Question:
There are $50$ students in a computer room. $36$ students work on tablets, and $20$ students work on laptops. There are $12$ students who work on none. A teacher chooses a student randomly. What is the probability of this student not working on laptop, given that she has worked on a tablet?
Solution:
n(only laptop) = $2$
n(only tablet) = $18$
n(both laptop and tablet) = $18$
n(neither) = $12$
P(not laptop given tablet has been used) = $\frac{P(\text{not laptop and tablet})}{P(\text{tablet})}$
P(not laptop) = $\frac{30}{50}$
P(tablet) = $\frac{36}{50}$
Therefore, P(not laptop given tablet has been used) should be $\frac{3}{5}$, but the answer in my textbook shows that it is $\frac{1}{2}$. How to do to get this answer?
Best Answer
The counting is correct. However, you need to write what these probabilities are. You have $\mathbb{P}[\text{not laptop} \wedge \text{tablet}] = \mathbb{P}[\text{only tablet}] = n(\text{only tablet}) / 50 = 18/50$.
Next, you need to find $\mathbb{P}[\text{tablet}]$ which is simply $n(\text{tablet})/50 = 36/50$.
Dividing both probabilities gives you $\mathbb{P}[\text{not laptop} \vert \text{tablet}] = 1/2$