Question

Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn from this lot. Find the mean and variance of the distribution.

Answer 1

Hint: First of all, we will consider the number of cases in which the event can take place. Then from that, we will consider each possibility, and based on that we will find the expression of a combination of oranges being drawn out of total oranges. And then solve that combination using the expression ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Then we will find the probability based on those answers and then will make the final distribution table of probability and number of events. After that use, the mean formula for the probability distribution, $\sum {{x_i}{P_i}} $ to calculate the mean. Then use the formula $\operatorname{var} \left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$ to calculate the variance.

Complete step-by-step solution:
First of we will assume that orange is drawn and it is bad, then we will denote the number of bad oranges as B.
Now, it may happen that when we draw an orange randomly then no bad orange is picked or only one orange is bad or all the two oranges are bad. So, here we will consider three cases.
So, in case 1 when an orange is picked randomly and the orange picked is not bad it can be given mathematically as,
$ \Rightarrow P\left( {B = 0} \right) = \dfrac{{{}^{16}{C_2}}}{{{}^{20}{C_2}}}$
Now, on solving the expression by using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ we will get,
$ \Rightarrow P\left( {B = 0} \right) = \dfrac{{\dfrac{{16!}}{{2!\left( {16 - 2} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}$
Simplify the terms,
\[ \Rightarrow P\left( {B = 0} \right) = \dfrac{{\dfrac{{16 \times 15 \times 14!}}{{2!14!}}}}{{\dfrac{{20 \times 19 \times 18!}}{{2!18!}}}}\]
Cancel out the common factors,
\[ \Rightarrow P\left( {B = 0} \right) = \dfrac{{4 \times 3}}{{19}}\]
Multiply the terms,
\[ \Rightarrow P\left( {B = 0} \right) = \dfrac{{12}}{{19}}\]..................….. (1)
Now, in case 2, only one orange drawn is bad, which means 1 orange out of 4 bad oranges and 1 orange out of 16 good ones are drawn. Here, the total number of oranges remains the same so, the expression can of probability be given as,
$ \Rightarrow P\left( {B = 1} \right) = \dfrac{{{}^4{C_1} \times {}^{16}{C_1}}}{{{}^{20}{C_2}}}$
Now, on solving the expression by using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ we will get,
$ \Rightarrow P\left( {B = 1} \right) = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{16!}}{{1!\left( {16 - 1} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}$
Simplify the terms,
\[ \Rightarrow P\left( {B = 1} \right) = \dfrac{{\dfrac{{4 \times 3!}}{{1!3!}} \times \dfrac{{16 \times 15!}}{{1!15!}}}}{{\dfrac{{20 \times 19 \times 18!}}{{2!18!}}}}\]
Cancel out the common factors,
\[ \Rightarrow P\left( {B = 1} \right) = \dfrac{{2 \times 16}}{{5 \times 19}}\]
Multiply the terms,
\[ \Rightarrow P\left( {B = 1} \right) = \dfrac{{32}}{{95}}\]................….. (2)
Now, in case 3, both oranges drawn are bad. Here, the total number of oranges remains the same so, the expression can of probability be given as,
$ \Rightarrow P\left( {B = 2} \right) = \dfrac{{{}^4{C_2}}}{{{}^{20}{C_2}}}$
Now, on solving the expression by using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ we will get,
$ \Rightarrow P\left( {B = 2} \right) = \dfrac{{\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}$
Simplify the terms,
\[ \Rightarrow P\left( {B = 2} \right) = \dfrac{{\dfrac{{4 \times 3 \times 2!}}{{2!2!}}}}{{\dfrac{{20 \times 19 \times 18!}}{{2!18!}}}}\]
Cancel out the common factors,
\[ \Rightarrow P\left( {B = 1} \right) = \dfrac{3}{{5 \times 19}}\]
Multiply the terms,
\[ \Rightarrow P\left( {B = 1} \right) = \dfrac{3}{{95}}\]............….. (3)
Thus, the probability distribution can be given as,

B	0	1	2
P(B)	$\dfrac{{12}}{{19}}$	$\dfrac{{32}}{{95}}$	$\dfrac{3}{{95}}$

Now find the mean of the distribution by the formula,
$\bar x = \sum {{x_i}{P_i}} $
Substitute the values and find a summation,
$ \Rightarrow \bar x = 0 \times \dfrac{{12}}{{19}} + 1 \times \dfrac{{32}}{{95}} + 2 \times \dfrac{3}{{95}}$
Multiply the terms,
$ \Rightarrow \bar x = 0 + \dfrac{{32}}{{95}} + \dfrac{6}{{95}}$
Add the terms,
$ \Rightarrow \bar x = \dfrac{{38}}{{95}}$
Cancel out the common factors,
$\therefore \bar x = \dfrac{2}{5}$
Thus, the mean is $\dfrac{2}{5}$.
Now find the value of $E\left( {{X^2}} \right)$ for calculating the variance.
$E\left( {{X^2}} \right) = \sum {{x_i}^2{P_i}} $
Substitute the values and find a summation,
$ \Rightarrow E\left( {{X^2}} \right) = {0^2} \times \dfrac{{12}}{{19}} + {1^2} \times \dfrac{{32}}{{95}} + {2^2} \times \dfrac{3}{{95}}$
Square the terms,
$ \Rightarrow E\left( {{X^2}} \right) = 0 \times \dfrac{{12}}{{19}} + 1 \times \dfrac{{32}}{{95}} + 4 \times \dfrac{3}{{95}}$
Multiply the terms,
$ \Rightarrow E\left( {{X^2}} \right) = 0 + \dfrac{{32}}{{95}} + \dfrac{{12}}{{95}}$
Now, add the terms,
$ \Rightarrow E\left( {{X^2}} \right) = \dfrac{{44}}{{95}}$
The formula of variance is,
$\operatorname{var} \left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$
Substitute the values,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{44}}{{95}} - {\left[ {\dfrac{2}{5}} \right]^2}$
Square the term,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{44}}{{95}} - \dfrac{4}{{25}}$
Take LCM on the right side,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{220 - 76}}{{475}}$
Subtract the values,
$\therefore \operatorname{var} \left( X \right) = \dfrac{{144}}{{475}}$

Thus, the variance is $\dfrac{{144}}{{475}}$.

Note: We generally make mistakes while considering the number of possible cases for an event to happen. We have to take all the possible cases otherwise the distribution function will be wrong. There is a good way to check if the distribution function is correct or not. If we add all the probabilities you should always get one. Probability of an event to occur or not occur will always lie between 0 and 1.