December 27th, 2018, 02:49 AM  #1 
Member Joined: Apr 2017 From: India Posts: 49 Thanks: 0  Bayes theorem
There is a question that came into Probability and Statistics exam related to Bayes theorem. Here it is: Three bags I, II and III contain 1 white, 2 black and 3 red balls; 2 white, 1 black, and 1 red balls; and 4 white, 5 black and 3 red balls, respectively. One bag is selected at random and two balls are drawn. These happen to be one white and one red. Find the probability that these balls were drawn from bag I. I am highly confused with this question. Please solve and show the process involved in solving this question. 
December 27th, 2018, 10:44 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281 
$A=\{\text{1 white and 1 red ball were drawn}\}$ $B=\{\text{bag I was chosen}\}$ $P[AB]P[B] = P[BA]P[A]$ $P[BA] = \dfrac{P[AB]P[B]}{P[A]}$ $P[AB] = \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} = \dfrac 1 5$ $P[A] = \dfrac 1 3 \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} + \dfrac 1 3 \dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}} + \dfrac 1 3 \dfrac{\dbinom{4}{1}\dbinom{3}{1}}{\dbinom{12}{2}} = \dfrac{118}{495}$ $P[B] = \dfrac 1 3$ $P[BA] = \dfrac{\dfrac 1 5 \dfrac 1 3}{\dfrac{118}{495}} = \dfrac{33}{118}$ 
December 27th, 2018, 11:53 AM  #3 
Newbie Joined: Dec 2018 From: Japan Posts: 16 Thanks: 1 
I think it's 6,67%, or 1/15 There are three bags, so the probability of I being chosen is 1/3 In bag I, there is one white ball and 3 red balls, so there are 3 possible combinations of one white and one red ball (see Fundamental Principle of Counting) The number of possible 2 ball combinations using all six balls inside bag 1 is C(6,2) = 15 So the answer to your question is ( 1/3 ) * ( 3 / 15 ) = 1/15 = 6,67% Last edited by Light Yagami; December 27th, 2018 at 11:58 AM. 
December 27th, 2018, 01:05 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281  Quote:
I believe you are mistaken.  
December 27th, 2018, 01:10 PM  #5 
Newbie Joined: Dec 2018 From: Japan Posts: 16 Thanks: 1  
December 27th, 2018, 01:23 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281  
December 27th, 2018, 01:55 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
The probability of picking bag k is $\dfrac{1}{3}.$ If bag I is picked, the number of ways to pick 2 balls is $\dbinom{6}{2} = \dfrac{6!}{2! * (6  2)!} = \dfrac{6 * 5}{2} = 15.$ The number of ways to pick 1 white ball and 1 red ball is 3. So the probability of picking 1 white and 1 red ball given having picked bag I is $\dfrac{3}{15} = \dfrac{1}{5}.$ And therefore the probability of picking bag I and 1 white and 1 red ball is indeed $\dfrac{1}{3} * \dfrac{1}{5} = \dfrac{1}{15} \approx 6.67\%.$ But that is not the only possibility. If bag II is picked, the number of ways to pick 2 balls is 6. The number of ways to pick 1 white and 1 red is 2. So the probability of picking 1 white and 1 red ball given that bag II was picked is $\dfrac{2}{6} = \dfrac{1}{3}.$ And the probability of picking bag II and 1 white and 1 red ball is $\dfrac{1}{3} * \dfrac{1}{3} = \dfrac{1}{9}.$ If bag III is picked, the number of ways to pick 2 balls is 66. The number of ways to pick 1 white and 1 red ball is 12. So the probability of picking 1 white and 1 red ball given that bag III was picked is $\dfrac{12}{66} = \dfrac{2}{11}.$ And the probability of picking bag III and 1 white and 1 red ball is $\dfrac{1}{3} * \dfrac{2}{11} = \dfrac{2}{33}.$ But, under the conditions contemplated, picking bag I or II or III are mutually exclusive and exhaustive. Therefore, the probability of picking 1 white and 1 red ball is $\dfrac{1}{15} + \dfrac{1}{9} + \dfrac{2}{33} = \dfrac{1}{3} * \left ( \dfrac{1}{5} + \dfrac{1}{3} + \dfrac{2}{11} \right ) =$ $\dfrac{1}{3} * \left (\dfrac{33}{165} + \dfrac{55}{165} + \dfrac{30}{165} \right ) = \dfrac{1}{3} * \dfrac{118}{165} = \dfrac{118}{3 * 5 * 33}.$ Now the probability of having picked bag I given that 1 white and 1 red ball were picked is, by definition, the probability of picking bag I and 1 white and 1 red ball divided by the probability of picking 1 white and 1 red ball. So we have $\dfrac{\dfrac{1}{15}}{\dfrac{118}{3 * 5 * 33}} = \dfrac{1}{15} * \dfrac{3 * 5 * 33}{118} = \dfrac{33}{118} \approx 27.97\%.$ In other words, romsek was, unsurprisingly, correct. Last edited by JeffM1; December 27th, 2018 at 01:58 PM.  
December 31st, 2018, 04:52 PM  #8 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 
I struggled with this type of question. romsek did a great job. You don't normally do the same thing to both sides of an equation in probability, but it works.

December 31st, 2018, 06:41 PM  #9 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  
January 1st, 2019, 10:34 AM  #10  
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281  Quote:
What did I do to both sides of an equation?  

Tags 
bayes, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Bayes' Theorem  amphinomos  Probability and Statistics  2  January 20th, 2015 11:43 PM 
Bayes’ theorem  Plus  Advanced Statistics  5  July 15th, 2013 02:34 PM 
Bayes theorem  FreaKariDunk  Algebra  2  September 21st, 2012 08:23 AM 
Bayes' theorem  Arkan  Advanced Statistics  4  September 20th, 2012 01:19 PM 
bayes theorem  parmar  Advanced Statistics  1  November 16th, 2010 10:13 AM 