Question 1090258
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This is a binomial probability problem.  We can represent the fact that each bomb has a 1/5 probability of hitting (Y = yes) the bridge and a 4/5 probability of not hitting the bridge (N = no) with a "probability vector":
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{{{((1/5)*Y+(4/5)*N)}}}
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Then the probabilities of hitting the bridge with 0, 1, ..., 5, or 6 of the bombs will be the coefficients of the expansion of
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{{{((1/5)Y+(4/5)N)^6}}}
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Use the binomial theorem to expand that expression and find the appropriate coefficients.  Since the bridge will be destroyed if it is hit by 2, 3, 4, 5, or 6 bombs, you could solve the problem by calculating those 5 coefficients.  But it is much faster to calculate the probability that the bridge is NOT destroyed by calculating the probabilities that the bridge is hit by either 0 or 1 bomb; then the probability that the bridge is destroyed is 1 minus that probability.
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{{{P(0)=C(6,0)(1/5)^0(4/5)^6=0.26214}}}
{{{P(1)=C(6,1)(1/5)^1(4/5)^5=0.39322}}}

The probability that the bridge will be destroyed is
{{{1-(0.26214+.39322) = 0.34464}}}