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 at random from this lot. Find the mean and variance of the distribution.

Answer 1

Hint:
Here we need to find the probability distribution of the number of bad oranges. So, for that, we will find the probability for each random variable and then we will draw the probability distribution table. Then we will use the formula of mean and the variance to find its required value for the given probability distribution table.

Complete Step by Step Solution:
Let \[X\] denote the random variable. As it is given that two oranges are drawn at random. Then \[X = 0,1,2\]
We know that 16 oranges are good oranges and 4 oranges are bad oranges.
Now in case 1, we will find the value of \[P\left( {X = 0} \right)\] which here means the probability that the orange picked is not a bad orange.
Therefore, we get
\[P\left( {X = 0} \right) = \dfrac{{{}^{16}{C_2}}}{{{}^{20}{C_2}}}\]
Now, using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and simplifying, we get
\[ \Rightarrow P\left( {X = 0} \right) = \dfrac{{\dfrac{{16!}}{{2!\left( {16 - 2} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}\]
On further simplifying the terms, we get
\[ \Rightarrow P\left( {X = 0} \right) = \dfrac{{16! \times 18!}}{{14! \times 20!}}\]
Computing the factorials, we get
\[ \Rightarrow P\left( {X = 0} \right) = \dfrac{{16 \times 15 \times 14! \times 18!}}{{14! \times 20 \times 19 \times 18!}}\]
Multiplying and dividing the terms, we get
\[ \Rightarrow P\left( {X = 0} \right) = \dfrac{{12}}{{19}}\] ……….. \[\left( 1 \right)\]
Now, in case 2, only one orange drawn is bad, which means one orange out of 4 bad oranges and one orange out of 16 oranges are drawn.
Therefore, its probability will be denoted by \[P\left( {X = 1} \right)\]
\[P\left( {X = 1} \right) = \dfrac{{{}^4{C_1} \times {}^{16}{C_1}}}{{{}^{20}{C_2}}}\]
Now, using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and simplifying, we get
\[ \Rightarrow P\left( {X = 1} \right) = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{16!}}{{1!\left( {16 - 1} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}\]
On further simplifying the terms, we get
\[ \Rightarrow P\left( {X = 1} \right) = \dfrac{{4! \times 16! \times 18! \times 2!}}{{20! \times 15! \times 3!}}\]
Computing the factorials, we get
\[ \Rightarrow P\left( {X = 1} \right) = \dfrac{{4 \times 3! \times 16 \times 15! \times 18! \times 2 \times 1}}{{20 \times 19 \times 18! \times 15! \times 3!}}\]
Multiplying and dividing the terms, we get
\[ \Rightarrow P\left( {X = 1} \right) = \dfrac{{16 \times 2}}{{5 \times 19}}\]
Again multiplying the terms, we get
\[ \Rightarrow P\left( {X = 1} \right) = \dfrac{{32}}{{95}}\] ……….. \[\left( 2 \right)\]
Now, in case 3, both the oranges drawn are bad, which means two oranges out of 4 bad oranges are drawn.
Therefore, its probability will be denoted by \[P\left( {X = 2} \right)\].
\[P\left( {X = 2} \right) = \dfrac{{{}^4{C_2}}}{{{}^{20}{C_2}}}\]
Now, using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and simplifying, we get
\[ \Rightarrow P\left( {X = 2} \right) = \dfrac{{\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}}\]
On further simplifying the terms, we get
\[ \Rightarrow P\left( {X = 2} \right) = \dfrac{{\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}}} = \dfrac{{4! \times 18!}}{{2! \times 20!}}\]
Computing the factorial, we get
\[ \Rightarrow P\left( {X = 2} \right) = \dfrac{{4 \times 3 \times 2! \times 18!}}{{2! \times 20 \times 19 \times 18!}}\]
On further simplification, we get
\[ \Rightarrow P\left( {X = 2} \right) = \dfrac{3}{{19 \times 5}}\]
On multiplying the terms, we get
\[ \Rightarrow P\left( {X = 2} \right) = \dfrac{3}{{95}}\] ……….. \[\left( 3 \right)\]
Now, we will draw the probability distribution table.

\[X\]	0	1	2
\[P\left( X \right)\]	\[\dfrac{{12}}{{19}}\]	\[\dfrac{{32}}{{95}}\]	\[\dfrac{3}{{95}}\]

Now, we will find the mean of the distribution by using the formula \[\overline x = \sum {{x_i}{P_i}} \] .
Now, we will substitute the values and we will find the sum. Therefore, we get
\[ \Rightarrow \overline x = 0 \times \dfrac{{12}}{{19}} + 1 \times \dfrac{{32}}{{95}} + 2 \times \dfrac{3}{{95}}\]
On multiplying the terms, we get
\[ \Rightarrow \overline x = 0 + \dfrac{{32}}{{95}} + \dfrac{6}{{95}}\]
On adding the terms, we get
\[ \Rightarrow \overline x = \dfrac{{38}}{{95}} = \dfrac{2}{5}\]
Now, we will calculate the value of by \[E\left( {{X^2}} \right)\] using the formula, \[E\left( {{X^2}} \right) = \sum {{x_i}^2{P_i}} \].
Now, we will substitute the values and we will find the sum. Therefore, we get
\[E\left( {{X^2}} \right) = {0^2} \times \dfrac{{12}}{{19}} + {1^2} \times \dfrac{{32}}{{95}} + {2^2} \times \dfrac{3}{{95}}\]
On multiplying the terms, we get
\[ \Rightarrow E\left( {{X^2}} \right) = 0 + \dfrac{{32}}{{95}} + \dfrac{{12}}{{95}}\]
On adding the terms, we get
\[ \Rightarrow E\left( {{X^2}} \right) = \dfrac{{44}}{{95}}\]
Now, we will find the variance of the distribution by using the formula;
\[{\mathop{\rm var}} \left( X \right) = E\left( {{X^2}} \right) - {\left( {\overline x } \right)^2}\]
Now, we will substitute the known values in the above equation. Therefore, we get
\[ \Rightarrow {\mathop{\rm var}} \left( X \right) = \dfrac{{44}}{{95}} - {\left( {\dfrac{2}{5}} \right)^2}\]
On applying the exponents on the base, we get
\[ \Rightarrow {\mathop{\rm var}} \left( X \right) = \dfrac{{44}}{{95}} - \dfrac{4}{{25}}\]
Subtracting the terms, we get
\[ \Rightarrow {\mathop{\rm var}} \left( X \right) = \dfrac{{220 - 76}}{{475}}\]
On further simplification, we get
\[ \Rightarrow {\mathop{\rm var}} \left( X \right) = \dfrac{{144}}{{475}}\]

Hence, the value of the variance is equal to \[\dfrac{{144}}{{475}}\].

Note:
We generally make mistakes while considering the number of possible cases. We need to take all the possible cases otherwise we will get the probability distribution incorrect. If we add all the probabilities of a probability distribution, we get the answer. Probability is defined as the ratio of number of desired or favorable outcomes to the total number of possible outcomes. In addition, the value of probability cannot be greater than 1 and the value of probability cannot be negative. Also remember that the probability of a sure event is always one.