Question

Suppose that P(A|B)=0.2, P(A|B')=0.3, and P(B)=0.7. What is the P(A)? Round your answer to two decimal places (e.g. 98.76).

Answer 1

Answer:
`P(A) = 0.23`

Explanation:

Given :

`P(A/B) = 0.2`

`P(A/B^(1))=0.3`

`P(B)= 0.7`

`P(A) = ?`

From rules of probability :

`P(A) = P(AnB) + P(A n B^(1))` ........................... equation *

`P(A n B)` can be written as
`P(A/B)` x
`P(B)`

Also ,
`P(A/B^(1))` can be written as
`P(A/B^(1))` x
`P(B^(1))`

substituting into equation * , we have

`P(A) = P(A/B)`
`P(B) + P(A/B^(1))P(B^(1))`

since P(B) = 0.7, then
`P(B^(1)) = 1 - P(B) = 0.3`

so , substituting each values , we have

`P(A) = (0.2)(0.7) + (0.3)(0.3)`

`P(A) = 0.14 + 0.09`

`P(A) = 0.23`