figure this out but i would like to how to use combinations to get the answer here i.e. i can get the denominator (10C3) but what about the numerator, not sure how to get that...

Although, the method I found was pretty simple & logical but I would still like to know another way to tackle this question, as it will improve my odds of figuring out such questions.

Anyways here it is in case it helps others too,

consider the prob of harry becoming either the sec or treasurer VS the prob that harry becomes the treasurer.

p(sec or tres) = 1/10 + 1/10 = 1/5 Vs p(tres) = 1/10

as you can see the prob of harry acquiring one position is much greater than acquiring any one of the positions.

So the answer must be greater than 1/10. eliminate a,b,c.

between d & e... e is the correct answer because p(a or b) = p(a) + p(b) - P(both) = 1/10 +1/10 +0* =1/5

*a person cannot hold two positions

Hope this helps others. BUT IF ANYONE KNOWS ANOTHER METHOD INVOLVING COMBINATIONS PLS LET ME KNOW, THANKS!

boysangur Junior | Next Rank: 30 Posts
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Sat Aug 28, 2010 1:47 pm
This is a question I just messed up on. The way I figured was that Harry needed NOT to be selected for the position of president out of 10 people (probability 9/10), AND (x) be selected as the SECRETARY (1/9) OR (+) TREASURER (1/8). So what i ended up with was 9/10 x (1/9 + 1/8). I know this answer is wrong but I can't figure out why. This equation basically says that he is NOT chosen to be the president AND he is chosen to be a secretary OR a treasurer. Isn't this what the question is asking?

mitzwillrockgmat Master | Next Rank: 500 Posts
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Mon Jun 21, 2010 3:18 am
hi, look at the letters above each no, maybe that will help you

p=president , s = sec , t = treasurer

note: when harry is sec or treasurer we put 1 in that position.

case 1
P S T
9*1*8 = 72 ways

case 2
P S T
9*8*1 = 72 ways

kvcpk Legendary Member
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Fri Jun 18, 2010 3:25 am
I am not sure if your method is right.. but looks simpler

mitzwillrockgmat Master | Next Rank: 500 Posts
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Fri Jun 18, 2010 3:27 am
kvcpk wrote:
I am not sure if your method is right.. but looks simpler
really? explain why?

kvcpk Legendary Member
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Fri Jun 18, 2010 3:29 am
President can be chosen from amongst 10 members in 10 ways
Secretary can be chosen from amongst 9 members in 9 ways and
Treasurer can be chosen from amongst 8 members in 8 ways.
Hence total number of possibilities = 10.9.8 = 720

Case I : When Harry is Secretary. It is important that he is not the President. Hence possibility = 9*1*8 = 72 ways
Case II: When Harry is Treasurer. It is important that he is neither President nor Secretary = 9*8*1 = 72 ways

Hence probability that he is either secretary or treasurer = 72 + 72 / 720 = 1/5

Hope this helps!!

mitzwillrockgmat Master | Next Rank: 500 Posts
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Fri Jun 18, 2010 7:09 am
kvcpk wrote:
President can be chosen from amongst 10 members in 10 ways
Secretary can be chosen from amongst 9 members in 9 ways and
Treasurer can be chosen from amongst 8 members in 8 ways.
Hence total number of possibilities = 10.9.8 = 720

Case I : When Harry is Secretary. It is important that he is not the President. Hence possibility = 9*1*8 = 72 ways
Case II: When Harry is Treasurer. It is important that he is neither President nor Secretary = 9*8*1 = 72 ways

Hence probability that he is either secretary or treasurer = 72 + 72 / 720 = 1/5

Hope this helps!!
ok, great thanks! so clearly combinations with the regular formulas is not going to work in this question! thanks!

jube Master | Next Rank: 500 Posts
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Fri Jun 18, 2010 7:47 am
Probability that Harry WON'T be chosen for president = 9/10
Probability that Harry WON'T be secretary or teasurer: (8/9)(7/8) = 7/9
Probability that Harry WILL be the secretary or treasurer = (9/10)[1-(7/9)]=(9/10)(2/9)=1/5

### GMAT/MBA Expert

Stuart Kovinsky GMAT Instructor
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Fri Jun 18, 2010 10:03 am
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

Sometimes we make questions way more complicated than they need to be. Common sense and logic go a long way on the GMAT.

Here we have 10 members competing for 2 positions and each member has an equal chance to be selected for a position. What's Harry's chance of getting 1 of the positions? 2/10 = 1/5

How do we know this?

Well, if Harry had a greater than 2/10 chance, then one of the remaining 9 people would have to have a less than 2/10 chance (since the total probabilities have to add to 20/10 - 2 jobs, hence 200% total probability when we add their probabilities together).

If Harry had a less than 2/10 chance, then one of the remaining 9 people would have to have a greater than 2/10 chance.

Therefore, since all 10 people must have the same chance, Harry's probability is 2/10 = 1/5.

