Question

There are 5 positive numbers and 6 negative numbers. Three numbers are chosen at random and multiplied. The probability that the product is a negative number is:

• A. $\frac{11}{34}$
• B. $\frac{17}{33}$
• C. $\frac{16}{35}$
• D. $\frac{15}{34}$
• E. $\frac{16}{33}$

Hint:

The sign of a product depends solely on the number of negative factors. An odd number of negative factors (1 or 3) results in a negative product, while an even number (0 or 2) results in a positive product.

Answer 1

Answer: (E)

Concept: The product of three numbers is negative if:
• All three numbers are negative ($(- \times - \times -) = -$).
• One number is negative and two are positive ($(- \times + \times +) = -$).

Step 1: Calculate the total number of ways to choose 3 numbers.
Total numbers $= 5 + 6 = 11$. Total outcomes $= \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 11 \times 5 \times 3 = 165$.

Step 2: Count the favorable cases.
Case 1: 3 Negatives \[ \text{Ways} = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] Case 2: 1 Negative and 2 Positives \[ \text{Ways} = \binom{6}{1} \times \binom{5}{2} = 6 \times 10 = 60 \] Total favorable outcomes $= 20 + 60 = 80$.

Step 3: Calculate the probability.
\[ P(\text{negative product}) = \frac{80}{165} \] Divide both numerator and denominator by 5: \[ P(\text{negative product}) = \frac{16}{33} \]