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: $P(A) = 0.23$

Step-by-step 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

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

$P(A) = 0.14 + 0.09$

$P(A) = 0.23$