Probability and Statistics Basic Probability and Statistics Math Forum

 December 27th, 2018, 02:49 AM #1 Member   Joined: Apr 2017 From: India Posts: 71 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: 2,463 Thanks: 1340 $A=\{\text{1 white and 1 red ball were drawn}\}$ $B=\{\text{bag I was chosen}\}$ $P[A|B]P[B] = P[B|A]P[A]$ $P[B|A] = \dfrac{P[A|B]P[B]}{P[A]}$ $P[A|B] = \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} = \dfrac 1 5$ $P[A] = \dfrac 1 3 \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} + \dfrac 1 3 \dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}} + \dfrac 1 3 \dfrac{\dbinom{4}{1}\dbinom{3}{1}}{\dbinom{12}{2}} = \dfrac{118}{495}$ $P[B] = \dfrac 1 3$ $P[B|A] = \dfrac{\dfrac 1 5 \dfrac 1 3}{\dfrac{118}{495}} = \dfrac{33}{118}$ Thanks from EvanJ 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: 2,463
Thanks: 1340

Quote:
 Originally Posted by Light Yagami 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%
to be polite....

I believe you are mistaken. December 27th, 2018, 01:10 PM   #5
Newbie

Joined: Dec 2018
From: Japan

Posts: 16
Thanks: 1

Quote:
 Originally Posted by romsek to be polite.... I believe you are mistaken.
Why so? December 27th, 2018, 01:23 PM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,463
Thanks: 1340

Quote:
 Originally Posted by Light Yagami Why so?
because the solution is much more complicated, see post #2, than you seem to realize. December 27th, 2018, 01:55 PM   #7
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by Light Yagami 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%
This post is quite wrong.

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: 645 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: 551

Quote:
 Originally Posted by EvanJ .... You don't normally do the same thing to both sides of an equation in probability ....
Would you care to give some examples of this proposition? January 1st, 2019, 10:34 AM   #10
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,463
Thanks: 1340

Quote:
 Originally Posted by EvanJ 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.
Thank you very much but I confess I don't know what you are referring to.

What did I do to both sides of an equation? Tags bayes, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post amphinomos Probability and Statistics 2 January 20th, 2015 11:43 PM Plus Advanced Statistics 5 July 15th, 2013 02:34 PM FreaKariDunk Algebra 2 September 21st, 2012 08:23 AM Arkan Advanced Statistics 4 September 20th, 2012 01:19 PM parmar Advanced Statistics 1 November 16th, 2010 10:13 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      