Probability that a 2 was rolled, given an even number has been rolled I've been learning conditional probability and encountered the problem statement given in the title. We roll the die only a single time:
My method of solving the question: A: Event that the die rolled a 2
B: Event that the die rolled an even number
P(A) = 1/6
P(B) = 1/2
Then we have P(A|B) = P(A,B)/P(B) = (1/6)*(1/2)/(1/6) = 1/2. In some places, the answer was given as 1/3. Is this method correct? Also how can we find P(A,B) when we only have a single throw of a die.
A: Had you not been provided with the knowledge that the die rolled an even number i.e. in case of simple probability, the probability of rolling 2 on single throw of die would have been $1/6$ as there are $6$ possible outcomes. But in this case of conditional probability, there are only three possible outcomes viz. $2,4,6$ once you are told that the die rolled an even number. This makes the required probability $1/3$.
Also the formula for conditional probability $ P(A/B)=P(A\cap B)/P(B)$ gives $P(A/B)= (1/6)/(1/2)=1/3$ as the probability of rolling an even number is $1/2$.
My earlier answer had $P(A)$ in the numerator which was erroneous although that yielded the correct answer as $P(B)$ and $P(A\cap B)$ were same.
