If A and B are two events, then which of the following is true?

To find the probability of event 'A or B', 1.e., $P(A\cup B)$. If S is sample space for tossing of three coins, then

$\begin{array}{l}\mathrm{S}=\{\mathrm{HHT},\mathrm{HHH},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\}\\ \text{ Let }\mathrm{A}=\{\mathrm{HHT},\mathrm{HTH},\mathrm{THH}\}\text{ and }\mathrm{B}=\{\mathrm{HTH},\mathrm{THH},\mathrm{HHH}\}\end{array}$

be two events associated with 'tossing of a coin thrice'.

Clearly, $\mathrm{A}\cup \mathrm{B}=\{\mathrm{HHT},\mathrm{HTH},\mathrm{THH},\mathrm{HHH}\}$

Now, $\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)+\mathrm{P}\left(\mathrm{HHH}\right)$

If all the outcomes are equally likely, then

$P(A\cup B)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}$

Also, $\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)=\frac{3}{8}$

and $\mathrm{P}\left(\mathrm{B}\right)=\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)+\mathrm{P}\left(\mathrm{HHH}\right)=\frac{3}{8}$

Therefore, $P\left(A\right)+P\left(B\right)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}$

It is clear that $\mathrm{P}(\mathrm{A}\cup \mathrm{B})\ne \mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)$

The points HTH and THH are common to both A and B. In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of $A\cap B$ are included twice. Thus, to get the probability of $P(A\cup B)$ we have to subtract the probabilities of the sample points in

$A\cap B$ from $\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)$.

i.e., $\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\sum \mathrm{P}\left({\omega }_{\mathrm{i}}\right),\mathrm{\forall }{\omega }_{\mathrm{i}}\in \mathrm{A}\cap \mathrm{B}$

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

Thus, we observe that,

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