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short cuts...

by aschaves » Mon Sep 03, 2007 10:01 am
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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Re: short cuts...

by aim-wsc » Mon Sep 03, 2007 5:06 pm
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.
aschaves 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%
making it a separate thread, so that members could help you :)
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by ri2007 » Fri Sep 07, 2007 12:56 pm
Does any one know a short cut or has any one found a post related to this?

Its quite easy to get the ans - just time consuming so a shortcut would be great :)

thanks
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by samirpandeyit62 » Sat Sep 08, 2007 12:33 am
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.
Regards
Samir
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by ri2007 » Sat Sep 08, 2007 6:51 am
Thanks again Samir
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