Question

From a set containing four positive numbers and four negative numbers, four numbers are chosen at random and they are multiplied. The probability that the obtained product is positive is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{19}{35} \)
• D. \( \frac{23}{35} \)

Hint:

Remember that a positive product requires an even number of negative factors. Breaking it down systematically into distinct cases (0, 2, or 4 negative numbers) ensures you don't miss any valid combinations.

Answer 1

The Correct Option is C

Concept: The total number of ways to pick 4 numbers out of 8 is \( \binom{8}{4} \). For the product of 4 numbers to be strictly positive, the number of negative numbers chosen must be even (either 0, 2, or 4 negative numbers).

Step 1: Calculating the total number of combinations.
\[ \text{Total Sample Space } (N) = \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

Step 2: Counting favorable combinations based on sign distribution.

• Case 1: 0 negative numbers and 4 positive numbers. \[ \text{Ways} = \binom{4}{0} \times \binom{4}{4} = 1 \times 1 = 1 \]

• Case 2: 2 negative numbers and 2 positive numbers. \[ \text{Ways} = \binom{4}{2} \times \binom{4}{2} = 6 \times 6 = 36 \]

• Case 3: 4 negative numbers and 0 positive numbers. \[ \text{Ways} = \binom{4}{4} \times \binom{4}{0} = 1 \times 1 = 1 \]
Total favorable choices \( = 1 + 36 + 1 = 38 \).

Step 3: Finding the final probability.
\[ \text{Probability} = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{38}{70} = \frac{19}{35} \] This matches option (C) perfectly.