question related to Poisson Distribution - refer details?

Question

Razor blades are produced and packed in packets of 10 each. 500 such packets were sold. There is a small probability of 0.05 for manufactured blade to be defective. Find how many packets contain: (1) No defectives, (2) more than one defective?

Answer 1

P(x)=e^(-λ)λ^x/x! λ=.05 for a blade in a packet

for 0 defects we have

e^(-.05)*(.05)^0/0!=.95 so for 500 we would have .95*500=475

While it's semantics, more than one means 2 or more so it's easier to calculate 1-P(0)-P(1). We already know the probability of 0 so let's calculate the probability of 1.

e^(-.05)*(.05)^1/1!=.0476

1-.9512-.0476=.0012

so for 500 packets we'll have 500*.0012 or .60 so less than 1.