ezhilkumarank wrote:

Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)

B) \((1-x^2) (1-y^2)\)

C) \((1-(1-x)^2) (1- (1-y)^2)\)

D) \((1-(1-x)^2) ((1-y)^2)\)

E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.

The probability James misses both shots is x*x = x^2. So the probability James makes at least one shot is 1 - x^2. Similarly, the probability Colleen makes at least one shot is 1 - y^2. To find the probability they both make at least one shot, we multiply the probability that each makes at least one shot: (1 - x^2)(1 - y^2).

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