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. 11/34
• B. 17/33
• C. 16/35
• D. 15/44
• E. 16/33

Answer 1

Answer: (E)

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