Question 968975
this uses binomial probability
p = .02, q = .98, n = 100
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binomial probability formula is
Probability ( X = k ) = (combination of n things taken k at a time) * p^k * q^(n-k)
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Probability ( X > or = 1 ) = 1 - P( X = 0 ) 
P( X = 0 ) = combination of 100 light bulbs taken 0 at a time) * (.02)^0 * (.98)^100 = (( 100! / (0! * (100 - 0)!)) * 1 * 0.132619556
P( X = 0 ) = 1 * 0.132619556 approx 0.13
Probability ( X > or = 1 ) = 1 - 0.13 = 0.87 
therefore,
probability that they get at least one that is defective is 0.87