Every shell fires by the M16 has a 0.85 probability of destroying its target.
(a)The M16 fires one shall at each of the 3 targets. What is the probability that all the 3 targets are destroyed?
(b) The M16 can fire a maximum of 3 shells at a single target. Wha t is the probability that the target is destroyed?

Ok, these are independent probabilities.

I will give you an example.

An unbaised coin is tossed thrice.
The probability at each toss is 0.5.
The probability of having 3 heads is (0.5 x 0.5 x 0.5) = 0.125. [or (0.5)^3]

Each time you fire, you have the probability 0.85 of destroying your target.

For the second part it is different. The outcomes are numerous. You can for example destroy it on the first shot, on the second shot, on the third shot or not at all.


P(first shot destroyed) = 0.85
P(second shot destroyed) = P(first shot missed) x P(second shot destroyed)

and so on. Then, you add them up because it is either the first case, the second case or the third case.

An easier way to look at it instead of finding 3 different probabilities and adding them up, is to see what is the probability of not destroying your target, then remove that from one. In either methods, you get the same answer.

Post your answer :)