Question

An urn contains five balls. Two balls are drawn and are found to be white. The probability that all the balls are white is \[\dfrac{1}{k}\] ​. Find the value of k.

Answer 1

Hint: Formula of Bayes theorem :
\[P\left( {A|B} \right){\rm{ }} = \dfrac{{\;P\left( A \right){\rm{ }}P(B|A)}}{{P\left( B \right)}}\]
how often A happens given that B happens, written P(A|B)
how often B happens given that A happens, written P(B|A)
and how likely A is on its own, written P(A)
and how likely B is on its own, written P(B)

Complete step-by-step answer:
Let \[{A_i}\left( {i = 1,2,3,4} \right)\;\] be the event that the urn contains 2,3,4 or 5 white balls and the event that two white balls are drawn.
We have to find $P{A_4,B}$ .
Since the four events \[{A_1},{A_2},{A_3},{A_4}\]​ are equally likely,
we have \[P({A_1}) = P({A_2}) = P({A_3}) = P({A_4}) = \dfrac{1}{4}\] ​.
\[P(B/{A_1})\;\]is the probability of the event that the urn contains 2 white balls and both have been drawn.
Hence \[P(B/{A_1}) = \dfrac{{{}^2{C_2}}}{{{}^5{C_2}}} = \dfrac{1}{{10}}\] ​.
Similarly, \[P(B/{A_2}) = \dfrac{{{}^3{C_2}}}{{{}^5{C_2}}} = \dfrac{3}{{10}}\] ​.
\[P(B/{A_3}) = \dfrac{{{}^4{C_2}}}{{{}^5{C_2}}} = \dfrac{6}{{10}} = \dfrac{3}{5}\]​. and \[P(B/{A_4}) = \dfrac{{{}^5{C_2}}}{{{}^5{C_2}}} = 1\].
  ∴ By Baye's theorem
\[P({A_4}/B) = {\rm{ }}\dfrac{{P({A_4})P(B/{A_4})}}{{\sum\limits_{i = 1}^4 {P({A_1})P(B/{A_i})} }}\]
\( = \dfrac{{\dfrac{1}{4}.1}}{{\dfrac{1}{4}\left( {\dfrac{1}{{10}} + \dfrac{3}{{10}} + \dfrac{3}{5} + 1} \right)}} = \dfrac{1}{2}\)

Note: It can be solved by another process.
We have a bag containing 3 balls, which previously held 5 balls before two were drawn from the bag and found to be white balls. The question is what is now our probability that the bag initially held 5 white balls? The only information we have is that each ball is either white or not white, and we assume that we can make that determination unambiguously after the ball is revealed. We also assume that the balls in the bag had equal probability of being selected from the bag at each stage of drawing.
If we knew the proportion of white balls in the batch of balls from which the 5 in the bag were drawn this would be a straight forward application of the Binomial distribution, but we have not such information. But, with the information we do have after drawing two balls we can estimate conditional probabilities for each possible starting condition and apply those to determine the probability of five white balls originally in the bag, given that two white balls have been drawn.
Having drawn two white balls the range of possibilities for the balls remaining in the bag is [0, 1, 2, 3], four possibilities for three balls. The probability requested is the one with 5 balls originally in the bag which corresponds to the case of 3 balls remaining. Let’s concentrate on the balls remaining to see what we can infer about the original number of white balls in the bag.
The probability of drawing two white balls given 2 to start is (2/5)x(1/4) = 0.1
The probability of drawing two white balls given 3 to start is (3/5)x(2/4) = 0.3
The probability of drawing two white balls given 4 to start is (4/5)x(3/4) = 0.6
The probability of drawing two white balls given 5 to start is (5/5)x(4/4) = 1.0
These probabilities indicate the relative likelihood that the bag contained the given number of white balls before any were drawn. Since the set of starting conditions is mutually exclusive and collectively exhaustive, a probability distribution can be generated by scaling by the likelihood ratios. Since these results sum to 2.0, scaling by ½ will give a probability mass distribution for the number of balls originally in the bag that were white. The case of interest has a probability of 0.5.