Stats Problem Please Help!?

Question

In a state 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) 6,7,8,8

b) 7, 7, 8, 8

The answers are 0.0012, 0.0006. Please give a full explanation for how you calculated it.

Answer 1

So for part (a)... we need to look at how many POSSIBLE outcomes total. Lets call this Omega.

Omega is a set where each element in the set is a combination of 4 numbers 0-9. Using a little combinatorics, we can find the total number or possible combinations.

For example Omega = {(0,0,0,0); (0,0,0,1); (0,0,0,2); ... ; (4,7,3,6); ... ; (9,9,9,9)}

And so the number of possible combinations we can choose is 10*10*10*10 = 10^4 = 10000

Now what we need to look at is our state space. Lets call it V.

V happens to be a set where each element is a combination of the numbers we have chosen. For example

V = {(6,7,8,8); (6,8,7,8); (6,8,8,7); ... ; (8,7,6,8)}

And again, using a little combinatorics we can see that:

Choose one of the 4 numbers

Then we can only choose 3 numbers

Then 2 of the numbers

And finally we are down to the last number

But we need to note, since 8 occurs twice, so we divide out the double digit...

So the total number of possible outcomes here happens to be:

(4*3*2*1) / (2*1) = 24 / 2 = 12

And as the way probability goes... take your V set and divide it by Omega:

12/10000 = 0.0012

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For (b), we have 7 occuring twice AND 8 occuring twice.

So now when we find total number of ways V can produce combinations, we need to divide out not only the double occurance of 8, but also the double occurance of 7:

(4*3*2*1) / (2*1) *(2*1) = 24 / (2*2) = 6

And again the way prob goes:

6 / 10000 = 0.0006