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...
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% ...
Hi,
Thanks for your reply.
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...
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%
Please show working:
1) What is the chance that at least one player catches a...