I am confused with what numbers I should put on top of my probability fraction. I have three numbers: # of ways for someone to choose 7 #s from 80 = C(80,7), # of ways for the computer to choose 11 numbers from 80 = C(80,11), and # of ways someone could have chosen the 7 from the 11 = C(11, 7). I think I understand that the number C(11, 7) is the total number of lottery numbers that are winners. And that the total number of lottery number is C(80, 11), but what do I do with C(80,7)?
~James Gilmore
I'm not sure if this is right, but my answer was C(11,7)/C(80,7). -Zoe
So how come we don't even use the C(80,11)? It just seems like that number is important--Krwade 16:41, 17 February 2009 (UTC)
It does not with how many choices the computer has, what really matters is the number 11 I think, because we only pick 7 numbers out of 80.
I guess I still have the same question as Krwade. Where does the number C(80,11) come into play? It does seem like it will be an important number. CMD
Start off by examining a simpler problem, lets examine the three integers chosen from the first five. The probability to correctly choose all of these will be given by
1/C(5,3) = 1/10
since there will be 10 arrangements:
123 124 125 134 135 145 234 235 245 345
now, let's examine the probability we could correctly choose two numbers from a set of three:
1/C(3,2) = 1/3
giving us a final probability: 3/10
it is easy to see from the above values that there is a three in ten probability that any two numbers chosen randomly from this set of five will be in a winning set so:
P = C(3,2)/C(5,3)
I hope this helped. ---mturczi
I'm still confused on this problem. In the example above, why do you need to use the C(3,2)? Can someone explain this part a little better? --elhalsme
