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Question
The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly 2 will miss the target.
Solution
Let X = number of bombs hitting the target.
p = probability that bomb will hit the target
∴ p = 0.8 = `8/10 = 4/5`
∴ q = 1 - p = `1 - 4/5 = 1/5`
Given: n = 10
∴ X ~ B `(10, 4/5)`
The p.m.f. of X is given as :
P[X = x] = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^10C_x (4/5)^x (1/5)^(10 - x)`
P (exactly 2 bombs will miss the target)
= P (exactly 8 bombs will hit the target)
= P[X = 8] = p(8)
`= "^10C_8 (4/5)^8 (1/5)^(10 - 8)`
`= "^10C_2 (4/5)^8 (1/5)^2 ....[because "^nC_x = "^nC_(n - x)]`
`= (10 xx 9)/(1 xx 2) xx 4^8/5^10 = (45 xx 4^8)/5^10 = 45(2^16/5^10)`
Hence, the probability that exactly 2 bombs will miss the target = `45(2^16/5^10)`
Notes
The answer in the textbook is incorrect.
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