# Three shooters, probability that two of them hits

#### 8Julio

##### New Member
I have a little problem.

There are three shooters ( i= 1,2,3) and they hit the target with probabilities
P(A1) = 1/2, P(B1) = 2/3, P(A3) = 3/4. All men shoot at a same time and
results are independent.
- What is the probability that there will be two hits?

P(A1) intersection P(B1) intersection P(A3) =

or 1/2 + 2/3 + 3/4 / 3 = 0.32

or 1/2 * 2/3 * 3/4 = 0.25

or 1/2 * 2/3 * 3/4 / 2 = 0.125

or 1/2 * 2/3 * 3/4 ^2 = 0.0625

- What is the probability that one shooter missed?

P(A1) * P(B1) * P(A3) / union ...

1/2 * 2/3 * 3/4 / 3 = 0.083

#### BGM

##### TS Contributor
First list out all the possibilities/scenario that here will be exactly two hits. Afterward the calculation of the probability is easy because you can use the independent properties.

#### 8Julio

##### New Member

Possibilities and scenarios with two hits are;

011 = A1, A2, A3 = two hits A2, A3
101 = A1, A2, A3 = two hits A1, A3
110 = A1, A2, A3 = two hits A1, A2

And now the independent properties;

P(A) = A2, A3 = 2/3 * 3/4 = 1/2

P(A) = A1, A3 = 1/2 * 3/4 = 3/8

P(A) = A1, A2 = 1/2 * 2/3 = 1/3

2. What is the probability that one shooter missed?

P(A) = 1/2 * 2/3 * 3/4 / 3 = 1/12

Are these correct ( and formal enough )?

#### BGM

##### TS Contributor
Do you know the probability that A1/A2/A3 miss?

When calculating the probability of the event "A1 miss and A2 hit and A3 hit", you cannot miss the probability of A1 miss; it is not equal to the event "A2 hit and A3 hit".

Afterward, since all 3 scenarios are mutually exclusive, you can sum them up to get the probability.

For the 2nd one, it is equivalent to the 1st question? or I missed something?

#### 8Julio

##### New Member
"Do you know the probability that A1/A2/A3 miss?" - No.
"For the 2nd one, it is equivalent to the 1st question?" - I think it is a different question

Here once again;

Possibilities and scenarios with two hits are;

011 = A1, A2, A3 = two hits A2, A3
101 = A1, A2, A3 = two hits A1, A3
110 = A1, A2, A3 = two hits A1, A2

And now the independent properties;

P(A) = A2, A3 = 2/3 + 3/4 = 1 5/12

P(A) = A1, A3 = 1/2 + 3/4 = 1 1/4

P(A) = A1, A2 = 1/2 + 2/3 = 1 1/6

2. What is the probability that one shooter missed? ( Because we know that there was two hits )

P(A) = 1/2 + 2/3 + 3/4 = 1 11/12

Are these now correct?

#### BGM

##### TS Contributor
I think #3 is the closest one (although not correct).

An important fact to bear in mind: any probability is bounded between 0 and 1. If you obtain a value outside this range, that means something goes wrong.

IMHO I think you better learn these elementary stuff once again, rather than guessing the answers and ask whether these guesses are correct or not; even if you guess correctly but do not understand, than it is not meaningful. On the other hand, it is not efficient for us to teach/you to learn every elementary stuff in this forum. It will be much better by asking the questions with your tutors/professors.

I can point out something you missed here anyway.

Note that in this question, the shooter can either hit or miss. Exactly one of them will happen. And the event that the shooter miss is the complementary event of the shooter hit - and the probability of a pair of complementary events sum up to 1.
So you can always determine the probability of the complementary event when you know the probability of that event.

$$P(A) + P(A^c) = 1$$

The second thing is that if the collection of events are independent, so as the complementary events. So the independence properties still apply.

If $$A, B$$ are independent, then $$P(A \cap B) = P(A)P(B)$$

Also $$P(A^c\cap B) = P(A^c)P(B), P(A\cap B^c) = P(A)P(B^c), P(A^c\cap B^c) = P(A^c)P(B^c)$$

Another thing is that the probability of the union of the mutually exclusive events is equal to the sum of the probability of each event.

If $$A, B$$ are mutually exclusive, then $$A \cap B = \varnothing, P(A \cap B) = 0$$ and $$P(A \cup B) = P(A) + P(B)$$

In general if they are not mutually exclusive, by inclusion-exclusion principle,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

So think again which events are independent and which events are mutually exclusive. Do not blindly apply the summation/multiplication.

