Probability - 3 Independent Events

[jmc]

Homework Statement

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

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. :?

 

[Ray Vickson]

Homework Statement

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

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}

This is the probably that all three hit the target

b. To me, it seems it would have to be one of P{A+B}, P{B+C}, or P{A+C}.

Not quite: when counting hits from A and B you also need C to miss, etc. I'm not sure how to make the math 'choose' which one though, and what formula to use to do that.

It does not choose; all three possibilities can occur, so all need to be counted.

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

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

 

[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.

 

[haruspex]

With independent events, the word and implies multiplication of probabilities.

Agreed

The word or implies addition of probabilities.

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.

 

[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.

 

[PeroK]

Homework Statement

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

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. :?

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.

 

[Ray Vickson]

Homework Statement

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

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. :?

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

 

[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

 

[PeroK]

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

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.)

 

[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) .

 

[jmc]

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