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)?
