I roll a single fair die with 6 faces. What is the probability that I roll less than or equal to 3 given that I roll an even number?
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$\begingroup$ $$P(\text{roll $\le 3$} \mid \text{roll even}) = \frac{P(\text{roll $\le 3$ and even})}{P(\text{roll even})}$$ $\endgroup$– angryavianCommented May 7, 2017 at 1:23
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$\begingroup$ @angryavian, so numerator is : (1/6+1/6+1/6) = (3/6) ? And denominator also : (3/6) ? $\endgroup$– e4e5Commented May 7, 2017 at 1:25
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$$P(\text{roll $\leq 3$ } | \text{ roll even}) = \frac{P(\text{roll $\leq 3$} \cap \text{roll even})}{P(\text{roll even})} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$
answered May 7, 2017 at 1:24
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$\begingroup$ okay, so numerator is : 1/6 , because only even number 2 is occur , right? $\endgroup$– e4e5Commented May 7, 2017 at 1:29
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$\begingroup$ Yes, if a roll is $\leq 3$ and even, then it's a $2$, which occurs with probability $1/6$ $\endgroup$ Commented May 7, 2017 at 1:31