A convenience store customer buys product $A$ with probability $0.50$ and product $B$ with probability $0.75$.Given that she buys $A$ the probability that she buys $B$ is $0.8$. Then what is the probability that she buys $B$ Given that a customer buy only one item, is
What i try :: Here $P(A)=0.5$ and $P(B)=0.75$ and
$\displaystyle P\bigg(\frac{B}{A}\bigg)=.8\Longrightarrow \frac{P(A\cap B)}{P(B)}=0.8\Longrightarrow P(A\cap B)=0.04$
As i understand we have to find
$\underbrace{P\bigg(\frac{B}{A}\bigg)}_{\text{probability of item B given that item A}}+\underbrace{P\bigg(\frac{B}{B}\bigg)}_{\text{probability of item B given that item B}}=0.8+1=1.8$
Is my process is correct, If not please help me How do i solve it, Thanks
If she only buys one item, there are two possibilities: $A$ or $B$.
$$P(B|A \cup B) = \frac{P(B)}{P(A) + P(B)}$$
The case of purchasing two objects is irrelevant.
And you should have seen that your reasoning was wrong once you computed a probability of $1.8$.