SOLUTION: A factory manufacturing tennis balls determines that the probability that a single can of three balls will contain at least one defective ball is .025. what is the probability that

Question 256038: A factory manufacturing tennis balls determines that the probability that a single can of three balls will contain at least one defective ball is .025. what is the probability that a case of 48 cans will contain at least two cans with a defective ball?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Let d=.025 and g=1-.025=.975
We have to expand the binomial (g+d)^48
g^48=.975^48=.2966 Prob of no defective cans
48g^47d=48(.975)^47(.025)=48*.3042*.025=.3651 Prob one defective can.
1-.2966-.3651=.3473 prob of at lest 2 cans with a defective ball.
.
Ed
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