Question

How do you find the probability of $A$ or $B?$

Answer 1

Finding probability of $A$ or $B$:

For two mutually exclusive events $A$ and $B$, the probability of $A$or $B$ is given by:

$P(A\cup B)=P\left(A\right)+P\left(B\right)$

But if events$A$and $B$ are not mutually exclusive then probability of $A$ or $B$ is given by:

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

For eg. if $A$ is the event of getting head and $B$ is the event of getting tail when a coin is tossed. Then $A$ and $B$are mutually exclusive events.

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{2}$

$\begin{array}{rcl}P(A\cup B)& =& P\left(A\right)+P\left(B\right)\\ & =& \frac{1}{2}+\frac{1}{2}\\ & =& 1\\ & & \end{array}$

Hence, probability of $A$or $B$ is $1$.

Let $C$ denote the event of getting an even number and $D$ denote the event of getting $6$ when a dice is rolled.

$P\left(C\right)=\frac{3}{6}=\frac{1}{2}$ , $P\left(D\right)=\frac{1}{6}$, $P(C\cap D)=\frac{1}{6}$

$\begin{array}{rcl}P(C\cup D)& =& P\left(C\right)+P\left(D\right)-P(C\cap D)\\ & =& \frac{1}{2}+\frac{1}{6}-\frac{1}{6}\\ & =& \frac{1}{2}\end{array}$

Hence, probability of $C$or $D$ is $\frac{1}{2}$.

Thus, in this way probability of $A$or $B$ can be found out.