# Probability and Statistics

• June 1st 2008, 04:37 PM
gwen01
Probability and Statistics
Here's my problem,

A certain Lottery has 49 numbers, six of which are the winning numbers for a particular game. To play the game each participant choses six numbers. What is the probability of choosing exactly...

a) six correct numbers

b) five correct numbers

c) four correct numbers

now i know how to find the answer to a) which is 49C6 which equals 13,983,816...so the answer is 1/13,983,816. But what is the formula for finding b. and c....

i've tried 6C5/49C6 but that's not right, i don't need the answers but does someone know the formula for figuring these out? this would be a huge help. Thanks!(Rock)
• June 1st 2008, 05:06 PM
TheEmptySet
Quote:

Originally Posted by gwen01
Here's my problem,

A certain Lottery has 49 numbers, six of which are the winning numbers for a particular game. To play the game each participant choses six numbers. What is the probability of choosing exactly...

a) six correct numbers

b) five correct numbers

c) four correct numbers

now i know how to find the answer to a) which is 49C6 which equals 13,983,816...so the answer is 1/13,983,816. But what is the formula for finding b. and c....

i've tried 6C5/49C6 but that's not right, i don't need the answers but does someone know the formula for figuring these out? this would be a huge help. Thanks!(Rock)

This should work for both b and c. I will do b

$\frac{\binom{5}{5} \binom{44}{1}}{\binom{49}{6}}=\frac{1}{317814}$

There are 44 arrangments of picking 5 correct numbers.