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# 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 b−a?

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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+nC0Pn−1q1+......+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 ⇒b−a=297  Suggest Corrections  0      Similar questions  Related Videos   Geometric Progression
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