M = {6, 5, 4, 3, 2} T = { 2, 1, 0, 1, 2, 3, 4, . . .n}, n is a
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05 Jun 2024, 02:37
Since the product of the two integers should be negative, we need to select a positive integer from one set and a negative integer from the other set.
Since set M has all terms negative, we need to select the positive integer from set T.
Set T has (n + 3) terms, of which, n are positive.
Number of ways of selecting a negative integer from set \(M = C^5_1 = 5 \)
Number of ways of selecting a positive integer from set \(T =C^n_1 =n\)
Thus, number of favorable cases = 5 x n.
Total number of cases = (# of ways of selecting an integer from set M) x (# of ways of selecting an integer from set T)
\(= C^5_1 \times C^{n+3}_1 = 5 x (n + 3) \)
Thus, required probability
Number of favorable cases/Total number of cases
\(\frac{5n}{5(n+3)}\)
\(\frac{n}{n+3}\)
Thus, we have:
\(\frac{n}{n+3}>\frac{3}{5}\)
\(5n>3n+9\)
\(n>\frac{9}{2}=4.5\)
Among the options, the eligible values for n are 5, 7, 8, and 9.
The correct answers are options B, C, D and E.