Probability 3349

We have natural numbers 3, 4, 6, 10, and 12. Calculate the probability that the sum of three randomly selected three different numbers is less than 20.

Correct answer:

p =  0.4

Step-by-step explanation:

$$\begin{gathered} 3+4+6<20 \\ 3+4+10<20 \\ 3+6+10<20 \\ 3+4+12<20 \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ p=4/\left(\genfrac{}{}{0ex}{}{5}{3}\right)=4/10=0.4 \end{gathered}$$

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