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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 ($6*6$).

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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 $1/6*(5/6) + 1/6*(5/6) = 10/36$

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  • $\begingroup$ if A is a subset of B then the intersection should be P(A) $\endgroup$
    – user658422
    Apr 3 '19 at 15:07
  • $\begingroup$ ... corrected. Thanks. $\endgroup$
    – RandyF
    Apr 3 '19 at 15:10
  • $\begingroup$ I don't understand what you did on the second part of the answer. Why did you sum those two probabilities? $\endgroup$
    – user658422
    Apr 3 '19 at 15:14
  • $\begingroup$ and what does the $!$ mean? $\endgroup$
    – user658422
    Apr 3 '19 at 15:16
  • $\begingroup$ They can be summed because they are mutually exclusive. It's equivalent to 2 choose 1 * 1/6 * 5/6. $\endgroup$
    – RandyF
    Apr 3 '19 at 15:18

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