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by crackgmat007 » Tue Oct 06, 2009 5:49 pm
During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?
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by sunnyjohn » Tue Oct 06, 2009 7:47 pm
how about:

9C3/11C3 ==> 9.8.7/11.10.9 ==> 28/55
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by crackgmat007 » Tue Oct 06, 2009 9:18 pm
can you explain pls?
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by sunnyjohn » Tue Oct 06, 2009 10:40 pm
means -
we can not substitute those two players out of 11 players.

means we have to choose 3 players out of 9 players for substitute.

so our Possible Set contain = 9C3 possible selections.

Total possible selection can be = 11C3

so probability of not selecting two forward player will be :- 9C3/11C3
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by crackgmat007 » Wed Oct 07, 2009 4:03 pm
Since there are 2 forwards and question asks about what is the probability that none of the forwards will be substituted, will this not be a constraint?
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by gmatdriller » Sat Oct 02, 2010 6:36 am
I know am making some errors here, but could someone address where
I have gone off track....

Let F = forwarder.

Chance of not selecting F = 1 - chance of selecting an F

Ways of selecting an F: xFF; FxF; FFx........ where x ==> any other player
xFF = (9/11)(2/10)(1/9) = 1/55
FxF = (2/11)(9/10)(1/9) = 1/55
FFX = (2/11)(1/10)(9/9) = 1/55
xFF OR FxF OR FFx = 3(1/55) = 3/55

chance of NOT selecting a forwarder = 1 - 3/55 = 52/55 (WRONG)


Where have a gone wrong?
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by GMATGuruNY » Sat Oct 02, 2010 6:47 am
crackgmat007 wrote:During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?
1st substitution: P(not forward) = 9/11 (11 total players, 9 of them not forwards)
2nd substitution: P(not forward) = 8/10 (10 players left, 8 of them not forwards)
3rd substitution: P(not forward) = 7/9 (9 players left, 7 of them not forwards)

Since we need all of these events to happen together, we multiply the fractions:

9/11 * 8/10 * 7/9 = 28/55.
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by gmatdriller » Sun Oct 03, 2010 7:44 am
Thanks GmatGuru for the detailed explanations.

My confusion stems from the OG12 PS #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 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?
solution:
P(S or T) = (9/10)x(1/9)x(1) + (9/10)x(8/9)(1/8) = 1/10 + 1/10 = 1/5

I understand the explanations given, and i tried to follow the same concept
but failed.

I tried to distinguish the two questions (coupled with your analysis)...and
am getting clearer.
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by GMATGuruNY » Sun Oct 03, 2010 12:07 pm
gmatdriller wrote:Thanks GmatGuru for the detailed explanations.

My confusion stems from the OG12 PS #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 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?
solution:
P(S or T) = (9/10)x(1/9)x(1) + (9/10)x(8/9)(1/8) = 1/10 + 1/10 = 1/5

I understand the explanations given, and i tried to follow the same concept
but failed.

I tried to distinguish the two questions (coupled with your analysis)...and
am getting clearer.
The OG does not offer the most efficient approach. Here's how you should solve the problem about Harry:

There are 10 people in the room. Each has an equal chance of being elected secretary. So P(Harry is elected secretary) = 1/10.

Each of the 10 people also has an equal chance of being elected treasurer. So P(Harry is elected treasurer) = 1/10.

We want the probability that Harry is elected secretary OR treasurer. P(A or B) = P(A) + P(B). So we need to add the fractions:

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

The explanation in the OG suggests -- unwisely -- that we determine every separate event that must occur for Harry to be elected secretary or treasurer.

For Harry to be elected secretary, 2 events must happen together: he must not be elected president, then he must be elected secretary.

P(Harry is not elected president) = 9/10 (10 total people, 9 other than Harry who can be elected president, giving us 9 good outcomes)
P(Harry is elected secretary) = 1/9 (9 people left, and Harry must be elected, giving us 1 good outcome)
Since both of these events must happen together in order for Harry to be elected treasurer, we multiply the fractions:

9/10 * 1/9 = 1/10.

For Harry to be elected treasurer, 3 events must happen together: he must not be elected president, then he must not be elected secretary, then he must be elected treasurer.

P(Harry is not elected president) = 9/10 (10 total people, 9 other than Harry who can be elected president, giving us 9 good outcomes)
P(Harry is not elected secretary) = 8/9 (9 people left, 8 other than Harry who can be elected secretary, giving us 8 good outcomes)
P(Harry is elected treasurer) = 1/8 (8 people left, and Harry must be elected, giving us 1 good outcome)
Since all of these events must happen together in order for Harry to be elected secretary, we multiply the fractions:

9/10 * 8/9 * 1/8 = 1/10.

We want the probability that Harry is elected secretary OR treasurer. P(A or B) = P(A) + P(B). So we need to add the fractions:

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

Same answer, but a less efficient approach.

Hope this helps!
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by gmatdriller » Wed Nov 10, 2010 12:43 am
@Gmatguru
Thanks for your clear and concise explanations.
+1 thanks.
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by vijchid » Wed Nov 10, 2010 11:33 am
1st substitution: P(not forward) = 9/11 (11 total players, 9 of them not forwards)
2nd substitution: P(not forward) = 8/10 (10 players left, 8 of them not forwards)
3rd substitution: P(not forward) = 7/9 (9 players left, 7 of them not forwards)

Since we need all of these events to happen together, we multiply the fractions:

9/11 * 8/10 * 7/9 = 28/55.
Mitch-
I understand the logic above, but why is the probability for the 2nd substitution 8/10. Where does the problem say that you cannot substitute a player that has come into the game for the first substitution? If you consider that then the probability for the second substitution will still remain 9/11.
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by GMATGuruNY » Wed Nov 10, 2010 1:27 pm
vijchid wrote:
1st substitution: P(not forward) = 9/11 (11 total players, 9 of them not forwards)
2nd substitution: P(not forward) = 8/10 (10 players left, 8 of them not forwards)
3rd substitution: P(not forward) = 7/9 (9 players left, 7 of them not forwards)

Since we need all of these events to happen together, we multiply the fractions:

9/11 * 8/10 * 7/9 = 28/55.
Mitch-
I understand the logic above, but why is the probability for the 2nd substitution 8/10. Where does the problem say that you cannot substitute a player that has come into the game for the first substitution? If you consider that then the probability for the second substitution will still remain 9/11.
Good point. On the GMAT, the problem would make clear what I assumed in my solution: that a player can be substituted into the game only once.
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by Stuart@KaplanGMAT » Wed Nov 10, 2010 2:12 pm
vijchid wrote:Mitch-
I understand the logic above, but why is the probability for the 2nd substitution 8/10. Where does the problem say that you cannot substitute a player that has come into the game for the first substitution? If you consider that then the probability for the second substitution will still remain 9/11.
The problem also doesn't state that there's an equal chance of any player being chosen, so the flaw you point out isn't the only problem. There also aren't answer choices, usually a sign that the question is either:

1) made up; or
2) "remembered" from another source (which often includes transcription errors).
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