# Various chances of 5 players catching balls

1. Oct 12, 2012

### Back<2steps

There are 5 players and me.
If I throw numerous balls in the air and the chances of each individual player catching a ball is as follows:
player 1: 13%
player 2: 16%
player 3: 14%
player 4: 15%
player 5: 18%

1) What is the chance that at least one player catches a ball?
2) Chance that no one catches a ball?
3) Chance that two or more players catch a ball?
4) Chance that only one player catches a ball?

2. Oct 12, 2012

### chiro

Hey Back<2steps.

For the first one, if P1, P2, P3, P4, and P5 is probability of those catching a ball, then P(P1 v P2 v P3 v P4 v P5) is the probability of at least one catching a ball. Can you use the Kolmogorov Axioms to get this probability?

3. Oct 13, 2012

### Back<2steps

Hi,

I have looked up Kolmogorov Axioms, but I am unclear on what you are telling me. I would be grateful if you would show the working for me.

I am also unclear on what 'P' and 'v' is in 'P(P1 v P2 v P3 v P4 v P5)'.

Are you saying I should use P(AuB)=P(A)+P(B)-P(AnB) ?
I don't understand how to use this formula on more than 2 variables.

Sorry for my lack of knowledge. Appreciate any help.

4. Oct 13, 2012

### awkward

Try doing 2) first; it's the easiest. You should then be able to answer 1) using your result from 2).

5. Oct 13, 2012

### Back<2steps

Was kind of hoping for a worked through example rather than advice as to which one to attempt first... :)

As you do not wish to answer for me, maybe you would be kind enough to tell me if the following is correct:
Here is my attempt at 2:
(1-0.13)*(1-0.16)*(1-0.14)*(1-0.15)*(1-0.18) = 43% ??
Therefore 1)=57%

??

Thanks

6. Oct 13, 2012

### Mentallic

Think about this: Instead, we have two players and both of them have a 50% chance of catching the ball, so the chance of either of them catching it should be 75%, right? But if we used your method we get
(1-0.5)*(1-0.5) = 25%
But 25% is what is left over from the 75%, hence (this isn't a proof, but we know we're on the right track) 25% is the chance of neither player catching the ball. This is why awkward mentioned you should answer 2) first.

So if you haven't already been able to put these ideas together to answer 1), the answer to my question should be

1 - (1-0.5)*(1-0.5) = 75%

7. Oct 13, 2012

### Back<2steps

So this is correct for 'chance that no one catches ball':
((1-0.13)*(1-0.16)*(1-0.14)*(1-0.15)*(1-0.18)) = 43.80% ??

And Chance of at least one player catching ball = 1-((1-0.13)*(1-0.16)*(1-0.14)*(1-0.15)*(1-0.18)) = 56.20%

Still have no clue how to begin on 3) Chance that two or more players catch a ball. & 4) Chance that only one player catches a ball?
Is this correct?
If so thank you very much.

Still have no clue how to begin on 3) Chance that two or more players catch a ball, & 4) Chance that only one player catches a ball?

Last edited: Oct 13, 2012