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# M13 Q10

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27 Feb 2009, 13:40
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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?

(A) $$\frac{21}{55}$$
(B) $$\frac{18}{44}$$
(C) $$\frac{28}{55}$$
(D) $$\frac{28}{44}$$
(E) $$\frac{36}{55}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The right answer is 9C3/11C3 or 28/55. I do not understand how they went about getting the answer. Thanks
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07 Mar 2013, 07:07
3
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Out of 11 players 3 can be substituted in $$11C3$$ ways.

Since 2 players are forward players, and these two are not suppose to be replaced.
Now as per given condition, we have to substitute 3 players out of 9.
SO number of ways are $$9C3$$.

Probability = No of favorable ways/Total number of ways
= 9C3/11C3

On solving, $$Probability = 28/55$$
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07 Mar 2013, 12:23
2
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My approach:
probability of selecting both forwards(a) = 9C1/11C3
probability of selection any one forward(b) = (9C2x2C1)/11C3
probability of selecting no forward = 1-(a+b)

solving we get 28/55
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27 Feb 2009, 14:20
1
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The probablity is: How many ways that only players among the 9 are picked for the 3 spots
_______________________________________________________________(Divided By)
How many ways that any of the 11 players may be picked for 3 spots

So prob is: 9!/ (6!3!) - ways to rearrange 9 players into 3 spots, and order doesn't matter
___________________________________________________________________(Divided By)
11! (8!3!) - ways to rearrange 11 players into 3 spots, and order doesn't matter
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27 Feb 2010, 11:54
1
KUDOS
My 5-step method for probability of multiple events:

1) Lay out the number of events (3 substitutions): _ _ _

2) Label the events with one specific example of the desired outcome (3 non-forwards): _ _ _
NF NF NF

3) Assign the relevant probability of each event and multiply across (start with 11 players and 9 non-forwards): 9/11 8/10 7/9
_ _ _
NF NF NF
product = 28/55

4) Determine the number of ways in which we can have the desired outcome (only one way to have three NF) = 1

5) Multiply the result of step 3 by the result of step 4 (ie add the probabilities of each combination of desired outcome):
28/55 X 1 = 28/55

(C)
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26 Feb 2010, 06:32
Since 2 are forward players , the probability of the first substitution not being one of them will be (11-2)/11.
Now there are only 10 players left of which 8 are eligible to be substituted .. so the probability of that will be 8/10

and similarly the last substitution will be - 7/9

therefore the total probability will be .. 9/11 * 8/10 * 7/9 = 28/55

_______________________________________________
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26 Feb 2010, 07:39
RuthlessCA 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?

(A) $$\frac{21}{55}$$
(B) $$\frac{18}{44}$$
(C) $$\frac{28}{55}$$
(D) $$\frac{28}{44}$$
(E) $$\frac{36}{55}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The right answer is 9C3/11C3 or 28/55. I do not understand how they went about getting the answer. Thanks

Substitute non forward = 9C3
Total ways to substitute = 11C3
Probability = 9C3/11C3 = 28/55 hence C.
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26 Feb 2010, 11:07
Ans is C , P(of not replacing FW) = 9C3/11C3 = 28/55
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27 Feb 2010, 00:18
either way it should work

using combinations
total no. of players =11
so on. of ways replacing 11 with 3 sub.. players =11c3

total players not forwards =9 so replacing them 3 sub. players =9c3

so ans is 9c3/11c3 =28/55
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01 Mar 2010, 11:58
The ways to substitute non-forwards=9C3=9!/(9-3)!=9!/6!=9*8*7*6!/6!=9*8*7
The ways to substitute any of the players (or total ways of substituting players)=11C3=11!/(11-3)!=11!/8!=11*10*9*8!/8!=11*10*9

Now
Probability for substituting non-forwards=The ways to substitute non-forwards/The ways to substitute any of the players
==> Probability for substituting non-forwards=(9*8*7)/(11*10*9)=(8*7)/(11*10)=56/110=28/55
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03 Mar 2010, 07:16
propabilty that none of the forwards will be substituted= probabilty that other 9 players will be subtitiuted

probabilty that other 9 players will be subtitiuted with 3 players = 9c3/11c3

9*8*7/11*10*9 = 28/55
so ans is C
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08 Sep 2010, 01:59
Its C.

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03 Mar 2011, 13:29
P(Not F) * P (not F) * P (not F)

P(F) for sub 1 is 9/11
P(F) for sub 2 is 8/10
P(F) for sub 1 is 7/9

== C == 28/55
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03 Mar 2011, 21:21
Ramz45 wrote:
Since 2 are forward players , the probability of the first substitution not being one of them will be (11-2)/11.
Now there are only 10 players left of which 8 are eligible to be substituted .. so the probability of that will be 8/10

and similarly the last substitution will be - 7/9

therefore the total probability will be .. 9/11 * 8/10 * 7/9 = 28/55

_______________________________________________

Your method is correct. You can also solve by combination as given by few other people.
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04 Mar 2011, 23:12
This question is awkward. If the "team" is composed of eleven players then that should include the substitutes so the actual probability of of no forwards being selected on the first substitution is 11-3-2=6. Or, 6/11. If you assume the substitutes that have entered the match cannot be taken out then the probability should look like 6/11*5/10*4/9
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06 Mar 2012, 21:43
i believe its:

outcome where none are forwards / total outcomes
or
9c3 / 11c3
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21 Feb 2014, 08:44
this is my way of doing it.

probability of a forward being first sub: 2/11. p( not forward)= 9/11
probability of a forward being second sub: 2/10 p(not forward)= 8/10
prob of a frward being third sub:2/9 p(not forward)= 7/9

prob of no forwar din all 3 subs= 9/11*8/10*7/9 = 28/55
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27 Apr 2014, 00:46
No forward is substituted -> choose 3 persons to substitute from the remaining 9 players
-> P(No forward is substituted) = No of ways to substitute from 9 players/Total no of ways = 9C3/11C3=28/55=> C
Re: M13 Q10   [#permalink] 27 Apr 2014, 00:46
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# M13 Q10

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