probability

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[hasty briar]

iii isn't right, the others look fine

A' means not A

so that means (2+4+1)/12

mb lol

my fault you're good

not sure about the tables, but if they're just addition which seem to be the case, then this is right

but you have to find i, ii and iii

yes

ii would be (5+2+4)/15

= 11/15

which is the same as P(AUB) = P(A and B) + P(A' and B) + P(B' and A)