Balls drawn without replacement probability question

leaning

Involved In Discussions
#1
Hello! Can someone check me? My answer is close (.373) but not exact so I'm thinking there's another way this should be done. It doesn't say "exactly one" "or at least one", just "one". I appreciate your help!

A random sample of 5 balls is drawn without replacement from a box that contains 27 black and 3 red balls. What is the probability that the sample contains one red ball?
0.108
0.270
0.369
0.500
C

N= 30 balls
A= 3 red balls
B= 26 black
n=5

C(3,1) x C (27,4)/C(30,5) = 3 x 17750/142506 = .373. Answer should be .369
 

leaning

Involved In Discussions
#3
Mark,
I think every post I make ends with either 1) a diagram I've pieced together with this group's help that I attach or 2) some other kind of bottom line answer. Is there something in particular you were looking for?

Regards,
John
 

Corca

Starting to get Involved
#4
Hello! Can someone check me? My answer is close (.373) but not exact so I'm thinking there's another way this should be done. It doesn't say "exactly one" "or at least one", just "one". I appreciate your help!

A random sample of 5 balls is drawn without replacement from a box that contains 27 black and 3 red balls. What is the probability that the sample contains one red ball?
0.108
0.270
0.369
0.500
C

N= 30 balls
A= 3 red balls
B= 26 black
n=5

C(3,1) x C (27,4)/C(30,5) = 3 x 17750/142506 = .373. Answer should be .369
Answer

P = C(27,4) * C(3,1) / C(30,5)

P = 17550 * 3 / 142506

P = 0.369

So your calculation was correct just your answer is wrong, redo your maths.
 

Marc

Captain Nice
Staff member
Admin
#5
I think every post I make ends with either 1) a diagram I've pieced together with this group's help that I attach or 2) some other kind of bottom line answer. Is there something in particular you were looking for?
Just asking. Since you're putting together a training plan and people here are helping you with a lot of these, I think it will be good if you post (share) what you are writing.
 

leaning

Involved In Discussions
#6
Corca: Yep, I re-ran it through the calculator again and got your answer this time. :) I don't know what happened there.

Here's the final:
Use the combination equation: n!/r!(n-r)!
where N=30, A=3 red balls, B=27 black balls, n=5:
[3!/1!(3-1)! x 27!/4!(27-4)!] / [30!/5!(30-5)!]
(3 x 17550)/ 142506 = 52650/142506 = .3694.

Marc: Look at all my posts. I've shared everything, and I've acknowledged everyone who assisted. I really don't know what else I can do or what you are asking for. If there is some in particular you don't think I've been forthcoming about, please let me know. (???)

Regards,
leaning
 
Last edited:

leaning

Involved In Discussions
#7
Found it!

On the top post, I had entered 17750 instead of 17550.

If you do that, you get .373 vice .369.

HA! Cause of mistake identified: Me! :rolleyes:

Regards,
leaning
 

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