Thread: Bayes theorem (text problem)

Two cannons each shoot one shell at the target. The probabilty that the first cannon hits the target is 0.2, and the probability that the second cannon hits the target is 0.6. The target will be destroyed if both cannons hit the target, but if just one cannon hits the target, the probability of target being destroyed is 0.3. What is the probability that the target is destroyed after two shots?


I think, I have to use the Bayes theorem here.

First event is C1, first cannon hits the target.
Second event is C2, second cannon hits the target.
Third event is T, target is destroyed.




This is conditional probability:



is this OK?, because the text is a little complex here.

Originally Posted by Nforce

Two cannons each shoot one shell at the target. The probabilty that the first cannon hits the target is 0.2, and the probability that the second cannon hits the target is 0.6. The target will be destroyed if both cannons hit the target, but if just one cannon hits the target, the probability of target being destroyed is 0.3. What is the probability that the target is destroyed after two shots?

Warning: I am not fully convinced that I am reading this correctly.
It seems that we must assume independence:

To be destroyed we want HH or H,NH,D or NH,H,D

Probability is (0.6x0.2)+(0.2x0.4x0.3)+(0.8x0.6x0.3)=0.12+0.024+0 .144=0.288

I get it now.

But it doesn't make sense:

For example:

If we assume that the "first cannon hits the target" and "second doesn't" and "one of them hits the target",
it translates to:

0.2 * 0.4 * 0.3

that's what biffboy has done.

but in plain english it's not logical. We already said first hits, and second doesn't, we can't yet say and one of them hits the target.

there are only two possibilities for just one hitting. That is (1st hits and 2nd misses) or (1st misses and 2nd hits)

So P of just one hitting is (0.2x0.4)+(0.8x0.6)