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The probability that John goes to swimming pool on Monday is 0.8. If he goes to the pool on Monday the probability that he goes on Tuesday is 0.3. If John does not go to the swimming pool on Monday the probability he goes on Tuesday is 0.9. Find the probability that John goes exactly one of the two days.

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    $\begingroup$ You can draw a tree with these probabilities, and add the two braches which are (M, but not T) + (not M, but T) $\endgroup$ – Michael Paris Sep 29 '17 at 11:07
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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made some effort. $\endgroup$ – José Carlos Santos Sep 29 '17 at 11:07
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Let:

  • $A$ = {John goes to the pool on Monday},
  • $B$ = {John goes to the pool on Tuesday}.

In your post, we have the following pieces of information:

  • $\mathbb{P}(A) = 0.8$,
  • $\mathbb{P}(B \mid A) = 0.3$,
  • $\mathbb{P}(B \mid \overline{A}) = 0.9$ .

What you are looking for is the probability that he goes only once (either Monday or Tuesday). That is the probability of the event $(A \cap \overline{B}) \cup (\overline{A} \cap B)$. [$A \cap \overline{B}$ : if he goes on Monday, he won't go on Tuesday ; $\overline{A} \cap B$ : if he did not go on Monday, he will go on Tuesday].

We have:

$$ \begin{align*} \mathbb{P}\big( (A \cap \overline{B}) \cup (\overline{A} \cap B) \big) & = \mathbb{P}( A \cap \overline{B} ) + \mathbb{P}(\overline{A} \cap B) \\[2mm] & = \mathbb{P}(\overline{B} \mid A)\mathbb{P}(A) + \mathbb{P}(B \mid \overline{A})\mathbb{P}(\overline{A}) \\[2mm] & = \big( 1 - \mathbb{P}(B \mid A )\big) \mathbb{P}(A) + \mathbb{P}(B \mid \overline{A})\big( 1 - \mathbb{P}(A)\big) \\[2mm] & = (1 - 0.3) \times 0.8 + 0.9 \times 0.2 \\[2mm] & = 0.74. \end{align*} $$

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Let $M=1$ denote the event that he goes on monday and $T=1$ denote tuesday. $M=0$ not going on monday, same for $T=0$. We know $P(M=1)=0.8 \Rightarrow P(M=0)=0.2$, $P(T=1|M=1)=0.3 \Rightarrow P(T=0|M=1)=0.7$, $P(T=1|M=0)=0.9$ and we have to find the probability $P(M+T=1)$. Now utilize Kolmogorov's definition of conditional probability to compute \begin{align*} P(M+T=1)&=P(M=1,T=0)+P(M=0,T=1)\\&=P(T=0|M=1)P(M=1)+P(T=1|M=0)P(M=0)\\&=0.7\cdot 0.8 + 0.9\cdot 0.2 = 0.74 \end{align*} which is your answer

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    $\begingroup$ You found (correctly) that $P(M = 0) = 0.2$, but you used $P(M = 0) = 0.1$ when you did your calculation. $\endgroup$ – N. F. Taussig Sep 29 '17 at 11:27
  • $\begingroup$ Thank you, it should be correct now. $\endgroup$ – O. T. Sep 29 '17 at 11:29

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