A building is examined by policemen with four dogs that are trained to detect the scent of explosives. if there are explosives i
n 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?
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.
