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