If that is the case, then the probability would be 0.222 or 22.2%.

[needhelp83]

Ray and Mike go target shooting together. Both shoot at a target at the same time. Suppose Ray hits the target with probability of 0.7, and Mike hits the target with probability 0.4. Given that exactly one bullet hit the target, what is the probability that it was Mike's bullet?

Given that the target is hit, what is the probability that it was Mike's bullet

Unsure where to start on this problem.
Any help

 

[Filip Larsen]

It may help to list up the combinations of the 2 bullets hitting and not hitting, and then think if it is possible to associate a probability to each of the 4 outcomes. With such probabilities you can now look at the combinations for exactly one hit and calculate the answer.

 

[needhelp83]

It may help to list up the combinations of the 2 bullets hitting and not hitting, and then think if it is possible to associate a probability to each of the 4 outcomes. With such probabilities you can now look at the combinations for exactly one hit and calculate the answer.

Why would you need to list the combination if there is only one that hit. Not exactly sure what you mean

 

[Filip Larsen]

If you only look at Rays bullet, it can either hit or miss, the probability of which given by your problem text. So knowing the probability for Ray to hit you also know the probability for him to miss (you should calculate this probability).

The same thing can be done for Mike, which will give you four probabilities. Two for Rays hit and miss, and two for Mikes hit and miss.

Now look at the combined "experiment" where Ray and Mike fire at the same time, and write up all the possible combinations of hit and miss events by Ray and Mike. If we call the first combination "Ray miss and Mike miss", what are the other 3 combinations then?

When you have those combinations you are ready to think (and make an assumption) about if Rays and Mikes shot were independent or not. Provided they are independent, you can now calculate the probability of the four combinations mentioned above. Look in your textbook for how to calculate the probability of a combination of independent events.

Now, when that is done you can inspect the four combinations and pick the two that describe that exactly one bullet hit (e.g. Ray hit and Mike miss), and knowing that either of these two events occurred you should be able to calculate the probability that is was Ray scoring a hit. You may want to read about conditional probabilities in your textbook for theoretical background and equations.

 

[needhelp83]

I think I finally got the first part of the problem, but I don't see what exactly the second part is asking...

1. Given that exactly one bullet hit the target, what is the probability that it was Mike's bullet?

P(Ray hit, Mike miss) = .7 * .6 = .42
P(Ray miss, Mike hit) = .4*.3 = .12

P(ray miss, mike hit) + p (ray hit, mike miss) = .54

.12/.54= 0.222 that Mike hit

2. Now it asks given that target is hit, what is the probability that it was Mike's bullet?
How is this any different from first question?

 

[willem2]

This second question is rather unclear. They could mean "Given that the target is hit at least once, what is the probability that Mike's bullet hit the target". It's the only interpretation that I can think of that isn't the same as the first question.
It has the problem that the target might be hit twice and then "it was Mike's bullet"
doesn't make sense.

 

[needhelp83]

Hmmm...any more ideas?

 

[Filip Larsen]

I think I finally got the first part of the problem, but I don't see what exactly the second part is asking...

1. Given that exactly one bullet hit the target, what is the probability that it was Mike's bullet?

P(Ray hit, Mike miss) = .7 * .6 = .42
P(Ray miss, Mike hit) = .4*.3 = .12

P(ray miss, mike hit) + p (ray hit, mike miss) = .54

.12/.54= 0.222 that Mike hit

Looks good.

2. Now it asks given that target is hit, what is the probability that it was Mike's bullet?
How is this any different from first question?

I think you should read the part "probability that it was Mike's bullet" as "probability that Mike's bullet hit". At least, that would allow you to solve it using a straight forward extension of what you did in 1).