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by shashank.ism » Tue Feb 09, 2010 1:02 pm
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?

15%
20%
30%
35%
45%
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by sparky_paris » Thu Feb 11, 2010 5:52 pm
35%
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by money9111 » Thu Feb 11, 2010 8:42 pm
agreed 35%.... 90% chance * 10 = .9^10 =
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by vscid » Sat Feb 13, 2010 2:21 pm
money9111 wrote:agreed 35%.... 90% chance * 10 = .9^10 =
How do you quickly calculate this?
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by harsh.champ » Thu Feb 18, 2010 6:11 am
vscid wrote:
money9111 wrote:agreed 35%.... 90% chance * 10 = .9^10 =
How do you quickly calculate this?
Well,the soln. goes like (9/10) x (9/10) x (9/10) x......10 times
0.9
0.81
0.729
0.66 approx
0.59approx

I guess they would have approximated the subsequent values.But my doubt is when we have close by 3 options(30,35,40) is it okay to approximate??

The other technique I know is that of log technique by which u can solve this ques. quickly.
10 log(0.9) = log x

[Are log tables given in the exam??]


Also,is there any other shortcut technique??
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by ajith » Thu Feb 18, 2010 6:18 am
harsh.champ wrote:
vscid wrote:
money9111 wrote:agreed 35%.... 90% chance * 10 = .9^10 =
How do you quickly calculate this?
Well,the soln. goes like (9/10) x (9/10) x (9/10) x......10 times
0.9
0.81
0.729
0.66 approx
0.59approx
(9/10)^2 = 0.81
(9/10)^4 = 0.6561 =0.66
(9/10)^8 = .66^2 = .44
(9/10)^10 = .44*.81 = .3564

Thats the shortest I can imagine
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