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KEAM 2016 Math Paper

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Question : 85 of 120

Marks: +1, -0

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

11/34
17/33
16/35
15/44
16/33

Solution:

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\cdot6+20=60+20=80

Total number of cases

={}^{11}C_{3}

=\frac{11!}{3!\,8!}=\frac{11\times10\times9}{6}=165

Hence, the required probability

\frac{\text{Favourable number of cases}}{\text{Total number of cases}}=\frac{80}{165}=\frac{16}{33}

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