#### 8Julio

##### New Member
Hi. I am sorry. I studied and thought more. This is much better effort!

I thought again which events are independent and which events are mutually exclusive.

We have here three events P(A) = ½, P(B) = 2/3 , P(C) = ¾

011 = A1, A2, A3 = two hits A2, A3
101 = A1, A2, A3 = two hits A1, A3
110 = A1, A2, A3 = two hits A1, A2

A miss:

1 – P(A^c U B U C ) means A will NOT happen, B and C happen

( ½ * 2/3 * 3/4 = Independent events )

= 1- ( ½ * 2/3 * 3/4 ) = 3/4 = 0.75

B miss:

1 – P(A U B^c U C ) = B will NOT happen, A and C happen

= 1 - ( ½ * 1/3 * 3/4 ) = 7/8 = 0.875

C miss:

1 – P(A U B U C^c ) = C will NOT happen, A and B happen

= 1 - ( ½ * 2/3 * 1/4 ) = 11/12 = 0.916

And question was What is the probability that there will be two hits?

Answer is 0.75 + 0.875 + 0.916 / 3 = 0.847 ( 0.75, 0.875, 0.916 = mutually exclusive events )

2.

SORRY: THE QUESTION 2. WAS “- What is the probability that FIRST shooter missed?”

So that means A and therefore;

A miss:

1 – P(A^c U B U C )

= 1 - ( ½ + 2/3 + 3/4 ) = 3/4 = 0.75

#### BGM

##### TS Contributor
Ok very close now.

Why do you consider $$1 - P(A^c \cap B \cap C )$$?

$$A^c \cap B \cap C$$ is the event you want, not its complement.

And the probability of the union of these mutually exclusive scenario is the sum of them.

Why take average? It's wrong here. No need to divide by 3.

For the second question, do you know which is going to shoot first? Are they equally likely to shoot first? If you are able to do the first question, just apply the same technique here.

#### 8Julio

##### New Member
Good! Here are new results, I truly thought these.

A miss:

P(A^c U B U C ) means A will NOT happen, B and C happen

( ½ * 2/3 * 3/4 = Independent events )

= ½ * 2/3 * 3/4 = 3/4 = 0.25

B miss:

P(A U B^c U C ) = B will NOT happen, A and C happen

= ½ * 1/3 * 3/4 = 1/8 = 0.125

C miss:
P(A U B U C^c ) = C will NOT happen, A and B happen

= ½ * 2/3 * 1/4 = 11/12 = 0.083

And question was What is the probability that there will be two hits?

Answer is 0. 25 + 0. 125 + 0.083 = 0.458 (0.25,0. 125, 0.083 = mutually exclusive events )

2. “do you know which is going to shoot first?” No, but there is no need, because all men shoot at a same time.

So that naturally means A and therefore;

A miss:

P(A^c U B U C )

½ + 2/3 + 3/4 = 0.25

#### BGM

##### TS Contributor
Ok the first question is good.

For the second question, even 3 shooters shoot at the same time, I think the question you provided do mean that we ignore the other two shooters. So it is just 1/2 as you calculated. It did not mention the other two shooters hit or miss.

#### 8Julio

##### New Member
Great! I'm happy!

So the second question is:

P(A) = A2, A3 = 2/3 * 3/4 = 1/2

#### 8Julio

##### New Member
No, second question is:

P(A1) = 1/2 ( there surely is no need to calculate shooters 2 and 3, just shooter 1 ).

Now I got it!

#### 8Julio

##### New Member
One more thing:

Can you please confirm that these are now correct;

A miss:

P(A^c U B U C ) means A will NOT happen, B and C happen

( ½ * 2/3 * 3/4 = Independent events )

= ½ * 2/3 * 3/4 = 3/4 = 0.25

B miss:

P(A U B^c U C ) = B will NOT happen, A and C happen

= ½ * 1/3 * 3/4 = 1/8 = 0.125

C miss:
P(A U B U C^c ) = C will NOT happen, A and B happen

= ½ * 2/3 * 1/4 = 11/12 = 0.083

And question was What is the probability that there will be two hits?

Answer is 0. 25 + 0. 125 + 0.083 = 0.458 (0.25,0. 125, 0.083 = mutually exclusive events )

2. question is:

P(A1) = 1/2 ( this was really easy, just look - given information - what a probability is... )

#### BGM

##### TS Contributor
One last comment: In the second question you are asked for the probability of $$A_1$$ missed. So you are seeking $$P(A_1^c)$$, but incidentally this equal to $$P(A_1) = \frac {1} {2}$$