Question

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

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

Answer 1

The correct option is E $\frac{16}{33}$

Given, $5$ positive and $6$ negative numbers
$\therefore$ Total numbers are chosen at random and multiplied
$\therefore$ Favourable number of cases
$={{}^{5}C}_2.{{}^{6}C}_1+{{}^{6}C}_3$
$=10.6+20=60+20=80$
Total number of cases $=={{}^{11}C}_3$
$=\frac{11!}{3!8!}=\frac{11\times 10\times 9}{6}=165$
Hence, the required probability
$=\frac{Favourablecases}{Totalnumberofcases}=\frac{80}{165}=\frac{16}{33}$