What is the simplest way to calculate the solution below? It would take forever to calculate (9/10)^10
4. In a basketball contest, players must make 10 free throws. Assuming a player has 90% chance of making each of his shots, how likely is it that he will make all of his first 10 shots?
Ans: The probability of making all of his first 10 shots is given by
(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%
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- aim-wsc
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Welcome to the forums, aschaves,
Search some threads I think we already have a discussion related to your query.
If you dont find any already-present thread on this topic; start a new thread, next time onwards. This topic certainly not related to this thread.
Search some threads I think we already have a discussion related to your query.
If you dont find any already-present thread on this topic; start a new thread, next time onwards. This topic certainly not related to this thread.
making it a separate thread, so that members could help youaschaves wrote:What is the simplest way to calculate the solution below? It would take forever to calculate (9/10)^10
4. In a basketball contest, players must make 10 free throws. Assuming a player has 90% chance of making each of his shots, how likely is it that he will make all of his first 10 shots?
Ans: The probability of making all of his first 10 shots is given by
(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%
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I think all depends upon the answer choices
here we need to find 9^10/10^10
i.e (9^2) ^ 5 / 10 ^ 10
i.e 81 ^ 5 /10 ^ 10
we can say 81 ~ 80
so it is 80 ^ 5/10 ^ 10
hence we have 8^ 5 /10 ^5
i.e 2^15 /10 ^5
now 2 ^ 10 =1024
so we can easily calculate 2^ 15 as 327... something/100000
so prob is .327.. (i.e.probabilty always lies between 0 & 1)
now our actual probalilty must be something little greater than .327...
so answer choices less than this can be eliminated
so it must be .35 if all other answer choices are greater than .4's
In my opinion in such problems the GMAT requires us to answer the questions with plausible certainity & not complete certainity (better but will consume more time)
so if you can find either the Min or Max extreme of the answer choice, the
other extreme would be taken care by the answer choices.
here we need to find 9^10/10^10
i.e (9^2) ^ 5 / 10 ^ 10
i.e 81 ^ 5 /10 ^ 10
we can say 81 ~ 80
so it is 80 ^ 5/10 ^ 10
hence we have 8^ 5 /10 ^5
i.e 2^15 /10 ^5
now 2 ^ 10 =1024
so we can easily calculate 2^ 15 as 327... something/100000
so prob is .327.. (i.e.probabilty always lies between 0 & 1)
now our actual probalilty must be something little greater than .327...
so answer choices less than this can be eliminated
so it must be .35 if all other answer choices are greater than .4's
In my opinion in such problems the GMAT requires us to answer the questions with plausible certainity & not complete certainity (better but will consume more time)
so if you can find either the Min or Max extreme of the answer choice, the
other extreme would be taken care by the answer choices.
Regards
Samir
Samir