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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 _____________
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$P(A ∩ B') = P(A) - P(A ∩ B)$

$\implies P(A)$ $=$  $P_{2}+P_{3}$
by Junior (883 points)
edited
+2
Thats good one. Event A can occur with B and with B' and since B and B' cannot occur together adding the probabilities of both will give P(A).
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can it be answered by conditional probability?Is the below approach correct?

p(A)=p(A/B)p(B)+p(A/B')p(B')

=p2p1+p3(1-p1)
+9

$P(A \cap \sim B)$=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.

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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??

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@srestha-No. P(A)=p2+p3

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$P(A)$ needs $P_{2}$ and $P_{3}$

right??

Not any one of them can do $P(A)$ fully

right??

+1 vote
Use Law of Total Probability:

$P(\text{A}) = P(\text{A and B}) +P(\text{P and Not B})$

$\Rightarrow P(\text{A}) = P(A \cap B) +P(P \cap \overline{B})$

$\Rightarrow P(\text{A}) = P_2 + P_3$
by Junior (849 points)

3p2+p1-p2

by Boss (14.3k points)
0
0
Wrong

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