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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} 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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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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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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Please advise if this is the correct approach .
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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} 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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Ans is C , P(of not replacing FW) = 9C3/11C3 = 28/55
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Manager
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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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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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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= 28/55So the answer is C
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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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Its C.
9C3/11C3 is the correct answer.
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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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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
_______________________________________________
Please advise if this is the correct approach . Your method is correct. You can also solve by combination as given by few other people.
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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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i believe its: outcome where none are forwards / total outcomes or 9c3 / 11c3
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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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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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