Question

Suppose that coloured balls are distributed in 3 indistinguishable boxes as follows:

Boxes: B1, B2, B3
Red 2, 4, 3
White 3, 1, 4
Blue 5, 3, 3

A box is selected at random from which a ball is drawn at random.

(i) Find the probability that the ball is red.

My Ans: 1/3

(ii) Given that the ball is red, what is the probability that box 3 was selected?

My Ans: 1/30

Are my answers correct?

Answer 1

@welshfella what about ii? I have a hunch that it is wrong.

Answer 2

Oh Hold on!
I can't have woke up this morning yet!

Answer 3

Forget what I've said already

Answer 4

I'm confused as to what the numbers against the colors mean
red 2 , 4 and 3 - does that mean there are 2 reds in box b1, 4 in box b2 and 3 in box b3?

Answer 5

I guess it must mean that

Answer 6

Then
probability is
1/15 + 1/6 + 1/10 = 1/3

Yes

Answer 7

I know i is likely to be right too, just ii I'm not sure @welshfella

Answer 8

- that last one was for part (i)

Answer 9

I think you apply Bayes Theorem to part (ii)
Now I've always had trouble with that... Have you studied Bayes theorem?

Answer 10

??

Answer 11

@welshfella our teacher did teach us but I'm confused about it.

Answer 12

yea I was confuse too - many years ago lol

in this instance , if its applicable the formula would be

P(B3|R) = P(R|B3) * P(B3)
-----------
P(R)

Answer 13

where P(B3|R) means the probability of B3 being chosen given that a red is drawn

Answer 14

But I'm not sure what these values are
well P(B3) is 1/3 but i'm not sure about the others

Answer 15

if P(R|B3) = 1/10 and P(R) = 1/3 ( we have found that out already)

then the required probability = 1/10 * 1/3
---------
1/3

= 1/10


but i'm not sure

Answer 16

is that correct @mathmate? My statistics sis pretty suspect !!

Answer 17

Part I is correct. I am looking at part II.

Answer 18

Gtg, will be back later.

Answer 19

Thank you @mathmate , look forward to your response!

Answer 20

Yeah, Baye's theorem is right.
That would make:
\(\Large P(B3|R)=\frac{P(R|B3)*P(B3)}{P(R|B1)*P(B1)+P(R|B2)*P(B2)+P(R|B3)*P(B3)}\)
\(\Large=\frac{\frac{2}{3}\frac{2}{3}}{\frac{1}{3}+\frac{1}{3}+\frac{2}{3}}\)
\(=\Large \frac{1}{2}\)

Answer 21

Thank you @mathmate so ii) is 1/2, I don't really get how bayes theorem differs from conditional probability, would you mind enligten me on that?

Answer 22

Hmm wait @mathmate wouldn't P(B3) be 1/3? you put 2/3

Answer 23

Hmm @mathmate @welshfella my new calculation is 1/10 My steps are below for ii)

P (B3|R) * P (R) = P (B3 and R) = P (R | B3) * P (B3)
P(B3|R) * 1/3 = 1/10 * 1/3
P(B3|R) = 1/10

I don't get why @mathmate answer is 1/2 though

I'm closing this question as I need to ask a new question but this question isn't resolved yet.

Answer 24

@snowpolar
My bad, I used the wrong data (from memory).
The real data is:
P(R|B1)=2/10
P(R|B2)=4/8
P(R|B3)=3/10
P(B1)=P(B2)=P(B3)=1/3
Bayes theorem has the advantage over straight definition of conditional probability, which is
P(B3|R)=P(B3 and R)/P(R)
We do not directly have information on P(R) so Baye's theorem comes to our rescue, which says
\(\Large P(B3|R)=\frac{P(R|B3)*P(B3)}{P(R|B1)*P(B1)+P(R|B2)*P(B2)+P(R|B3)*P(B3)}\)
all of which are known, so substituting
\(\Large =\frac{\frac{3}{10}\frac{1}{3}}{\frac{2}{10}\frac{1}{3}+\frac{4}{8}\frac{1}{3}+\frac{3}{10}\frac{1}{3}}\)
\(\Large =\frac{\frac{3}{10}}{\frac{2}{10}+\frac{4}{8}+\frac{3}{10}}\)
\(\Large =\frac{\frac{3}{10}}{\frac{8+20+12}{40}}\)
\(\Large =\frac{3}{10}\)

Alternatively, since P(R)=1/3 has been calculated in part A based on Baye's theorem, we can also simply write
P(B3|R)=P(B3 and R)/P(R)=P(R|B3)*P(B3)/P(R)=[3/10*(1/3) ] / (1/3) = 3/10
using P(R)==1/3 from part A.

Answer 25

@mathmate Hmm Thanks, now I got to figure out how I get 1/10...

Answer 26

I think you forgot to divide by P(R), which is 1/3.

Answer 27

3/10 * 1/3 = 1/10 not 3/10 the top part of ur equation @mathmate

Answer 28

Hmm Guess I'm wrong. Sorry @mathmate

Answer 29

@snowpolar
Yeah, there was a 1/3 lurking both top and bottom, so it could be confusing! lol