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Question
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
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Solution
Two tickets are drawn from 11 tickets numbered 1 to 11 with replacement
∴ n(S) = 11C1 x 11C1 = 121
Let A be the event that the sum of two numbers is even.
The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn.
There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11.
∴ n(A) = 6 × 6 + 5 × 5 = 36 + 25 = 61
∴ P(A) = `("n"("A"))/("n"("S")) = 61/121`
Let B be the event that the numbers tickets drawn are odd
∴ n(B) = 6 × 6 = 36
∴ P(B) = `("n"("B"))/("n"("S")) = 36/121`
Since 6 odd numbers are common between A and B
∴ n(A ∩ B) = 6 × 6 = 36
∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 36/121`
∴ Probability of both the numbers are odd, given that sum is even, is given by
`"P"("B"/"A") = ("P"("A" ∩ "B"))/("P"("A")`
= `(36/121)/(61/121)`
= `36/61`
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