# Permutation of lottery?

## In a state of lottery four digits are drawn at random one at a time with replacement from 0 to 9. Suppose that you win if any permutation of your selected integers is drawn. Give the probability of winning if you select a.6,7,8,9 b.6,7,8,8 c.7,7,8,8 d.7,8,8,8

See below:

#### Explanation:

With a permutation, the order of the draw matters. Since we are looking at draws with replacement, each digit has a $\frac{1}{10}$ probability of being drawn. This means then for each of the selections, we have a:

1/10xx1/10xx1/10xx1/10=1/(10,000)=.01%

probability of our number being drawn.

If, however, the question is saying that with the four drawn numbers they can be rearranged into any permutation, then what we are really talking about is combinations (where order of the draw doesn't matter). These combinations are again done with replacement, and so we need to look at each case separately.

a

There is a $\frac{4}{10}$ probability of drawing the 6, 7, 8, or 9 on the first draw. Then a $\frac{3}{10}$ probability of drawing one of the remaining 3 numbers in the second draw. And so on. This gives:

4/10xx3/10xx2/10xx1/10=(4!)/10^4=24/(10,000)=.24%.

b

There is a $\frac{3}{10}$ probability of drawing either a 6,7, or 8 on the first draw:

$\frac{3}{10} \times \left(\ldots\right)$

If we drew an 8 on the first draw (and there is a 50% chance of doing so), then the second, third and fourth draws will be at probabilities of $\frac{3}{10} , \frac{2}{10} \mathmr{and} \frac{1}{10}$.

However, the other 50% of the time we'll draw either the 6 or the 7. If we do so, we then have to look to look a little further for our calculation:

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{3}{10} \times \frac{2}{10} \times \frac{1}{10}\right) + \frac{1}{2} \left(\ldots\right)\right)$

With the second draw (after drawing either a 6 or a 7), we can draw either an 8 (which will happen $\frac{2}{3}$ of the time) or the other non-8 number (which will happen the other $\frac{1}{3}$).

If we drew an 8, the third and fourth draws will be at probabilities at $\frac{2}{10} \mathmr{and} \frac{1}{10}$. However, if we drew the other non-8 number, we need to do a bit more work:

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{3}{10} \times \frac{2}{10} \times \frac{1}{10}\right) + \frac{1}{2} \times \left(\frac{2}{3} \times \left(\frac{2}{10} \times \frac{1}{10}\right) + \left(\frac{1}{3} \times \left(\ldots\right)\right)\right)\right)$

For the third and fourth draws and only 8s remaining, there is a $\frac{1}{10}$ probability of drawing that as a third and a fourth number:

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{3}{10} \times \frac{2}{10} \times \frac{1}{10}\right) + \frac{1}{2} \times \left(\frac{2}{3} \times \left(\frac{2}{10} \times \frac{1}{10}\right) + \left(\frac{1}{3} \times \left(\frac{1}{10} \times \frac{1}{10}\right)\right)\right)\right)$

Let's evaluate:

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{3}{10} \times \frac{2}{10} \times \frac{1}{10}\right) + \frac{1}{2} \times \left(\frac{2}{3} \times \left(\frac{2}{10} \times \frac{1}{10}\right) + \left(\frac{1}{3} \times \frac{1}{100}\right)\right)\right)$

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{3}{10} \times \frac{2}{10} \times \frac{1}{10}\right) + \frac{1}{2} \times \left(\frac{4}{300} + \frac{1}{300}\right)\right)$

$\frac{3}{10} \times \left(\frac{1}{2} \times \left(\frac{6}{1000}\right) + \frac{5}{600}\right)$

$\frac{3}{10} \times \left(\frac{6}{2000} + \frac{5}{600}\right)$

$\frac{3}{10} \times \left(\frac{18}{6000} + \frac{50}{6000}\right)$

3/10xx68/6000=68/20000=34/10000=.34%

c

There is a $\frac{2}{10}$ probability of drawing either a 7 or an 8:

$\frac{2}{10} \times \left(\ldots\right)$

If we drew a 7 (50% chance), then on the second draw if we draw an 8 ($\frac{2}{3}$ chance), the third and fourth draws will be at $\frac{2}{10} \mathmr{and} \frac{1}{10}$ probabilities. We have the same situation if we flip flop 7 for 8 and 8 for 7. And so:

$\frac{2}{10} \times \left(2 \times \frac{1}{2} \times \frac{2}{3} \times \frac{2}{10} \times \frac{1}{10} + \ldots\right)$

If we drew a 7 on both the first and second ($\frac{1}{3}$ chance) draws, we then can only draw 8s for the third and fourth draws. Again, this is true if we draw 8s on the first and second draws - we can only draw 7s for the third and fourth draws:

$\frac{2}{10} \times \left(2 \times \frac{1}{2} \times \frac{2}{3} \times \frac{2}{10} \times \frac{1}{10} + 2 \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{10} \times \frac{1}{10}\right)$

And evaluate:

2/10xx(4/300+1/300)=10/3000=0.bar3%

d

On the first draw, we can only draw a 7 or 8, with a probability of $\frac{2}{10}$:

$\frac{2}{10} \times \left(\ldots\right)$

If we drew a 7 (a $\frac{1}{4}$ chance), then we can only draw 8s for the second, third, and fourth draws.

If we drew an 8, we need to look further:

$\frac{2}{10} \times \left(\frac{1}{4} \times \frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} + \frac{3}{4} \times \ldots\right)$

On the second draw (after the first draw of an 8), we can draw either a 7 or 8.

If we drew a 7 ($\frac{1}{3}$ chance), the third and fourth draws have to be 8s.

If we drew an 8, the third and fourth draws will be at $\frac{2}{10} \mathmr{and} \frac{1}{10}$:

$\frac{2}{10} \times \left(\frac{1}{4} \times \frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} + \frac{3}{4} \times \left(\frac{1}{3} \times \frac{1}{10} \times \frac{1}{10} + \frac{2}{3} \times \frac{2}{10} \times \frac{1}{10}\right)\right)$

Let's evaluate:

$\frac{2}{10} \times \left(\frac{1}{4} \times \frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} + \frac{3}{4} \times \left(\frac{1}{300} + \frac{4}{300}\right)\right)$

$\frac{2}{10} \times \left(\frac{1}{4000} + \frac{5}{400}\right)$

2/10xx51/4000=51/20000=.255%