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by manulath » Sun Sep 28, 2008 7:12 am
similar question was posted earlier. See the link

https://www.beatthegmat.com/arrows-proba ... 18789.html
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by manulath » Sun Sep 28, 2008 7:22 am
This question is slightly diff as it asks prob for ATLEAST 2 times
find probalility of no hit + probability of exactly 1 hit
then subtract the sum from 1

probability of hit = 1/4
probability of miss = 3/4

5 try - no hits = (3/4)^5

1 hit= 5C1*(1/4)*(3/4)^4 = 5*3^4/4^5

At least two hits = 1 - (3/4)^5 - 5*3^4/4^5

= (4^5 - 3^5 - 5*3^4) / 4^5

= 47/128
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by manulath » Sun Sep 28, 2008 7:23 am
This question is slightly diff as it asks prob for ATLEAST 2 times
find probalility of no hit + probability of exactly 1 hit
then subtract the sum from 1

probability of hit = 1/4
probability of miss = 3/4

5 try - no hits = (3/4)^5

1 hit= 5C1*(1/4)*(3/4)^4 = 5*3^4/4^5

At least two hits = 1 - (3/4)^5 - 5*3^4/4^5

= (4^5 - 3^5 - 5*3^4) / 4^5

= 47/128
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by gmatguy16 » Sun Sep 28, 2008 8:06 am
I agree with the method,i am getting answer as 79/160.
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by manulath » Sun Sep 28, 2008 8:55 am
well just try and add the probabilities for 2 hits, 3 hits, 4 hits and 5 hits

i still had 47/128

(4^5 - 3^5 - 5*3^4) / 4^5

-3^5 - 5*3^4 can be simplified to

-3^4(3+5) = 8*3^4

hence numerator becomes 4^5 - 8*3^4
4^5 = 2^10 = 2^3*2^5 = 8*2^5 = 8*128
3^4 = 81

8*128 - 8*81 = 8*47

ans= 8*47 / 4^5 = 8*47/ 8*128 = 47/128

--------------------------------------
= 47/128
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