A Pair of Dice is thrown. Find the Probability That the number Appearing on atleast one die is 3

Question3. A pair of dice is thrown. it is given that the sum of numbers appearing on both dice is an even number. find the probability that the number appearing on at least one die is 3.
Solution:

$\mathrm{E} = 5 \mathrm{has} \mathrm{appeared} \mathrm{atleast} \mathrm{once}$
$\mathrm{E} = \left\{\right(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6\left)\right\}$
$\mathrm{F} = \mathrm{sum} \mathrm{of} \mathrm{number} \mathrm{on} \mathrm{die} \mathrm{is} 7$
$\mathrm{F} = \left\{\right(1,6),(2,5), (3,4), (4,3), (5,2), (6,1\left)\right\}$
$\mathrm{E}\cap \mathrm{F} = \left\{\right(2,5),(5,2\left)\right\}$
$\mathrm{n}(\mathrm{E}\cap \mathrm{F}) = 2$
$\mathrm{P}(\mathrm{E} | \mathrm{F}) = \frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)} = \frac{\mathrm{n}(\mathrm{E}\cap \mathrm{F})}{\mathrm{n}\left(\mathrm{F}\right)} = \frac{2}{6} = \frac{1}{3}$

Question 4A pair of dice is thrown. it is given that the sum of numbers appearing on both dice is an even number. Find the probability that the number appearing on at least one die is 3.

$\mathrm{E} = \mathrm{Sum} \mathrm{of} \mathrm{numbers} \mathrm{on} \mathrm{two} \mathrm{dice} \mathrm{is} \mathrm{even}$
$\mathrm{E} = \left\{\right(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4), (4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6\left)\right\}$
$\mathrm{F} = \mathrm{number} \mathrm{apearing} \mathrm{on} \mathrm{one} \mathrm{die} \mathrm{is} \mathrm{atleast} 3$
$\mathrm{F} = \left\{\right(1,3),(2,3), (3,3), (4,3), (5,3), (6,3), (3,1),(3,2),(3,4),(3,5),(3,6\left)\right\}$
$\mathrm{E}\cap \mathrm{F} = \left\{\right(1,3),(3,1),(3,3),(3,5\left)\right\}$
$\mathrm{n}(\mathrm{E}\cap \mathrm{F}) = 4$
$\mathrm{P}(\mathrm{E} | \mathrm{F}) = \frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)} = \frac{\mathrm{n}(\mathrm{E}\cap \mathrm{F})}{\mathrm{n}\left(\mathrm{F}\right)} = \frac{4}{18} = \frac{2}{9}$