Q.

Two integers are selected at random from integers 1 to 11. If the sum is even then the probability that both numbers are odd is

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a

611

b

35

c

25

d

45

answer is B.

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Detailed Solution

Sum even = both even or both odd n(A)= 5C2+6C2 n(A∩B) = 6C2 P(B/A) = n(A∩B)n(A)                    = 6C25C2+6C2                    = 1510+15 = 35 Required Probability is  35
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Two integers are selected at random from integers 1 to 11. If the sum is even then the probability that both numbers are odd is