Question 1166965: The probability that a bomber hits its target on any particular run is .80. If three bombers are sent after the same target, what then is the probability that at least one of the three bombers will hit the target
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the probability that at least one of the bombers will hit its target is equal to 1 minus the probability that none of the bombers will hit its target.
the probability that each bomber will hit the target is equal to .8.
the probability that each bomber will miss the target is equal to 1 minus .8 = .2
the probability that all 3 bombers will miss the target is .2 * .2 * .2 = .008.
the probability that at least one of the bombers will hit the target is therefore equal to 1 minus .008 = .992.
this can be seen in the following table from excel.
the formula used is p(x) = p^x * q^(n-x) * c(n,x)
c(n,1) = n! / (x! * (n-x)!)
p is equal to the probability of a hit from each plane.
q is equal to the probability of a miss from each plane.
when x = 0, p(x) becomes p(0) = .8^0 * .2^(3-0) * c(3,0).
this becomes p(0) = 1 * .008 * 1 = .008.
when x = 1, p(x) becomes p(0) = .8^1 * .2^(3-1) * c(3,1).
this becomes p(1) = .8 * .04 * 3 = .096
all the calculations for p(x) from x = 0 to x = 3 are shown in the spreadsheet.
the spreadsheet shows that the sum of the probabilities for p(x) when x = 0,1,2,3 is equal to 1, as it should be.
the spreadsheet also shows that the sum of the probabilities for p(x) when x = 1,2,3 is equal to .992.
the spreadsheet also shows that this is the same as the probability for 1 minus p(x) when x = 0.
that calculation was done on top when i showed you that .2 * .2 * .2 = .008 and that 1 minus .008 is equal to .992.
your solution is that the probability that at least one of the bombers will hit the target is .992.
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