Question

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?

Answer 1

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.