Question

There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is a negative number, is

• A. $\frac{496}{1001}$
• B. $\frac{505}{1001}$
• C. $\frac{490}{1001}$
• D. $\frac{504}{1001}$

Hint:

The product of numbers is negative if there is an odd number of negative factors. Break down the problem into mutually exclusive cases (e.g., 1 negative, 3 positive or 3 negative, 1 positive) and use combinations to count possibilities for each case.

Answer 1

The Correct Option is A

Step 1: Determine the total number of items and the total number of ways to choose 4 numbers. Total number of positive numbers $= 6$. Total number of negative numbers $= 8$. Total numbers available $= 6 + 8 = 14$. Total ways to choose 4 numbers from 14 is given by $\binom{14}{4}$. \[ \binom{14}{4} = \frac{14!}{4!(14-4)!} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 14 \times 13 \times \frac{1}{2} \times 11 = 7 \times 13 \times 11 = 1001 \]
Step 2: Determine the conditions for the product of four numbers to be negative. The product of four numbers is negative if and only if there is an odd number of negative numbers chosen. Case 1: 1 negative number and 3 positive numbers. Case 2: 3 negative numbers and 1 positive number.
Step 3: Calculate the number of ways for Case 1 (1 negative, 3 positive). Number of ways to choose 1 negative number from 8 negative numbers is $\binom{8}{1} = 8$. Number of ways to choose 3 positive numbers from 6 positive numbers is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. Number of ways for Case 1 $= \binom{8}{1} \times \binom{6}{3} = 8 \times 20 = 160$.
Step 4: Calculate the number of ways for Case 2 (3 negative, 1 positive). Number of ways to choose 3 negative numbers from 8 negative numbers is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. Number of ways to choose 1 positive number from 6 positive numbers is $\binom{6}{1} = 6$. Number of ways for Case 2 $= \binom{8}{3} \times \binom{6}{1} = 56 \times 6 = 336$.
Step 5: Calculate the total number of favorable outcomes. Total favorable outcomes = Number of ways for Case 1 + Number of ways for Case 2 Total favorable outcomes $= 160 + 336 = 496$.
Step 6: Calculate the probability. \[ \text{Probability} = \frac{\text{Total favorable outcomes{\text{Total number of ways = \frac{496}{1001} \]