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I came across the following question in a textbook (bear in mind that this is the only information given)-

There is a $50$ percent chance of rain today. There is a $60$ percent chance of rain tomorrow. There is a $30$ percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?

My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of $30$ percent. However, the answer given is $40$ percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.

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  • $\begingroup$ if they were independent you multiply for example you flip three coins the probability they all end up heads is $({1 \over 2})^3={1\over 8}$ dependent probabilities aren't so simple. $\endgroup$ – user451844 Aug 1 '17 at 13:37
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Yes, this is about independence. Or, better put, these events are not independent, which by common sense makes sense, since if it rains one day, it typically means there is a higher chance than normal that it rains the next day as well (we might well be in the 'rainy season')

You can also tell that the events are not independent given the numbers given to you. If the events were independent, then the probability of it not raining either day should have been $0.5 \cdot 0.4=0.2$, but you were told it is actually $0.3$.

What is always true, however (so you can always use this formua, whether the events are independent or not), is that:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

And from that you can derive your desired:

$$P(A \cap B) = P(A) +P(B) - P(A \cup B) = P(A) +P(B)-(1-P(A^C \cap B^C))=$$

$$0.5+0.6-(1-0.3) =1.1-0.7=0.4$$

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Yes, the multiplication rule only applies when events are independent, and there is no reason to assume the events of rain on each day are independent.

For this type of problem, it helps to make a Venn diagram. There are two circles, $A$ and $B$, representing the events of rain today and tomorrow. This gives four regions: $A\cap B$ (rain on both days), $A^c\cap B^c$ (no on both days), $A\cap B^c$ (rain today, but not tomorrow), and $A^c\cap B$. The given information, and the fact that the total probability is $1$, tells us that \begin{align} P(A)=P(A\cap B)+P(A\cap B^c)&=0.5\\ P(B)=P(A\cap B)+P(A^c\cap B)&=0.6\\ P(A^c\cap B^c)&=0.3\\ P(A\cap B)+P(A\cap B^c)+P(A^c\cap B)+P(A^c\cap B^c)&=1 \end{align} This is four equations equations in four unknowns, allowing you to solve for $P(A\cap B)$.

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When you have only two independent variables, sometimes it's easier just to make a table.

Let $X$ be the event that it rains today and $Y$ be the event that it rains tomorrow. We are given that $P(X)=0.5$, $P(Y)=0.6$, and $P(\overline X \cap \overline Y)=0.3$:

$$ % outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X) \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y) \\ \end{array} \end{array} $$

We know that $P(X)+P(\overline X)=1$ and likewise $P(Y)+P(\overline Y)=1$, so we can fill this in as follows:

$$ % outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)\color{red}{=0.5} \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y)\color{red}{=0.4} \\ \end{array} \end{array} $$

We know that the rows sum across and the columns sum down, so:

$$ % outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y)\color{red}{=0.1} \\ \hline \overline X & P(\overline X \cap Y)\color{red}{=0.2} & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array} $$

And finally:

$$ % outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y)\color{red}{=0.4} & P(X \cap \overline Y)=0.1 \\ \hline \overline X & P(\overline X \cap Y)=0.2 & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array} $$

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