# Probability - 3 Independent Events

1. Oct 30, 2014

### jmc

1. The problem statement, all variables and given/known data
Three tanks are sent to shoot a certain target. Each tank is to shoot one round. The
probabilities of a tank's round hitting the target are 1/2, 2/3 and 3/4 for tanks A, B, and C
respectively. Assuming independence of A, B and C hitting, find the probability that:
a. rounds from all three of the tanks hit the target.
b. rounds from exactly 2 tanks hit the target
c. the target is hit at least once

2. Relevant equations

I'm don't know where to start/which eqn's a relevant for this type of problem.

3. The attempt at a solution
a. P{A}=0.5, P{B}=2/3, P{C}=0.75. Since all are independent events, P{hitting the target}=P{A}*P{B}*P{C}

b. To me, it seems it would have to be one of P{A+B}, P{B+C}, or P{A+C}.
I'm not sure how to make the math 'choose' which one though, and what formula to use to do that.

c. Not sure how to proceed to think about solving it. :?

2. Oct 30, 2014

### Ray Vickson

Try looking at the complementary event that all three miss the target.

3. Oct 31, 2014

### RUber

With independent events, the word and implies multiplication of probabilities. The work or implies addition of probabilities.
So in b, you have (A and B and notC) or (A and notB and C) or (notA and B and C).
For c, it is definitely much easier to say that at least one hits the target is equivalent to not none of them hit the target.

4. Oct 31, 2014

### haruspex

Agreed
No, that's for mutually exclusive events, which are definitely not independent.
If A and B are independent then to get the rule for OR apply the AND rule to not A, not B, and complement the answer.

5. Oct 31, 2014

### RUber

Good catch. For part b, these are mutually exclusive options and should be summed. For part c, you cannot say p(a hits or b hits or c hits) = p(a hits )+ p( b hits ) + p(hits) since they are not mutually exclusive, and the sum would be more than 1.

6. Oct 31, 2014

### PeroK

Why not treat this as an experiment that you can repeat many times? Hint: 12 times, for example. Look at what happens on average. This is a technique called drawing a "probability tree". If you're stuck with probabilities, it's often a good idea to try this: it often gives you a good idea of what's happening and even the answer.

7. Oct 31, 2014

### Ray Vickson

Hello jmc: are you still here? Have you looked at any of the replies?

8. Nov 2, 2014

### jmc

Thank you
It makes sense to incl. the probability of not hitting the target in each case for part B. I don't know why I didn't catch on to that before.
For what I understand, it should be calculated as such:
Pr{A & B & Not C} + Pr{A & Not B & C} + Pr{Not A & B & C} = n(A*B*NotC)+n(A*NotB*C)+n(NotA*B*C) = 2/24 + 3/24 + 6/24 = 11/24

PeroK, I've tried to use a probability tree for Part C. I only found 8 'experiments' ? from the results from each experiment that meet the conditions (At least 1 hit), would I sum each of those probabilities to get the total?

Regards

9. Nov 2, 2014

### PeroK

Your probability tree looks spot on. That tells you everything about the problem, in fact. Yes, you can add up relevant probabilities.

(I made a mistake with "12". It should have been 24: everything is a multiple of 1/24. That's what I meant.)

10. Nov 2, 2014

### RUber

And, as a check, you want to make sure that all of your probabilities sum to 1. If there is only 1 outcome that doesn't satisfy your requirements, as in your tree, it is sometimes faster to subtract the odds of all missing from 1 rather than sum all the events where at least one hits.
1-p(none hit) = sum p(at least one hit) .

11. Nov 2, 2014

### jmc

Yes RUber, I realized that earlier as I was ensuring all probs summed to 1. Much faster method for solving!

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