Question

Find the probability distribution of the number of white balls drawn in a random draw of three balls without replacement, from a bag containing 4 white and 6 red balls.

Answer 1

Hint: First of all, we will consider the number of cases in which in how many ways the event can take place. Then from that we will consider each and every possibility and based on that we will find the expression of combination of balls been drawn out of total balls. And then solve that combination using the expression ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Then we will find the probability based on that answers and then will make final distribution table of probability and number of events.

Complete step by step solution:
In question it is given that three balls are drawn without replacement from a bag of 4 white and 6 red balls and we are asked to find the probability of number of distribution of balls being white balls. So, first of we will assume that a ball is drawn and it is white, then we will denote the number of white balls as B.
Now, it may happen that when we draw a ball randomly then no white ball is picked or only one ball is white or 2 balls are white or all the three balls are white. So, here we will consider four cases.
So, case 1 when a ball is picked randomly and ball picked is not white it can be given mathematically as,
$P\left( B=0 \right)=\dfrac{{}^{6}{{C}_{3}}}{{}^{10}{{C}_{3}}}$
${}^{6}{{C}_{3}}$ means 3 balls out of 6 red balls as, no ball was white so it means all the three balls were red and ${}^{10}{{C}_{3}}$ means 3 balls drawn out of total 10 balls i.e. $4\ \text{white}+6\ \text{red}=10\ \text{balls}$ total number of balls. Now, on solving the expression by using the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we will get,
$P\left( B=0 \right)=\dfrac{\dfrac{6!}{\left( 6-3 \right)!3!}}{\dfrac{10!}{\left( 10-3 \right)!3!}}$
$\Rightarrow P\left( B=0 \right)=\dfrac{\dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1}}{\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 3\times 2\times 1}}$ ($\because $as $\left( 6-3 \right)!=3!\ \text{and}\ \left( 10-3 \right)!=7!$)
On, cancelling out the common terms and simplifying the expression we will get,

