Question

There are $6$ positive numbers and $8$ negative numbers. Three numbers are chosen from them 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. $\frac{44}{91}$

Answer 1

Answer: (D)

Total number of ways $=^{14}C_{3}$
If three numbers are multiplied and the product is a negative number then it should be either $3$ negative numbers or $2$ positive and $1$ negative number
Total favourbale number of ways $=^{8}C_{3}+^{6}C_{2}^{8}C_{1}$
Hence, required probability $=\frac{^{8} C_{3} +^{6} C_{2}^{8} C_{1}}{^{14} C_{3}}$
$=\frac{56 + 120}{364}$
$=\frac{44}{91}$