Poisson random variable problem

[forty]

The children in a small town own slingshots. In a recent contest 4% of them were such poor shots that they did not hit the target even once in 100 shots. If the number of times a randomly selected child has hit the target is approximately a Poisson random variable, determine the percentage of children who have hit the target at least twice.

Just want to make sure my reasoning/logic for this is correct...

so P(X=x) = e^-λλⁿ /n!

so from the question P(X=0) = 0.04 = e^-λ => λ = 3.21887

Then P(X >= 2) = 1 - ( P(X=0) + P(X=1) ) = 1 - 0.04 - e^-3.218873.21887¹/1!

I get 0.83

does this seem reasonable?

 

[Mark44]

Yes, it does. I didn't check your numeric calculations, but how you set it up looks correct, and your answer seems reasonable. Since 4% of the kids didn't hit the target in 100 tries, that means 96% hit it one or more times. We would expect that the percentage of kids who hit the target two or more times would be less than 96%, which is what you have.

 

[forty]

Thanks, just wanted to make sure i had the right idea!