$\Rightarrow P\left( B=0 \right)=\dfrac{\dfrac{6\times 5\times 4}{3\times 2\times 1}}{\dfrac{10\times 9\times 8}{3\times 2\times 1}}=\dfrac{6\times 5\times 4}{10\times 9\times 8}$
$\Rightarrow P\left( B=0 \right)=\dfrac{6\times 5\times 4}{10\times 9\times 8}=\dfrac{360}{720}=\dfrac{1}{2}$ ………………(i)
Now, in case 2, only one ball drawn is white, which means 1 ball out of 4 white balls and 2 balls out of 6 red balls are drawn. Here, the total number of balls remains same as there is no replacement so, the expression can of probability can be given as,
$P\left( B=1 \right)=\dfrac{{}^{6}{{C}_{2}}\times {}^{4}{{C}_{1}}}{{}^{10}{{C}_{3}}}$
Now, on solving the expression by using the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we will get,
$P\left( B=1 \right)=\dfrac{\dfrac{6!}{\left( 6-2 \right)!2!}\times \dfrac{4!}{\left( 4-1 \right)!1!}}{\dfrac{10!}{\left( 10-3 \right)!3!}}$
\[\Rightarrow P\left( B=1 \right)=\dfrac{\dfrac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2\times 1}\times \dfrac{4\times 3\times 2\times 1}{3\times 2\times 1\times 1}}{\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 3\times 2\times 1}}\] ($\because $as $\left( 6-2 \right)!=4!,\ \left( 4-1 \right)!=3!\ \text{and}\ \left( 10-3 \right)!=7!$)
On, cancelling out the common terms and simplifying the expression we will get,
\[\Rightarrow P\left( B=1 \right)=\dfrac{\dfrac{6\times 5}{2\times 1}\times \dfrac{4}{1}}{\dfrac{10\times 9\times 8}{3\times 2\times 1}}=\dfrac{6\times 5\times 2}{10\times 3\times 4}\]
\[\Rightarrow P\left( B=1 \right)=\dfrac{60}{120}=\dfrac{1}{2}\] ……………………(ii)
In case 3, 2 white balls are drawn so, we can say that 2 balls out of 4 white and 1 ball out of 6 red balls were drawn and here also the total number of balls remains same, it can be given mathematically as,
$P\left( B=2 \right)=\dfrac{{}^{6}{{C}_{1}}\times {}^{4}{{C}_{2}}}{{}^{10}{{C}_{3}}}$
Now, on solving the expression by using the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we will get,
$P\left( B=2 \right)=\dfrac{\dfrac{6!}{\left( 6-1 \right)!1!}\times \dfrac{4!}{\left( 4-2 \right)!2!}}{\dfrac{10!}{\left( 10-3 \right)!3!}}$
\[\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\times 1}\times \dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}{\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 3\times 2\times 1}}\] ($\because $as $\left( 6-1 \right)!=5!,\ \left( 4-2 \right)!=2!\ \text{and}\ \left( 10-3 \right)!=7!$)
On, cancelling out the common terms and simplifying the expression we will get,
\[\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{6}{1}\times \dfrac{4\times 3}{2\times 1}}{\dfrac{10\times 9\times 8}{3\times 2\times 1}}=\dfrac{6\times 2\times 3}{10\times 3\times 4}\]
\[\Rightarrow P\left( B=2 \right)=\dfrac{36}{120}=\dfrac{3}{10}\] ……………..(iii)
In case 4, all the three balls drawn are white, so there will be zero red balls and 3 balls out of 4 white balls. Here also total number of balls remains same, the expression can be given as,
$P\left( B=3 \right)=\dfrac{{}^{4}{{C}_{3}}}{{}^{10}{{C}_{3}}}$
${}^{6}{{C}_{3}}$ means 3 balls out of 6 red balls as, no ball was white so it means all the three balls were red and ${}^{10}{{C}_{3}}$ means 3 balls drawn out of total 10 balls i.e. $4\ \text{white}+6\ \text{red}=10\ \text{balls}$ total number of balls. Now, on solving the expression by using the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we will get,
$P\left( B=3 \right)=\dfrac{\dfrac{4!}{\left( 4-3 \right)!3!}}{\dfrac{10!}{\left( 10-3 \right)!3!}}$
$\Rightarrow P\left( B=3 \right)=\dfrac{\dfrac{4\times 3\times 2\times 1}{1\times 3\times 2\times 1}}{\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 3\times 2\times 1}}$ ($\because $as $\left( 4-3 \right)!=1!\ \text{and}\ \left( 10-3 \right)!=7!$)
On, cancelling out the common terms and simplifying the expression we will get,
$\Rightarrow P\left( B=3 \right)=\dfrac{\dfrac{4}{1}}{\dfrac{10\times 9\times 8}{3\times 2\times 1}}=\dfrac{4}{10\times 3\times 4}$
$\Rightarrow P\left( B=3 \right)=\dfrac{1}{30}$ ………………..(iv)
Thus, the probability distribution of balls B can be given as,

B	Probability
0	$\dfrac{1}{2}$
1	$\dfrac{1}{2}$
2	$\dfrac{3}{10}$
3	$\dfrac{1}{30}$

Note: Here, in question it was mentioned that the balls are drawn from the bag without replacement so, we have considered the total number of balls as 10 in all the cases, if it is given that balls are being replaced then we have considered three different draws and find the probability for each draw and then find the final probability, for example if all the 3 balls drawn with replacement are white out of 4 balls, then the expression will be, ${{1}^{st}}\ \text{draw}\ \text{white}=\dfrac{4}{10},{{2}^{nd}}\ \text{draw}\ \text{white}=\dfrac{4}{10}\ \text{and }{{\text{3}}^{rd}}\ \text{draw}\ \text{white}=\dfrac{4}{10}$,
So, the total probability will become $P(B)=\dfrac{2}{5}\times \dfrac{2}{5}\times \dfrac{2}{5}=\dfrac{8}{125}$. In this way, we can calculate for other cases also.