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bharatv Junior | Next Rank: 30 Posts
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Fri Jun 18, 2010 11:27 am
kvcpk wrote:
Case I : When Harry is Secretary. It is important that he is not the President. Hence possibility = 9*1*8 = 72 ways
Case II: When Harry is Treasurer. It is important that he is neither President nor Secretary = 9*8*1 = 72 ways

Hope this helps!!
Can you break down further for case II ?
I could understand why 72 ways for Case I but not for Case II. If President and secretary are already chosen we would have 8 ways for Harry to be treasurer? Am I missing something here?

mitzwillrockgmat Master | Next Rank: 500 Posts
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Mon Jun 21, 2010 3:07 am
Stuart Kovinsky wrote:
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

Sometimes we make questions way more complicated than they need to be. Common sense and logic go a long way on the GMAT.

Here we have 10 members competing for 2 positions and each member has an equal chance to be selected for a position. What's Harry's chance of getting 1 of the positions? 2/10 = 1/5

How do we know this?

Well, if Harry had a greater than 2/10 chance, then one of the remaining 9 people would have to have a less than 2/10 chance (since the total probabilities have to add to 20/10 - 2 jobs, hence 200% total probability when we add their probabilities together).

If Harry had a less than 2/10 chance, then one of the remaining 9 people would have to have a greater than 2/10 chance.

Therefore, since all 10 people must have the same chance, Harry's probability is 2/10 = 1/5.
Thanks! I guess, 'logical' thinking is the way to go then!

vijaynaik Senior | Next Rank: 100 Posts
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Sat Aug 28, 2010 3:40 pm
boysangur wrote:
This is a question I just messed up on. The way I figured was that Harry needed NOT to be selected for the position of president out of 10 people (probability 9/10), AND (x) be selected as the SECRETARY (1/9) OR (+) TREASURER (1/8). So what i ended up with was 9/10 x (1/9 + 1/8). I know this answer is wrong but I can't figure out why. This equation basically says that he is NOT chosen to be the president AND he is chosen to be a secretary OR a treasurer. Isn't this what the question is asking?
In TREASURER case u need to see that harry is not already picked as SECRETARY.

so 9/10 * [ 1/9 + 8/9 * 1/8] = 1/5

diebeatsthegmat Legendary Member
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Sun Aug 29, 2010 10:56 am
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

Hi! Can someone help me out with this question? I eventually found a way to figure this out but i would like to how to use combinations to get the answer here i.e. i can get the denominator (10C3) but what about the numerator, not sure how to get that...

Although, the method I found was pretty simple & logical but I would still like to know another way to tackle this question, as it will improve my odds of figuring out such questions.

Anyways here it is in case it helps others too,

consider the prob of harry becoming either the sec or treasurer VS the prob that harry becomes the treasurer.

p(sec or tres) = 1/10 + 1/10 = 1/5 Vs p(tres) = 1/10

as you can see the prob of harry acquiring one position is much greater than acquiring any one of the positions.

So the answer must be greater than 1/10. eliminate a,b,c.

between d & e... e is the correct answer because p(a or b) = p(a) + p(b) - P(both) = 1/10 +1/10 +0* =1/5

*a person cannot hold two positions

Hope this helps others. BUT IF ANYONE KNOWS ANOTHER METHOD INVOLVING COMBINATIONS PLS LET ME KNOW, THANKS!
yeah , i also solved the problem as the way you did
P(T or S)=P(t)+ P(s)-P(t and s)
P(t)=1/10
P(s)=1/10
and since no one is chosen to get 2 positions both T and S thus P(t and S)=0
P(t or s)=1/5

KristenH88 Junior | Next Rank: 30 Posts
Joined
20 Feb 2013
Posted:
18 messages
1
Mon Nov 17, 2014 12:52 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

Hi! Can someone help me out with this question? I eventually found a way to figure this out but i would like to how to use combinations to get the answer here i.e. i can get the denominator (10C3) but what about the numerator, not sure how to get that...

Although, the method I found was pretty simple & logical but I would still like to know another way to tackle this question, as it will improve my odds of figuring out such questions.

Anyways here it is in case it helps others too,

consider the prob of harry becoming either the sec or treasurer VS the prob that harry becomes the treasurer.

p(sec or tres) = 1/10 + 1/10 = 1/5 Vs p(tres) = 1/10

as you can see the prob of harry acquiring one position is much greater than acquiring any one of the positions.

So the answer must be greater than 1/10. eliminate a,b,c.

between d & e... e is the correct answer because p(a or b) = p(a) + p(b) - P(both) = 1/10 +1/10 +0* =1/5

*a person cannot hold two positions

Hope this helps others. BUT IF ANYONE KNOWS ANOTHER METHOD INVOLVING COMBINATIONS PLS LET ME KNOW, THANKS!
I Don't Get it!
My approach was wrong but none of these look simple. It's not clicking.

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.

What I wrote on paper was P(S+T) = 1/9 + 1/8 = 17/72 (not there) SO I tried to divide that by 10 total members. 17/720 also not there. Where did I go wrong in this method; using what was in the question?
Thank you.

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GMATGuruNY GMAT Instructor
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Mon Nov 17, 2014 1:01 pm
Quote:
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.

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:
Quote:
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.
1/10 + 1/10 = 2/10 = 1/5.

Other problems that test this concept:

http://www.beatthegmat.com/beat-this-probability-qs-t185719.html
http://www.beatthegmat.com/probablity-ques-t60161.html
http://www.beatthegmat.com/manhattan-probability-t89481.html (2 posts)
http://www.beatthegmat.com/a-box-contains-some-blue-balls-and-9-red-balls-if-the-t151368.html

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