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Question

A bomber wants to destroy a bridge. Two bombs are sufficient to destroy it. If four bombs are dropped, what is the probability that it is destroyed, if the chance of a bomb hitting the target is 0.4,
If the answer is ab (HCF of a and b is 1), then find ba?

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Solution

Given,Bomber can destroy bridge with two bombs.
Probability of bomb hitting, P(B)=0.4
Probability of bomb not hitting, P(NB)=0.6
4 bombs are dropped,
Given, distribution is in the form of binomial proability distribution,
(P+q)n=nC0Pn+nC0Pn1q1+......+nCnqn
Probability of bridge destroyed = (Probability only two bomb hits the bridge) + (Probability three bomb hits the bridge)(Probability four bombs hits the bridge)
=P(X=2)+P(X=3)+P(X=4)
=4C2×(P(B))2×(P(NB))2)+4C3×(P(B))3×(P(NB))+4C4×(P(B))4
=6×(25)2×(35)2+4×(25)3×(35)+(25)4
=(216+96+16)625=328625
Compare with given ab, we have
a=328,b=625
ba=297


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