## anonymous one year ago 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?

1. anonymous

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

2. welshfella

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

3. welshfella

4. welshfella

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?

5. welshfella

I guess it must mean that

6. welshfella

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

7. anonymous

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

8. welshfella

- that last one was for part (i)

9. welshfella

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

10. welshfella

??

11. anonymous

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

12. welshfella

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)

13. welshfella

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

14. welshfella

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

15. welshfella

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

16. welshfella

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

17. mathmate

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

18. mathmate

Gtg, will be back later.

19. anonymous

Thank you @mathmate , look forward to your response!

20. mathmate

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

21. anonymous

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?

22. anonymous

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

23. anonymous

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.

24. mathmate

@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.

25. anonymous

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

26. mathmate

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

27. anonymous

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

28. anonymous

Hmm Guess I'm wrong. Sorry @mathmate

29. mathmate

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