Question

The events X and Y are independent. Calculate the P(X and Y) if P(X) = 0.64 and P(Y)=0.23. Hence find P(X or Y)

Answer 1

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The events $X$ and $Y$ are independent. Calculate the $P(X\cap Y)$ if $P(X)=0.64$ and $P(Y)=0.23$. Hence find $P(X\cup Y)$.

Solution. We can regard events $X$ and $Y$ as subsets of some probability space $\Omega$. Then the event “$X$ and $Y$” corresponds to the intersection $X\cap Y$, while the event “$X$ or $Y$” is the union $X\cup Y$ of these events.

Recall that events $X$ and $Y$ are called *independent* if

$P(X\cap Y)=P(X)\cdot P(Y).$

Thus in our case

$P(X\text{ and }Y)=P(X\cap Y)=P(X)\cdot P(Y)=0.64\cdot 0.23=0.1472.$

We should compute $P(X\cup Y)$. We claim that

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y).$

Indeed, recall that is the intersection $A\cap B$ of events $A$ and $B$ is empty, then

$P(A\text{ or }B)=P(A\cup B)=P(A)+P(B).$

Notice that $X\cup Y$ can be represented as a union of three disjoint subsets:

$X\cup Y=(X\setminus Y)\ \cup\ (X\cap Y)\ \cup\ (Y\setminus X),$

whence

$P(X\cup Y)=P(X\setminus Y)+P(X\cap Y)+P(Y\setminus X).$

Moreover,

$X=(X\setminus Y)\ \cup\ (X\cap Y),\qquad Y=(Y\setminus X)\ \cup\ (X\cap Y),$

whence

$P(X)=P(X\setminus Y)+P(X\cap Y),\qquad P(Y)=P(Y\setminus X)+P(X\cap Y).$

Therefore

$P(X\setminus Y)=P(X)-P(X\cap Y),\qquad P(Y\setminus X)=P(Y)-P(X\cap Y),$

and so

$P(X\cup Y)=P(X)-P(X\cap Y)+P(X\cap Y)+P(Y)-P(X\cap Y)=P(X)+P(Y)-P(X\cap Y).$

Thus

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)=0.64+0.23-0.1472=0.7228.$