I remember the probability must lie between 0 and 1, so it couldn't be 2 for total probability(1 would be an almost certain 5-0 season). Isn't that right?
Since each is 1/5 of the total, maybe .4X.2+ .7X.2, etc
=.08 + =.14 + etc?
I'd conclude nothing froma sample this small, especially sporting events. Regardless of %, I thought it would still be the equation
Pr(A+B+C)= Pr(A) X Pr(B) X Pr(C).
It's just when I calculate all 12 "50%, I come up with the worls worst probability of winning even one game, much less 12. To...
I'm a bit lost on use of mutually exclusive and independent rule on this. I know probability of winning a game is a loose term, but...
If I had a team's schedule of 12 games with probabilities listed of winning each game, would I add those probabilities, then divide sum by 12 for an average...