Productive Probability

Cool notation: $(\frac23 N + \frac13 P)(\frac23 N + \frac13 P)$ = $\frac49 NN + \frac29 NP + \frac29 PN + \frac19 PP$

$\frac49 NN + \frac19 PP$ for positive product = $\frac49 + \frac19$ = $\frac59$

Answer: $\boxed{\displaystyle \frac59}$

The answer is not so accurate. In fact, in the interval [-20;10] we have 31 numbers. 10 of them positive, 20 of them negative and 0. So the probability of picking two positive numbers is 100/961 and the probability of picking two negative ones is 400/961. Finally, the answer should be 500/961... Please tell me where I get wrong!

We must either pick two positive numbers or two negative numbers. There is a $\frac{2}{3}$ chance of picking a negative and therefore a $\frac{2}{3}\frac{2}{3} = \frac{4}{9}$ chance of getting a positive. Then there is also a $\frac{1}{3}$ chance of picking a positive and therefore a $\frac{1}{3}\frac{1}{3} = \frac{1}{9}$ chance of getting a positive. Add the two and you get $\frac{5}{9}$ .