A target is to be destroyed in a bombing exercise and there is a 75 \% chance that a bomb will hit the target. Assuming that two direct hits are required to destroy the target completely, the minimum…

Question

A target is to be destroyed in a bombing exercise and there is a 75 \% chance that a bomb will hit the target. Assuming that two direct hits are required to destroy the target completely, the minimum number of bombs to be dropped in order that the probability of destroying the target is not less than 99 \%, is

Accepted Answer

Correct answer: C. (c) Let probability of hit the target p=75 \% or 34 Hence, probability to not hit the target q=25 \% or 14 Now, let ' n ' is the minimum number of bombs that are dropped, then according to question probability of destroying the target 99100 (given) ⇒ P (X 2 ) 99100, (here, X represents the number of bombs dropped) ⇒ 1- [P (X=0 )+P (X=1 ) ] 99100 ⇒ P (X=0 )+P (X=1 ) 1- 99100 ⇒ ^nC_0p^0q^n+ ^nC_1p^1q^n-1 1100 ⇒ 14^n+n 34^n 1100 ⇒4^n 300n+100 ⇒n 6

Solution

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