Two thousand people each purchase one ticket for a local lottery. In this lottery, the tickets are numbered from 0 to 1999. What is the probability that the winning number has atleast 3 1s.
There are 2000 people and each purchases a lottery ticket with numbers from 0000 to 1999. The question to be answered is what is the probability that the winning number contains at least 3 1s. All the other information provided does not influence the result.
For a number to have at least 3 1s, only the fourth digit can be a number other than 1. If this is the 1st digit, there are 9 possible numbers. Similarly for the digit at the second, third and fourth position also there are 9 possible numbers each. In all, there are 36 numbers. In addition 1111 also satisfies the given criterion. This makes a total of 37 eligible numbers of the total 2000 numbers.
The probability that one of these is the winning number is 37/2000.
The required probability that the winning number has at least 3 1s is 37/2000.