Question

A bag contains $4$ balls. Two balls drawn at random without replacement and are found to be white. What is the probability that all balls are white?

Answer 1

In order to solve this type of questions first we have to look on cases carefully and find the probability of the events that can get by $= \dfrac{{{\text{Number of favourable outcome}}}}{{{\text{Total number of favourable outcome}}}}$. There may be Lots of possibilities of occurring. So we will see it carefully.

Complete step-by-step answer:
Given that bag contains $4$ balls. Two balls drawn at random without replacement and are found to be white.

Let we assume that A is the event of drawing two white balls from the bag which contains $4$ balls.

For the remaining $2$ balls has three options that is:
(A) Two balls are not white
(B) Among two remaining balls one is white while the other one is not a white ball.
(C) Both are white balls.

So let we take
${E_1}$event in the case when both remaining balls are not white.
${E_2}$event when one is white and the other one is not a white ball.
${E_3}$event when both remaining balls are white.

Since here if we take 1 ball from bag then probability of ${E_1}$event $P({E_1}) = \dfrac{1}{3}$
Likewise $P({E_2}) = \dfrac{1}{3}$ and $P({E_3}) = \dfrac{1}{3}$
So we can write it as $P({E_1}) = P({E_2}) = P({E_3}) = \dfrac{1}{3}$

Probability of drawing $2$ white balls from the bag contains $2$ white and $2$ not white = $P\left( {\dfrac{A}{{{E_1}}}} \right)$
Probability of drawing $2$ white balls from the bag containing $2$ white and $1$ non-white = $P\left( {\dfrac{A}{{{E_2}}}} \right)$
Probability of drawing $2$ white ball from bag containing $4$ white balls = $P\left( {\dfrac{A}{{{E_3}}}} \right)$

By calculating $P\left( {\dfrac{A}{{{E_1}}}} \right)$ we get
$= \dfrac{{{}^2{C_2}}}{{{}^4{C_2}}} \\\ (\because {}^n{C_n} = 1) \\\ {\text{So it would be }} = \dfrac{1}{{{}^4{C_2}}} \\\ \left( {\because {}^4{C_2} = \dfrac{{4!}}{{2!2!}} = \dfrac{{4 \times 3}}{2} = 6} \right) \\\ \Rightarrow \dfrac{1}{6} \\\$
By calculating $P\left( {\dfrac{A}{{{E_2}}}} \right)$ we get
$= \dfrac{{{}^3{C_2}}}{{{}^4{C_2}}} \\\ (\because {}^n{C_{n - 1}} = n) \\\ = \dfrac{3}{6} = \dfrac{1}{2} \\\$

Similarly for $P\left( {\dfrac{A}{{{E_3}}}} \right)$ we get
$= \dfrac{{{}^4{C_2}}}{{{}^4{C_2}}} \\\ = 1 \\\$

But in the question we have asked that the probability that all balls are white.
So we have to find $P\left( {\dfrac{{{E_2}}}{A}} \right)$ that is given as $\because P\left( {\dfrac{{{E_2}}}{A}} \right) = \dfrac{{P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right)}}{{P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right) + P({E_3})P\left( {\dfrac{A}{{{E_3}}}} \right)}}$

$\because P\left( {\dfrac{{{E_2}}}{A}} \right) = \dfrac{{P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right)}}{{P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right) + P({E_3})P\left( {\dfrac{A}{{{E_3}}}} \right)}}$

So by putting all values in the formula we get
$= \dfrac{{\dfrac{1}{3} \times \dfrac{1}{2}}}{{\dfrac{1}{3}\left( {\dfrac{1}{6} + \dfrac{1}{2} + 1} \right)}} \\\ = \dfrac{{\dfrac{1}{3} \times \dfrac{1}{2}}}{{\dfrac{1}{3}\left( {\dfrac{{1 + 3 + 6}}{6}} \right)}} \\\ = \dfrac{{\dfrac{1}{3} \times \dfrac{1}{2}}}{{\dfrac{1}{3}\left( {\dfrac{5}{3}} \right)}} \\\ = \dfrac{{\dfrac{1}{6}}}{{\left( {\dfrac{5}{9}} \right)}} \\\ = \dfrac{1}{6} \times \dfrac{9}{5} \\\ = \dfrac{3}{{10}} \\\$
So the probability of all balls being white is $\dfrac{3}{{10}}$.

Note: Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely it is that a proposition is true. The probability of an event is a number between $0$ and $1$, where, roughly speaking, $0$ indicates impossibility of the event and $1$ indicates certainty.