The probability of an event $B$ is $P_1$. The probability that events $A$ and $B$ occur together is $P_2$ while the probability that $A$ and $\bar{B}$ occur together is $P_3$. The probability of the event $A$ in terms of $P_1, P_2$ and $P_3$ is _____________
$P(A \cap \sim B)$=Something that is in A but not in B.
And Since, P(B)=$P_1$ and $P(A\cap B)=P_2$, the region which is only in B has probability $P_1-P_2$
so, $P(A)=P_3+P_2$
Solving using venn diagram looks easier.
@Ayush Upadhyaya
Then it will be $P\left ( A \right )=P_{2}\times P_{3}$, I mean AND inside those two Probability
not $P\left ( A \right )=P_{2}+ P_{3}$
right??
@srestha-No. P(A)=p2+p3
$P(A)$ needs $P_{2}$ and $P_{3}$
Not any one of them can do $P(A)$ fully
3p_{2}+p_{1}-p_{2}
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