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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 _____________

### 1 comment

we know ,P(B | A)+ P(B’|A) = 1

given,

P(B’|A) P(A) = p3

P(B|A) P(A) = p2...(1)

P(B|A) = 1-(p3/P(A))

So, P(B|A) P(A) = P(A) -p3……...(2)

and from 1 and 2

P(A) = p2+p3.

$P(A ∩ B') = P(A) - P(A ∩ B)$

$\implies P(A)$ $=$  $P_{2}+P_{3}$
by

$P(A)$ needs $P_{2}$ and $P_{3}$

right??

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

right??

Nothing is said about P(A) and P(B) being independent so saying P(A$\cap$B)  = P(A)P(B) is wrong.
Thanks
Use Law of Total Probability:

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

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

$\Rightarrow P(\text{A}) = P_2 + P_3$

Well, from venn diagram it looks pretty obvious that the

P(A) = P1 + P2

But, I tried another approach...

@Arjun sir, can you please check this out..

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

That is, the probability of occurrence of A is the sum of:

i) Probability that A occurs given that B has occurred: P(A/B)
ii) Probability that A occurs given that B' has occurred: P(A/B')

Since, B and B' totally make up sample space, the occurrence of A is also covered.

P(A/B) = P(A∩B)  =  P2
P(B)            P1

P(A/B') = P(A∩B')  =  P3
P(B')          1-P1

P(A) = P2   +   P3
P1       1-P1

by

### 1 comment

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

This part is wrong,

law of total probability is

$P(A) = P(B)*P(A/B) + P(B’)*P(A/B’)$ = $P(A\cap B) + P(A\cap B^{'})$

given,

A.B = p2

A.B’ = p3

so adding ,we get

A.B + A.B’ = p2+p3

or A = p2+p3

1
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