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 $=\frac{{}^8{C}_3{+}^6{C}_2^8{C}_1}{{}^{14}{C}_3}$
$=\frac{56+120}{364}$
$=\frac{44}{91}$