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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?
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella what about ii? I have a hunch that it is wrong.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Oh Hold on! I can't have woke up this morning yet!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Forget what I've said already

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1I'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?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1I guess it must mean that

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Then probability is 1/15 + 1/6 + 1/10 = 1/3 Yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know i is likely to be right too, just ii I'm not sure @welshfella

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1 that last one was for part (i)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1I think you apply Bayes Theorem to part (ii) Now I've always had trouble with that... Have you studied Bayes theorem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella our teacher did teach us but I'm confused about it.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yea I was confuse too  many years ago lol in this instance , if its applicable the formula would be P(B3R) = P(RB3) * P(B3)  P(R)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1where P(B3R) means the probability of B3 being chosen given that a red is drawn

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1But I'm not sure what these values are well P(B3) is 1/3 but i'm not sure about the others

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1if P(RB3) = 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

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1is that correct @mathmate? My statistics sis pretty suspect !!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Part I is correct. I am looking at part II.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Gtg, will be back later.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @mathmate , look forward to your response!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, Baye's theorem is right. That would make: \(\Large P(B3R)=\frac{P(RB3)*P(B3)}{P(RB1)*P(B1)+P(RB2)*P(B2)+P(RB3)*P(B3)}\) \(\Large=\frac{\frac{2}{3}\frac{2}{3}}{\frac{1}{3}+\frac{1}{3}+\frac{2}{3}}\) \(=\Large \frac{1}{2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm wait @mathmate wouldn't P(B3) be 1/3? you put 2/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm @mathmate @welshfella my new calculation is 1/10 My steps are below for ii) P (B3R) * P (R) = P (B3 and R) = P (R  B3) * P (B3) P(B3R) * 1/3 = 1/10 * 1/3 P(B3R) = 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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@snowpolar My bad, I used the wrong data (from memory). The real data is: P(RB1)=2/10 P(RB2)=4/8 P(RB3)=3/10 P(B1)=P(B2)=P(B3)=1/3 Bayes theorem has the advantage over straight definition of conditional probability, which is P(B3R)=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 \(\Large P(B3R)=\frac{P(RB3)*P(B3)}{P(RB1)*P(B1)+P(RB2)*P(B2)+P(RB3)*P(B3)}\) 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(B3R)=P(B3 and R)/P(R)=P(RB3)*P(B3)/P(R)=[3/10*(1/3) ] / (1/3) = 3/10 using P(R)==1/3 from part A.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate Hmm Thanks, now I got to figure out how I get 1/10...

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2I think you forgot to divide by P(R), which is 1/3.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03/10 * 1/3 = 1/10 not 3/10 the top part of ur equation @mathmate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm Guess I'm wrong. Sorry @mathmate

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@snowpolar Yeah, there was a 1/3 lurking both top and bottom, so it could be confusing! lol
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