Question

For the purposes of this problem, assume that every person is either retired or working(not both) and is an adult(man or woman). Two thousand people ea purchase oneticket for a local lottery. In this lottery, the tickets are numbered from 0 to 1999 , and one winning number is randomly selected. Nine hundred females have purchased a ticket, and of those 900, exactly half of them are retired, and the other half are not. Also, 640 woking males have entered the lottery. Wat is the probability that the number called contains at least three 1s?

Answer 1

this feels like a trick question <.< all the info about women/men, retired whatever isnt needed. There are only a couple of tickets that have at least 3 ones.

0111
1_11 (0, 2, 3, 4, 5, 6, 7, 8, 9) can go in the blank
11_1 same numbers in the blank
111_ same
1111

Answer 2

The men and women, working or not have no bearing.
Put three 1's ,and fill in the fourth digit.
If the fourth digit is a 1, then there is only one way. 1111
If it is a zero, it can go in the first place. 0111
For the second, third and fourth places, the numbers can be 0,2,3,4...9 for 9x3 = 27 ways.
In total there are 29 ways.