# An experiment consists in throwing a die twice

An experiment consists in throwing a die twice. Knowing that none of the launches provides the same result, calculate the probability that exactly one of the two launches gives as a result $$1$$.

So if I say $$A=$${the results of the two launches are different} and $$B=$${exactly one of the two launches gives as a result $$1$$} I have to calculate $$P(B|A)$$ which is equal to $$P(B∩A)/P(A)$$.

My problem is, how can I calculate the intersection between B and A? Regarding $$P(A)$$ I suppose it is equal to $$30/36$$ because $$30$$ are the favorable cases ($$6*5$$) and $$36$$ are the possible cases ($$6*6$$).

B is a subset of A because B states that it is impossible to have 2 ones thrown. so $$P(A \cap B) = P(B)$$. The probability that exactly one 1 is thrown is P(1 on first throw | !1 on second throw) + P(1 on second throw | !1 on first throw) or $$1/6*(5/6) + 1/6*(5/6) = 10/36$$
• and what does the $!$ mean? Apr 3 '19 at 15:16