Problem with my ball bag...
Problem with my ball bag...
Author
Discussion

Burty88

Original Poster:

140 posts

238 months

Tuesday 12th January 2010
quotequote all
...Can any mathsy people out there just confirm I got a question right on a stats paper I just took?

A bag contains 5 white balls, 2 blue balls and 5 red balls, the balls are taken from the bag in succession without being replaced.

Q. What is the probability that the second ball is red?

I got (5/12 * 4/11) + (7/12 * 5/11) = 5/12

Thanks in advance,

RemaL

25,102 posts

264 months

Tuesday 12th January 2010
quotequote all
not what I was expecting

Silverbullet767

11,311 posts

236 months

Tuesday 12th January 2010
quotequote all
This is not what I was expecting this thread to be about....

I think the answer is. Wednesday.

Project 644

37,069 posts

218 months

Tuesday 12th January 2010
quotequote all
A Blue Whale?

Gun

13,432 posts

248 months

Tuesday 12th January 2010
quotequote all
You really do have a problem with your ball bag if you've got all that in there.

zakelwe

4,449 posts

228 months

Tuesday 12th January 2010
quotequote all
Why does it mention 2 blue balls in the bag? Shirely that is irrelevant?

Andy

Edited by zakelwe on Tuesday 12th January 11:32

Kit80

4,764 posts

217 months

Tuesday 12th January 2010
quotequote all
I hope this get moves to health matters laugh

ccr32

1,983 posts

248 months

Tuesday 12th January 2010
quotequote all
you've got herpes.

tribbles

4,170 posts

252 months

Tuesday 12th January 2010
quotequote all
I think it's the probability that you have 2 red, or 1 non-red followed by a red:

(2/12 * 1/11) + (10/12 * 2/11)

= 2/132 + 20/132
= 22/132
= 2/12

= 1/6.

louiebaby

10,972 posts

221 months

Tuesday 12th January 2010
quotequote all
I get 15/44.

Which is .34 or 34%.

I look forward to the answer.

Stupidlikeafox

794 posts

208 months

Tuesday 12th January 2010
quotequote all
42.

robinhood21

31,141 posts

262 months

Tuesday 12th January 2010
quotequote all
Pain. Lots of it!

Burty88

Original Poster:

140 posts

238 months

Tuesday 12th January 2010
quotequote all
tribbles said:
I think it's the probability that you have 2 red, or 1 non-red followed by a red:

(2/12 * 1/11) + (10/12 * 2/11)

= 2/132 + 20/132
= 22/132
= 2/12

= 1/6.
But there's 5 red balls in my ball bag, so would it not be the same equation I got?:

(5/12 * 4/11) + (7/12 * 5/11)

Incorrigible

13,668 posts

291 months

Tuesday 12th January 2010
quotequote all
Burty88 said:
tribbles said:
I think it's the probability that you have 2 red, or 1 non-red followed by a red:

(2/12 * 1/11) + (10/12 * 2/11)

= 2/132 + 20/132
= 22/132
= 2/12

= 1/6.
But there's 5 red balls in my ball bag, so would it not be the same equation I got?:

(5/12 * 4/11) + (7/12 * 5/11)
No, there's either 5 red or 4 red after then first one's taken out

I think Tribbles is right

tribbles

4,170 posts

252 months

Tuesday 12th January 2010
quotequote all
Incorrigible said:
Burty88 said:
tribbles said:
I think it's the probability that you have 2 red, or 1 non-red followed by a red:

(2/12 * 1/11) + (10/12 * 2/11)

= 2/132 + 20/132
= 22/132
= 2/12

= 1/6.
But there's 5 red balls in my ball bag, so would it not be the same equation I got?:

(5/12 * 4/11) + (7/12 * 5/11)
No, there's either 5 red or 4 red after then first one's taken out

I think Tribbles is right
I'm wrong - I misread the colour ordering.

5 white, 2 blue, 5 red

(I read it as 5 white, 2 red, 5 blue).

ETA: Just put in the right numbers:

(5/12 * 4/11) + (7/12 * 5/11)

Which does indeed make 5/12.

Edited by tribbles on Tuesday 12th January 11:45

Devilstreak

8,088 posts

211 months

Tuesday 12th January 2010
quotequote all
God this takes me back, Not that I can remember any of it though frown

Burty88

Original Poster:

140 posts

238 months

Tuesday 12th January 2010
quotequote all
Incorrigible said:
Burty88 said:
tribbles said:
I think it's the probability that you have 2 red, or 1 non-red followed by a red:

(2/12 * 1/11) + (10/12 * 2/11)

= 2/132 + 20/132
= 22/132
= 2/12

= 1/6.
But there's 5 red balls in my ball bag, so would it not be the same equation I got?:

(5/12 * 4/11) + (7/12 * 5/11)
No, there's either 5 red or 4 red after then first one's taken out

I think Tribbles is right
I'm not disagreeing with his method, mine's the same. It's just that his equation relies on the assumption that there's two red balls and 10 non red balls when actually there's 5 red and 7 non red. After the first draw of ball there's either 5 red balls or 4 red balls left and 6 or 7 non reds respectively.

louiebaby

10,972 posts

221 months

Tuesday 12th January 2010
quotequote all
louiebaby said:
I get 15/44.

Which is .34 or 34%.

I look forward to the answer.
It's the number of red balls in the bag after one round, divided by 11.

Therefore there is a 5/12 chance of there being 4 red balls after the first turn, and a 7/12 chance of there being 5 red balls after the first turn. There are therefore 55/12 red balls in the bag after the first round. 55/12 /11 = 55/132 = 5/12.

My original answer was wrong, as I'm a wally.

V8mate

45,899 posts

219 months

Tuesday 12th January 2010
quotequote all
Green.


plasticpig

12,932 posts

255 months

Tuesday 12th January 2010
quotequote all
Not enough information to answer the question. Are the balls all the same size? Can the person removing the balls see the balls in the bag? If so what is the persons favourite color.....