# Homework Help: Probability question

1. Feb 15, 2010

### Sirsh

Six numbers are randomly selected from 45 numbers. Once a number is selected it is not replaced. Determine the probability of correctly predicting:

a) exactly four of the six numbers

b) at least four of the six numbers.

a) 45C4/45C6 = 1.83% chance

b) 45C4*4 / 45C6 = 7.317%

if anyone would be able to help, thanks alot.

Last edited: Feb 15, 2010
2. Feb 15, 2010

### CompuChip

You can also approach the question like this: suppose I write down 6 numbers and ask you to perform the draw. What is the probability that you get precisely 4 of them correct, and two of them wrong?

First let me ask you if you are positive about your answer, because I got something (much) smaller. If not, can you explain how you were led to this?

3. Feb 15, 2010

Sirsh, your counting in the numerator of a isn't correct; in b, think about what the phrase "at least 4" means in terms of the number correct.

4. Feb 18, 2010

### Sirsh

Six numbers are randomly selected from 45 numbers. Once a number is selected it is not replaced. Determine the probability of correctly predicting:

a) exactly four of the six numbers

b) at least four of the six numbers.

a) 4C4/45C6

b) 4C4+4C3+4C2+4C1+4C0 / 45C6

i'm completely puzzled..

5. Feb 18, 2010

### CompuChip

OK, so suppose you have these six numbers. Then the question is: if you pick six number from a total of 45, what is the probability that exactly four of them match four of the chosen ones.

1) what is the probability that the first number picked is one of them?
2) same question for the second one, if you indeed picked one
3) now what is the probability that you first pick four of the six chosen numbers
4) what is the probability that the next two numbers are not the remaining two?

6. Feb 18, 2010

### Sirsh

1) what is the probability that the first number picked is one of them?

6C1 out of 45C6?

with the other one's im unsure, my school has supplied me with virtually no materials to learn probability with neither which i can find on the internet, sorry.