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16 November, 10:13

A building is examined by policemen with four dogs that are trained to detect the scent of explosives. if there are explosives in a certain building, and each dog detects them with probability 0.6, independently of other dogs, what is the probability that the explosives will be detected by at least one dog?

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  1. 16 November, 10:53
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    The probability of a dog finding the explosive is. 6

    The probability of a dog will not find the explosive is. 4

    The probability that no dogs will find the explosive is. 4^4

    The probability that at least 1 dog will find the explosive is 1 -.4^4.

    this equals 1 -.0256 which equals. 9744

    the probability that at least 1 dog will find the explosive is. 9744.

    if you wanted to find the probability that exactly x out of 4 will find the explosives, then you would use the following formula:

    p (0) = 4c0 *.6^0 *.4^4

    p (1) = 4c1 *.6^1 *.4^3

    p (2) = 4c2 *.6^2 *.4^2

    p (3) = 4c3 *.6^3 *.4^1

    p (4) = 4c4 *.6^4 *.4^0

    4c0 and 4c4 = 1 each

    4c1 and 4c3 = 4 each

    4c2 = 6

    probabilities become:

    p (0) =.0256

    p (1) =.1536

    p (2) =.3456

    p (3) =.3456

    p (4) =.1296

    sum of the probabilities equals 1 as it should.

    combination formula of ncx = n! / ((n-x) ! x!)

    if you add up the probabilities for 1 and 2 and 3 and 4 dogs finding the explosives, you will see that it is equal to 1 - probability that no dogs find the explosives.
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