Probability that a 2 was rolled, given an even number has been rolled
By James Williams
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.
$\endgroup$1 Answer
$\begingroup$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.
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