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प्रश्न
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.
उत्तर
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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