# What is the probability of drawing a $13$ in the Lottery?

The Powerball lottery has the following rules. There are $$69$$ white balls (numbered from $$1$$ to $$69$$) and $$26$$ red balls (numbered from $$1$$ to $$26$$). At each drawing five white balls and a red ball are chosen uniformly at random. The result of the drawing is the five white numbers are listed in decreasing order and the red number.

Find the probability that the number $$13$$ shows up among the drawn six numbers.

So far I have that the following. There is a $$\frac{1}{69}$$ chance of drawing a $$13$$ numbered white ball in the first white ball being drawn. Then in the second there is a $$\frac{1}{68}$$ chance, assuming we did not draw it in the first one. And we proceed with the white balls similarly. Then for the red ball, we have a $$\frac{1}{26}$$ chance of drawing the ball numbered $$13$$.

Then the total probability of the number $$13$$ showing up is equal to $$\frac{1}{26}+\frac{1}{69}+\frac{1}{68}+\frac{1}{67}+\frac{1}{66}+\frac{1}{65} \approx 0.113.$$

Is the above reasoning correct? If not, could someone please point out the mistake. Thank you in advance.

Among the white balls the chance of getting $$13$$ is just $$\frac 5{69}$$ because you choose five balls of $$69$$. To do it the way you did, the chance of getting $$13$$ on the second draw is not $$\frac 1{68}$$, but $$\frac {68}{69}\cdot \frac 1{68}$$ because you have to not get it on the first draw to get it on the second. This is just $$\frac 1{69}$$ and each of the other balls has the same chance.
The chance of getting $$13$$ on the red ball is indeed $$\frac 1{26}$$, but the events of getting it on a white ball and red ball are not disjoint. You have counted the times you get it on both colors twice, so need to subtract it once.
• Taking the approach suggested first, would the following calculation of the probabilty in question be correct? We have $\frac{5}{69} \cdot \frac{25}{26}+ \frac{68}{69} \cdot \frac{1}{26} + \frac{5}{69} \cdot \frac{1}{26}.$ – G the Stackman Jan 31 '19 at 22:43
• Yes, that would be fine. You have three disjoint events, so you can add their probabilities. The calculation I described would be $\frac 5{69}+\frac 1{26}-\frac 5{69}\cdot \frac 1{26}$ because the chance of getting both is already included in each of the first two. Sometimes one approach will be easier, sometimes the other. – Ross Millikan Jan 31 '19 at 22:46