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Difficult Math Question #4

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Difficult Math Question #4

by 800guy » Mon Aug 28, 2006 7:27 pm
try answering this one--OA to come later

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?
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by aim-wsc » Wed Aug 30, 2006 4:35 am
if u hv two throws:
ur chances of getting a right shot ll be 0.9 X 0.9=0.81
for 3: 0.9^3=0.729
... ...
SO ON for ten throws
probability=0.9 ^ 10

present the options pls.
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by 800guy » Thu Aug 31, 2006 6:16 pm
OA:

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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by aim-wsc » Thu Aug 31, 2006 7:39 pm
yeah thats what i said.

but how can you expect a student know the value of 0.9 ^ 10 be 0.37 or something??

no i wont take this////
you gotta come up with options.

anyway
ya i know they hv not provided options..just answers right sigh!
no problem

hey
800guy,
you hv got 123 problems ...( it would take TWO MONTHS to finish...)

so be fast post 4-5 everyday instead....
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by naveen1003 » Thu Jun 09, 2011 2:25 am
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?
Chance of making a shot = 90%
Probability of making a shot = 0.9
Thus probability of making all of first 10 shots = 10C10*(0.9)^10 = [spoiler](0.9)^10[/spoiler]
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by [email protected] » Thu Jun 09, 2011 2:36 am
I do not think you see sums like these on the gmat i.e .9 ^ 10. Do you really?
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