At the Counterfeiters' Club...
The Counterfeiters' Club is having a lottery. There are two prizes. Every member of the Club submits exactly one entry, and these entries are stored in a box.
After this is done, Fred the Forger forges 6 extra entries with his name on them and sneaks them into the box. He is the only member of the Club to sneak in extra entries for himself.
The chances of Fred winning both prizes, and his forgery thus being discovered, is exactly 7%.
How many people are members of the Counterfeiters' Club?
The answer is 19.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Fred has submitted a total of 7 entries. Let there be x members of the Club other than Fred. Each of them has submitted exactly one entry, so there are x additional entries.
The chance of Fred winning both prizes is ( 7 / ( 7 + x ) ) ∗ ( 6 / ( 6 + x ) ) = 0 . 0 7
4 2 / ( ( 7 + x ) ( 6 + x ) ) = 0 . 0 7
1 / ( ( 7 + x ) ( 6 + x ) ) = 1 / 6 0 0
( 7 + x ) ( 6 + x ) = 6 0 0
x = 18
So there are 18 additional members of the Club, plus Fred makes a total of 1 9 members.