Two integers are selected at random.
The first choice has 11 options from the 11 integers, and the second choice has 10 options from the remaining 10 integers.
Let P(A) be the probability of choosing both numbers odd.
Let P(B) be the probability of choosing the numbers to yield an even number.
Sample space of
B = {(1,3),(1,5),(1,7),(1,9),(1,11),(3,5),(3,7),(3,9),(3,11),(5,7),(5,9),(5,11),(7,9),(7,11),(9,11), (2,4),(2,6),(2,8),(2,10),(4,6),(4,8),(4,10),(6,8),(6,10),(8,10)}
Let P(A ∩ B) be the probability of getting both odd numbers giving an even sum.
∴(A∩B) = {(1,3),(1,5),(1,7),(1,9),(1,11),(3,5),(3,7),(3,9),(3,11),(5,7),
(5,9),(5,11),(7,9),(7,11),(9,11),}
\(\therefore P(A \cap B)=\frac{15}{110}\)
The probability of getting both numbers odd given that sum is even:
