Probability Homework.. help!

[falconfighter]

I just can't seem to get how to do this problem...
a bag has 500 marbles in it. 495 are red, 5 are green. What is the prob. of drawing at least 1 green if you draw 50 marbles?
If you draw 75 marbles?
I have no idea where to go at this. Help!

 

[vtqanh]

If I remember how to do this problem correctly:

Probability of red is 495/500, probability of green is 5/500
P (Drawing at least 1 green) = 1 - P (Drawing no green) = 1 - P (Drawing all 50 red)
= 1 - (495/500)^50 * (5/500)^ 0 ~ 40%

The 75 case should be similar

 

[Modeps]

Originally posted by: vtqanh
Probability of red is 495/500, probability of green is 5/500
P (Drawing at least 1 green) = 1 - P (Drawing no green) = 1 - P (Drawing all 50 red)
= 1 - (495/500)^50 * (5/500)^ 0 ~ 40%

The 75 case should be similar

He's never gonna learn if you do it for him!

 

[Lonyo]

Use your calculator.
There's a function for this.
pdfbin or similar (probability something binomial blah)
Or you might want cdfbin or similar (cumulative blah binomial somesuch).
And then put in 3 numbers or 2 numbers or something.
And make the calculator work it out for you.

<-- is glad he got 95% on his Stats in January so he could forget it ALL.

 

[TheChort]

Originally posted by: vtqanh
If I remember how to do this problem correctly:

Probability of red is 495/500, probability of green is 5/500
P (Drawing at least 1 green) = 1 - P (Drawing no green) = 1 - P (Drawing all 50 red)
= 1 - (495/500)^50 * (5/500)^ 0 ~ 40%

The 75 case should be similar

you forgot the fact that the probability changes for each marble he draws

 

[Mrvile]

iirc this is pre-algebra seventh grade?

 

[falconfighter]

No.... I got how he did that. Thank you!

 

[falconfighter]

Originally posted by: Mrvile
iirc this is pre-algebra seventh grade?

No this is me forgetting pre-algebra in hs prob class.

 

[Atomicus]

(495 choose 49) times (5 choose 1)

that divided by 500 choose 50

for the 75 draw case, change numbers accordingly

 

[oog]

Originally posted by: vtqanh
If I remember how to do this problem correctly:

Probability of red is 495/500, probability of green is 5/500
P (Drawing at least 1 green) = 1 - P (Drawing no green) = 1 - P (Drawing all 50 red)
= 1 - (495/500)^50 * (5/500)^ 0 ~ 40%

The 75 case should be similar

that's not right, is it? if you're drawing 50 marbles, the probably of drawing each successive red changes a bit. it may start at 495/500, but the very next one will be 494/499. this shifting probability with each successive marble throws off your (495/500)^50 term. i'll leave it to the OP to determine what to do next.

 

[konichiwa]

Depends on whether you put the marbles back after you draw them or not...