Question

The CASH 6 lottery in Xanadu lets you pick six numbers in {1 . . 49}. How many ways are there for a player to get 4 or more number correctly.

Answer 1

To win one need to get all six numbers correct. Do you agree with this?

Answer 2

@nicole1234

Answer 3

i suppose..but the question asks how many ways to get 4 correct...

Answer 4

Yeah we want to get 4 or more
So either we get 4 or 5 or 6 correct

Answer 5

Do you know what's
\[^aC_b?\]

Answer 6

oh, that's right sorry. misread the problem

Answer 7

this would be a combination with replacement problem, right?

Answer 8

i'm not sure of the notation you used?

Answer 9

\[^aC_b=\frac{a!}{b!(a-b)!}\]
No of ways one can choose b items from a items

Answer 10

Do you know this?
It's not a replacement problem?

Answer 11

ah, that's what i thought. Combination 'without' replacement is what i meant

Answer 12

thanks for your help!

Answer 13

Cool, Did you get it?
You can solve it, can't you?

Answer 14

i know how to find the number of ways to choose 6 items out of the 49, but what about the option of needing at least 4 correct?

Answer 15

\[Total\ no.\ of\ ways=Ways\ to\ get\ 4\ correct+\ to\ get\ 5\ correct\ +\ to\ get\ 6\ correct\]

Answer 16

Okay, so to double check, it would be the number of ways to choose 4 out of 49 plus the number of ways to choose 5 plus the number of ways to choose 6?

Answer 17

Yeah correct:D

Answer 18

No wait

Answer 19

Thanks so much! I tend to get confused with word problems :)

Answer 20

You have to choose like this, there are 6 correct and 43 incorrect
To get 4 correct= ( choose 4 out of 6) \(\times\) ( choose the other two from the 43 incorrect)
To get 5 correct= ( choose 5 out of 6) \(\times\) ( choose one from the 43 incorrect)
To get 6 correct= ( choose 6 out of 6) \(\times\) ( choose none from the 43 incorrect)

Answer 21

Did you understand?

Answer 22

i think so...i was just trying to work it out

Answer 23

great, let me know if you get stuck. I'll help you in figuring out this:)

Answer 24

so to find 4 correct I find (the number of ways to choose 4 out of 6)*(the number of ways to choose 2 out of 49) and then add that to the results of getting 5 correct and 6 correct?

Answer 25

Yeah:) Whenever you have or it's equivalent to +
\[or =>+\]

Answer 26

makes sense :)

Answer 27

@nicole1234 I forgot
\[\large Welcome\ to\ Open\ Study:)\]

Answer 28

thanks :)

Answer 29

i'm getting 14707... does that sound about right?

Answer 30

Let me check!!!

Answer 31

6c4=15
43C2=903
6c5=6
43c1=43
6c6=1
43c0=1
\[15*903+6*43+1*1\]
Did you get this?

Answer 32

yes..but i accidently multiplied 903 by 16 instead of 15...ill fix it

Answer 33

ok 13,804...

Answer 34

Cool:D

Answer 35

thank you so much!

Answer 36

Glad to help:D:D