Question
The probability of a bomb hitting a bridge is 12 and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge beeing destroyed is greater then 0.9, is
Answer: Option A
:
A
Let n be the least number of bombs required and x the number of bombs that hit the bridge. Then x follows a binomial distribution with parameter n and p=12
Now, P(X≥2)>0.9⇒1−P(X<2)>0.9
⇒P(X=0)+P(X=1)<0.1
⇒nC0(12)n+nC1(12)n−1(12)<0.1⇒10(n+1)<2n
This gives n≥8
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:
A
Let n be the least number of bombs required and x the number of bombs that hit the bridge. Then x follows a binomial distribution with parameter n and p=12
Now, P(X≥2)>0.9⇒1−P(X<2)>0.9
⇒P(X=0)+P(X=1)<0.1
⇒nC0(12)n+nC1(12)n−1(12)<0.1⇒10(n+1)<2n
This gives n≥8
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