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Leave X alone an irregular variable addressing the number shows up on  the primary pass on  what's more, Y be an arbitrary variable addressing the number shows up on  the subsequent bite the dust  the normal amount of the numbers that show up on two dice = E(X)+E(Y) Given that  P(X=3) = 2* P(X=1) = 2* P(X=2) = 2* P(X=4) = 2* P(X=5)  Let P(X=3)=q  Then  P ... See the full answer