In a precision bombing attack, there is a `50%` chance that any one bomb will strick the target. Two direct hits are required to destroy the target co

Question

In a precision bombing attack, there is a `50%` chance that any one bomb will strick the target. Two direct hits are required to destroy the target co

In a precision bombing attack, there is a `50%` chance that any one bomb will strick the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a `99%` chance or better of completely destroying the target can be
A. 12
B. 11
C. 10
D. 13

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A::B::D
We have, the probability that the bomb stricks the target is `p=1//2.` Let n be the number of boombs which should be dropped to ensure `99%` chance or better of completely destroying the target. Then the probability that out of n bombs at least two bombs strike the target is greater than `0.99.` Let X denote the number of bombs striking the target. Then
`P(X=r)=""^(n)C_(r)((1)/(2))^(n-r)=""^(n)C_(r)((1)/(2))^(n),r0,1,2,....,n`
We should have
`P(Xge2)ge0.99`
`implies{1-P(Xlt2)}ge0.99`
`implies1-{(1+n)(1)/(2^(n))}ge0.99`
`implies0.001ge(1+n)/(2^(n))`
`2^(n)gt100+100n`
`impliesnge11`
Thus, the minimum number of bombs is 11.