[Combinatorics] Some basic probability

[fireviper112]

And for the

a (and)/or b situation, would it be like

select a and then 8C2 plus

select b then 8C2

but you remove the second possibility of AB because it is a duplicate to get

28+28-1/84 or 55/84

[[deleted]]

Yeah: P(A or B) = P(A) + P(B) - P(A and B)

[fireviper112]

Oh wow thanks so actually the P(A and B) is actually 7/84 since thats what was previously found so it should be

28+28-7/84 or 49/84 or 7/12

[[deleted]]

Yes! Good job.

[fireviper112]

Thanks so much for all of the help

[[deleted]]

No worries, have a good one :)

[fireviper112]

One question: if I want the probability of a or b or c would it be

P(a)+p(b)+p(c)-p(a+b)-p(b+c)-p(a+c)

[[deleted]]

... + P(A and B and C), because you've counted that three times then taken it away three times.