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December 27th, 2018, 02:49 AM   #1
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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.
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December 27th, 2018, 10:44 AM   #2
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$A=\{\text{1 white and 1 red ball were drawn}\}$
$B=\{\text{bag I was chosen}\}$

$P[A|B]P[B] = P[B|A]P[A]$

$P[B|A] = \dfrac{P[A|B]P[B]}{P[A]}$

$P[A|B] = \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} = \dfrac 1 5$

$P[A] =
\dfrac 1 3 \dfrac{\dbinom{1}{1}\dbinom{3}{1}}{\dbinom{6}{2}} +
\dfrac 1 3 \dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}} +
\dfrac 1 3 \dfrac{\dbinom{4}{1}\dbinom{3}{1}}{\dbinom{12}{2}} = \dfrac{118}{495}$

$P[B] = \dfrac 1 3$

$P[B|A] = \dfrac{\dfrac 1 5 \dfrac 1 3}{\dfrac{118}{495}} = \dfrac{33}{118}$
Thanks from EvanJ
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December 27th, 2018, 11:53 AM   #3
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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.
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December 27th, 2018, 01:05 PM   #4
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Quote:
Originally Posted by Light Yagami View Post
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.
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December 27th, 2018, 01:10 PM   #5
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Quote:
Originally Posted by romsek View Post
to be polite....

I believe you are mistaken.
Why so?
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December 27th, 2018, 01:23 PM   #6
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Quote:
Originally Posted by Light Yagami View Post
Why so?
because the solution is much more complicated, see post #2, than you seem to realize.
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December 27th, 2018, 01:55 PM   #7
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Quote:
Originally Posted by Light Yagami View Post
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.
Thanks from romsek and Joppy

Last edited by JeffM1; December 27th, 2018 at 01:58 PM.
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December 31st, 2018, 04:52 PM   #8
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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.
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December 31st, 2018, 06:41 PM   #9
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Quote:
Originally Posted by EvanJ View Post
.... 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?
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January 1st, 2019, 10:34 AM   #10
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Quote:
Originally Posted by EvanJ View Post
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?
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