In a lottery game run by a certain state, for every game ticket purchased, a player can pick any 4 numbers from 1 through 43. If the numbers match those drawn by the state, the player wins. If Martin buys 3 tickets, what is his probability of winning?

Question

anonymous · Answer

Answer Choices:
4/185,115
1/52,890
3/123,410
1/123,410

ranga · Answer

How many 4 digit numbers can be formed out of 43 numbers?
Total Possibilities = 43C4 = 43! / (39! * 4!) = ?      

If 3 tickets were purchased then the probability of winning is 3 / Total Possibilities

anonymous · Answer

Wow. Who would have thought that my Honors Algebra 2 teacher would mix the next unit in with the normal distribution unit? Could you tell me the standard form of that formula? Also how you created that 43 C 4 and if there is another thing like that and similar.

ranga · Answer

If there are n distinct objects and you pick r objects out of it then the total number of ways you can do it is nCr  (C is Combinations)$$\Large ^{n}C _{r} = \frac{ n! }{ (n-r)! r! }$$

anonymous · Answer

So in this case the answer is C. Wow, that seems a bit complex, yet easy.

ranga · Answer

In the pick-4 lottery, 4 numbers are picked from a total of 43 numbers.
n = 43, r = 4 
Plug that into the formula to find the total number of ways to pick 4 digit numbers from 43 different numbers.  

Yeah, C.

ranga · Answer

Total possibilities = 43! / (39! * 4!) = 43 x 42 x 41 x 40 / (4 x 3 x 2 x 1) = 123410

If 3 tickets are bought the probability of winning is = 3 / 123410

anonymous · Answer

Question, when you are writing it that way, you don't seem to be including the (n-r)!--r!--that part. Are you just doing the part in the parenthesis? @ranga