Two numbers are drawn simultaneously from the set of integers from 1 to 12. If it is known that the sum of drawn two numbers is odd, then the probability that only one of the two numbers is a prime number, is
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Sum of two integers is odd \(\iff\) one is even and one is odd.
List prime numbers carefully (1 is not prime).
For conditional probability \(P(B|A) = n(A \cap B)/n(A)\).
Set S = \{1, ..., 12\}. Even: E=\{2,4,6,8,10,12\} (6 numbers). Odd: O=\{1,3,5,7,9,11\} (6 numbers).
Primes P=\{2,3,5,7,11\} (5 numbers). Non-primes NP=\{1,4,6,8,9,10,12\} (7 numbers).
Condition A: Sum is odd. This means one number is Even, one is Odd.
Number of ways for A: \(n(A) = \binom{6}{1}\binom{6}{1} = 6 \times 6 = 36\).
Event B: Only one of the two numbers is prime.
We need \(P(B|A) = n(A \cap B) / n(A)\).
\(A \cap B\): Sum is odd (one Even, one Odd) AND only one is prime.
Case 1: Even Prime (EP) and Odd Non-Prime (ONP).
EP = \{2\} (1 number). ONP = \{1, 9\} (2 numbers: Odd numbers that are not prime).
Pairs: (2,1), (2,9). Number of pairs = \(1 \times 2 = 2\). Sums are 3, 11 (odd).
Case 2: Odd Prime (OP) and Even Non-Prime (ENP).
OP = \{3, 5, 7, 11\} (4 numbers). ENP = \{4, 6, 8, 10, 12\} (5 numbers: Even numbers that are not prime).
Number of pairs = \(4 \times 5 = 20\). All sums are odd.
Total pairs for \(A \cap B\) = \(2 + 20 = 22\).
So, \(P(B|A) = \frac{22}{36} = \frac{11}{18}\).
\[ \boxed{\frac{11}{18}} \]
