Interesting thing about archers hitting the target

[Дьявол]

Hello, again!

I got one very interesting question.

We got three archers, and the probability of the ones to hit the target is:
A₁, A₂, A₃.

What if the task is to find the probability that the target will be hit at least from one archer.

So at least one archer to hit the target.

Is it P(A₁ U A₂ U A₃) = A₁ + A₂ +A₃ ?

Or $$(1- P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1 - \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}$$, where $$\bar{A}$$ is opposite of A?

Or maybe, both are valid?

Thanks in advance.

 

[HallsofIvy]

Hello, again!

I got one very interesting question.

We got three archers, and the probability of the ones to hit the target is:
A₁, A₂, A₃.

What if the task is to find the probability that the target will be hit at least from one archer.

So at least one archer to hit the target.

Is it P(A₁ U A₂ U A₃) = A₁ + A₂ +A₃ ?

No. This is wrong. $P(A_1\cup A_2)= P(A_1)+ P(A_2)- P(A_1\cap A_2)$ and that extends to 3 events: $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)$$- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)$.

Or $$(1- P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1 - \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}$$, where $$\bar{A}$$ is opposite of A?

Yes, this is correct.

Or maybe, both are valid?

Thanks in advance.

 

[Дьявол]

Thanks for the post, HallsofIvy.

In this case, the shootings of the archers are independent cases. So that's why we do not need $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)$
since $P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)=0-0-0-0=0$. That's why I said independent cases.

But what if P(A₁)=0.8, P(A₂)=0.9, P(A₃)=0.75

In that case the sum $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)=0.8+0.9+0.75=2.45$

This is strange.

 

[mXSCNT]

The fact that two cases A and B are independent doesn't mean P(A n B) = 0. It means that P(A n B) = P(A)P(B).

 

[Дьявол]

If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?

 

[statdad]

If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?

No,

$$A \cap B = \emptyset$$

if the two sets are disjoint , which is not the same as independent.

 

[Tibarn]

If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?

Actually, that would mean that they're not independent. Two events are independent if the occurrence of one does not influence the occurrence of the other, i.e. P(A|B) = P(A). If $$A \cap B = \emptyset$$, then P(A|B) = 0, so the events are not independent.

With regards to the original question, it would be easiest to take the probability that every archer misses and subtract it from one.

 

[Дьявол]

Thanks for the replies.
@Tibarn, in this case the occurrence of one does not influence the occurrence of the other.
$$P(A/B)=\frac{m_{A\cap B} }{ m_{B}}$$
out of there
$$P(A/B)=\frac{\frac{m_{A\cap B}}{n}}{\frac{m_{B}}{n}}$$
and
$$P(A/B)=\frac{P(A\cap B)}{ P(B)}$$

The cases are independent if the occurrence of one does not influence the occurrence of the other. So, if two cases are independent, then P(A/B)=P(A). Out of there P(A n B)=P(A)*P(B)

Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint.

P(A U B)=P(A) + P(B) - P(A n B)

In this case $m_{A\cap B}=0$ because the cases:
I - the 1st archer will hit the target
II - the 2nd archer will hit the target

DO NOT have something in common.

$$P(A U B) = \frac{m_A+m_B-m_{A\cap B}}{n}$$

$$P(A U B) = \frac{m_A+m_B-0}{n}=P(A)+P(B)$$

Is this true?

Are those cases disjoint??

 

[CRGreathouse]

Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint.

P(A U B)=P(A) + P(B) - P(A n B)

In this case $m_{A\cap B}=0$ because the cases:
I - the 1st archer will hit the target
II - the 2nd archer will hit the target

DO NOT have something in common.

$$P(A U B) = \frac{m_A+m_B-m_{A\cap B}}{n}$$

$$P(A U B) = \frac{m_A+m_B-0}{n}=P(A)+P(B)$$

Is this true?

Are those cases disjoint??

You're saying that if one archer hits the target the other one always misses?

 

[Дьявол]

Are those cases disjoint?

 

[Tibarn]

In this case $$m_{A \cap B}=0$$ because the cases:
I - the 1st archer will hit the target
II - the 2nd archer will hit the target

DO NOT have something in common.

Think about this intuitively. If A and B are disjoint, then event A and event B cannot occur simultaneously. So, if A is the first archer hitting the target and B is the second archer hitting the target, $$P(A \cap B) = 0$$ means that both archers cannot simultaneously hit the target. If we know that A hit the target, then it would follow that B missed the target, so the events are not independent (unless A or B always misses).

If both archers take one shot, then we have four possible events:
1. Both miss
2. Archer A hits, B misses
3. Archer A misses, B hits
4. Both hit.

In this case, $$A \cap B$$ is case 4, where both archers hit. If you're going to do probability by cases, it's important that you get all of them.

 

[Дьявол]

Now, I understood. Thank you very much for the help.

Regards.

 

[HallsofIvy]

You may be confusing "mutually exclusive" with "independent". "Independent" means what happens in one case does not affect what happens in the other- $P(A\cap B)= P(A)P(B)$ or, equivalently, P(A|B)= P(A). "Mutually exclusive" means $P(A\cap B)= 0$ so that P(A|B)= 0. Not at all the same thing!

 

[Дьявол]

Yes, you're right. I did a little research, and find out that the events in this case aren't exclusive, but they are independent. Because if they are exclusive P(A) or P(B) will be equal 0 so that P(A n B)=0.