Question 256038
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.
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Ed