- #1

Phys12

- 351

- 42

**Problem statement:**In a lottery, players win a large prize when they pick four digits that match, in the correct order,four digits selected by a random mechanical process. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize?

**Solution:**The total number of choosing 4 digits are: 10 * 10 * 10 * 10 = 10,000. Ways of choosing three digits correctly out of 4 is the same choosing one incorrectly. Ways of choosing the first digit incorrectly are 9, same for second, third and fourth. You add the total number of ways to get 9 + 9 + 9 + 9 = 36 and then divide by the total number of ways of choosing 4 digits to get the probability: 36/10000

How would you solve this problem by not considering the different ways of getting the digit incorrect, but by considering the ways of getting three digits correct?

**My attempt:**If I think about getting the digits correct, then for getting the first three digits correctly, I have 9 ways of picking it since the 4th digit can be any digit except for the correct one, similarly for the other three digits and another and then one more time. Giving us again 9 + 9 + 9 + 9 = 36 ways. Is that a/the correct way of thinking about it?