The number a is selected randomly from Set A; the number b is selected randomly from Set B.
Therefore, total number of outcomes = \(5_C_1\) * \(4_C_1\) = 20. The denominator of the required probability will be 20.
This helps us eliminate answer options B and D since the number 20 can never be simplified to yield 3 or 9 since neither of these are factors of 20.
To find out the probability that the product of ab is positive, we can use the strategy,
Probability (ab > 0) = 1 – Probability (ab ≤ 0)
Probability (ab ≤ 0) = Probability (ab < 0) + Probability (ab = 0)
The product ab can only be made zero by making a = 0 and multiplying it with any of the four numbers in Set B.
Therefore, favourable outcomes for (ab = 0) = 4
The product ab can be made negative by making one of the numbers positive and the other number negative.
a can be – 3 or -8 and b can be 2 or 4; therefore, a total of 4 cases where the product ab is negative.
a can be 1 or 6 and b can be -1 or -4; therefore, a total of 4 cases where the product ab is negative.
Therefore, favourable outcomes for (ab < 0) = 8
Hence, Probability (ab ≤ 0) =\(\frac{ 8 }{ 20}\) + \(\frac{4 }{ 20}\) = \(\frac{12 }{ 20}\) = \(\frac{3 }{ 5}\) and so,
Probability (ab > 0) = 1 – \(\frac{3 }{ 5}\) = \(\frac{2 }{ 5}\).
The correct answer option is C. _________________
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