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Arrows probability

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by Stockmoose16 » Wed Sep 17, 2008 10:02 am
If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2


What's with the 5C2?
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by gabriel » Wed Sep 17, 2008 12:16 pm
5c2 is used to calculate the number of ways the arrows hits the target. For eg. it can be the 1st and 2nd time or the 3rd and 4th time. It is similar to choosing 2 things out of 5.

Check out Stuarts post in the GMAT math section. He has given an excellent explanation of a similar problem.
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by chayanika » Wed Sep 17, 2008 12:21 pm
The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
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by Stockmoose16 » Wed Sep 17, 2008 3:22 pm
chayanika wrote:The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
Why wouldn't the number of ways it can be done be 5!, instead of 5C3?
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Re: Arrows probability

by sudhir3127 » Wed Sep 17, 2008 3:44 pm
Stockmoose16 wrote:If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2




What's with the 5C2?

Binomial Theroem would be the best way to do such problems

( X+ Y)^n = n C k X^n-k y^k

(https://en.wikipedia.org/wiki/Binomial_theorem)

Thus it will be
5C2 (1/4)^2 ( 1-1/4)^5-2

5C2 * (1/4)^2 * (3/4)^3

Hope that helps...
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Re: Arrows probability

by Stockmoose16 » Wed Sep 17, 2008 3:46 pm
sudhir3127 wrote:
Stockmoose16 wrote:If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2




What's with the 5C2?

Binomial Theroem would be the best way to do such problems

( X+ Y)^n = n C k X^n-k y^k

(https://en.wikipedia.org/wiki/Binomial_theorem)

Thus it will be
5C2 (1/4)^2 ( 1-1/4)^5-2

5C2 * (1/4)^2 * (3/4)^3

Hope that helps...
Is the reason it's 5C2 rather than 5! because, in this example, there are 3 hits and 2 misses, which would be redundant. So, it's really

5!/3!*2! = 5C2 (The 3!*2! takes out the redundancy of 3 hits and 2 misses)?
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by pseudononymous » Wed Sep 17, 2008 6:08 pm
Stockmoose16 wrote:
chayanika wrote:The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
Why wouldn't the number of ways it can be done be 5!, instead of 5C3?
The C2 eliminates duplicates since order doesn't matter.
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