Chances of choosing 5 right numbers out of 6 out of 41

Question

I have been stuck on this probability problem for almost one entire day. Here it is: There is a game, a form of lottery lets suppose, where you must choose the correct series of 6 numbers out of 41 possible numbers. To find the probability of effectively winning, I did:

$$\frac{41!}{(41-6)!\cdot6!}$$

which gave me the total combination s -numbers cannot be repeated, and the order is not important- of series of 6 numbers that can be made using 41 numbers: 4496388 . Our chance of winning, then, is 1 out of 4496388 . That is right answer, according to my teacher.

Now, I'm asked which are the chanches of choosing 5 correct numbers out of the 6. I tried just replacing 6 with 5, but that was not the correct answer. I really cannot see which way I must take to solve this, could someone give me at least a tip, so I can try to tackle it by myself?

Lots of thanks.

Answer 1

You are looking for the hypergeometric distribution. There are $\binom{41}{6}$ possibilities to choose $6$ numbers out of $41$. There are $\binom{6}{5}$ possibilities to choose $5$ right numbers out of $6$ numbers and there are $\binom{35}{1}$ possibilities to choose $1$ wrong number out of $35$ wrong numbers, thus there are $\binom{6}{5}\cdot \binom{35}{1}$ possibilities to choose $5$ out of $6$ out of $41$. The probabiliy for that is $$\frac{\binom{6}{5}\binom{35}{1}}{\binom{41}{6}} $$

Answer 2

Hint: Call a number "good" if it is one of the six numbers that need to be selected to win the lottery. Call a number "bad" if it is one of the other $41 - 6 = 35$ numbers that could have been selected by the player. Since the player selects six numbers, to correctly select exactly $5$ of the $6$ "good" numbers, a player must select five of the six "good" numbers and one of the $35$ "bad" numbers.