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$\begin{array}{|c|c|c|c|} \hline & X = a & X = b & \text{Total} \\ \hline Y = c & 20 & 70 & 90 \\ \hline Y = d & 15 & 45 & 60 \\ \hline \text{Total} & 35 & 115 & 150 \\ \hline \end{array}$

$P(X = a\hspace{.5 em} OR\hspace{.5 em} Y \ne c)$

Since X can be and Y can be not c at the same time, I'm guessing they are not mutually exclusive so I cannot use the following formula.

$P(X = a\hspace{.5 em} OR\hspace{.5 em} Y \ne c) = P(X = a) + P(Y \ne c) = \frac{95}{150}$

I just need to verify this is wrong or right.

Daniele Tampieri
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Talha Abid
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    You are right about that way being invalid because of the overlap. But using DeMorgan's rule your event becomes $X \neq a$ AND $Y=c$ which is only one box, and can get that easily. – coffeemath Oct 29 '21 at 23:15
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    $\frac{95}{150}$ is wrong and you are correct that you cannot use that expression. $P(X = a\text{ OR } Y \ne c) = P(X = a \text{ AND } Y \ne c) + P(X = a \text{ AND } Y = c) + P(X \not= a \text{ AND } Y \ne c) $ would involve mutually exclusive events – Henry Oct 30 '21 at 00:32

1 Answers1

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The addition law of probability states that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

So the general formula should be $$P(X=a \cup Y\not=c) = P(X=a)+P(Y\not=c) - P(X=a \cap Y\not=c)$$

The two events are mutually exclusive if they cannot happen together. This means the probabilty of both events happening is 0: $$P(X=a \cap Y\not=c)=0$$ which falls back to the formula you correctly stated for the addition probability of mutual exclusive events: $$P(X=a \cup Y\not=c) = P(X=a)+P(Y\not=c)$$ Each of the probability terms can be readily computed from the numbers in the appropriate cells of the contingency table.

Yusuf N
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