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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?

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  1. anonymous
    • one year ago
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    @welshfella what about ii? I have a hunch that it is wrong.

  2. welshfella
    • one year ago
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    Oh Hold on! I can't have woke up this morning yet!

  3. welshfella
    • one year ago
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    Forget what I've said already

  4. welshfella
    • one year ago
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    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
    • one year ago
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    I guess it must mean that

  6. welshfella
    • one year ago
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    Then probability is 1/15 + 1/6 + 1/10 = 1/3 Yes

  7. anonymous
    • one year ago
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    I know i is likely to be right too, just ii I'm not sure @welshfella

  8. welshfella
    • one year ago
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    - that last one was for part (i)

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

  10. welshfella
    • one year ago
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    ??

  11. anonymous
    • one year ago
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    @welshfella our teacher did teach us but I'm confused about it.

  12. welshfella
    • one year ago
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    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
    • one year ago
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    where P(B3|R) means the probability of B3 being chosen given that a red is drawn

  14. welshfella
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    is that correct @mathmate? My statistics sis pretty suspect !!

  17. mathmate
    • one year ago
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    Part I is correct. I am looking at part II.

  18. mathmate
    • one year ago
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    Gtg, will be back later.

  19. anonymous
    • one year ago
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    Thank you @mathmate , look forward to your response!

  20. mathmate
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Hmm wait @mathmate wouldn't P(B3) be 1/3? you put 2/3

  23. anonymous
    • one year ago
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    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
    • one year ago
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    @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
    • one year ago
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    @mathmate Hmm Thanks, now I got to figure out how I get 1/10...

  26. mathmate
    • one year ago
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    I think you forgot to divide by P(R), which is 1/3.

  27. anonymous
    • one year ago
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    3/10 * 1/3 = 1/10 not 3/10 the top part of ur equation @mathmate

  28. anonymous
    • one year ago
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    Hmm Guess I'm wrong. Sorry @mathmate

  29. mathmate
    • one year ago
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    @snowpolar Yeah, there was a 1/3 lurking both top and bottom, so it could be confusing! lol

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