Question

Suppose
P
(
A
)
=
0.29 and
P
(
B
)
=
0.48. If
P
(
A
∣
B
)
=
0.146, what is
P
(
A
∩
B
)
?

Answer 1

Answer:

$P(\text{A}\cap \text{B}) = 0.07008$.

Step-by-step explanation:

The vertical bar here reads "given." $P(\text{A}|\text{B})=0.146$ means that is the probability that A is true given that B is true is 0.146.

Consider the formula for conditional probabilities:

$\displaystyle P(\text{A}|\text{B}) = \frac{P(\text{A}\cap\text{B})}{P(\text{B})}$.

Multiply both sides by $P(\text{B})$, the probability of B:

$\displaystyle P(\text{A}|\text{B})\cdot P(\text{B}) = \frac{P(\text{A}\cap\text{B})}{P(\text{B})} \cdot P(\text{B})$.

$\displaystyle P(\text{A}|\text{B})\cdot P(\text{B}) = {P(\text{A}\cap\text{B})}$.

Both $P(\text{A}|\text{B})$ and $P(\text{B})$ are given, and the question is asking for $P(\text{A}\cap\text{B})$.

$P(\text{A}\cap\text{B}) = P(\text{A}|\text{B})\cdot P(\text{B}) = 0.146 \times 0.48=0.07008$.