Question

Five defective bolts are accidentally mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer 1

Hint: Here, it is given that defective bolts are mixed with good bolts. We have to take four bolts from the lot. So we have to find the probability of the number of defective bolts. For that, take each number of defective bolts at once and find the frequency distribution one by one using the formula of combinations.

Complete step-by-step solution:
Given is,
The number of defective bolts \[ = 5\]
The number of good bolts \[ = 20\]
Total number of bolts \[ = 20 + 5 = 25\]
The number of bolts to be drawn \[ = 4\]
Let the number of defective bolts drawn be \[x\]
\[x\] Can be either \[0,1,2\]\[,3\] or \[4\]
The total possible ways of selecting \[4\] bolts \[ = {}^{25}{C_4}\]
Now, taking each defective bolt into consideration, we get the frequency distribution as follows:
\[P\left( {X = 0} \right)\], which means selecting no defective bolt.
\[P\left( {X = 0} \right)\]\[ = \dfrac{{{}^{20}{C_4}}}{{{}^{25}{C_4}}}\]
Expanding the expression, we get;
$\Rightarrow$\[P\left( {X = 0} \right) = \dfrac{{20 \times 19 \times 18 \times 17}}{{25 \times 24 \times 23 \times 22}}\]
Multiplying the above expression, we get;
$\Rightarrow$\[P\left( {X = 0} \right) = \dfrac{{969}}{{2530}}\]
Now,
\[P\left( {X = 1} \right)\], which means selecting \[1\] defective bolt and \[3\] good bolts.
$\Rightarrow$\[P\left( {X = 1} \right) = \dfrac{{{}^5{C_1} \times {}^{20}{C_3}}}{{{}^{25}{C_4}}}\]
Expanding the expression, we get;
$\Rightarrow$\[P\left( {X = 1} \right)\]\[ = \dfrac{{\dfrac{{5!}}{{4!}} \times \dfrac{{20!}}{{17!3!}}}}{{\dfrac{{25!}}{{21!4!}}}}\]
Further expanding the expression into simple terms, we get;
$\Rightarrow$\[P\left( {X = 1} \right)\]\[ = \dfrac{{5 \times 4 \times 20 \times 19 \times 18}}{{25 \times 24 \times 23 \times 22}}\]
Multiplying the above expression, we get;
$\Rightarrow$\[P\left( {X = 1} \right) = \dfrac{{114}}{{253}}\]
Now,
\[P\left( {X = 2} \right)\], which means selecting \[2\] defective bolt and \[2\] good bolts.
\[P\left( {X = 2} \right) = \dfrac{{{}^5{C_2} \times {}^{20}{C_2}}}{{{}^{25}{C_4}}}\]
Expanding the expression, we get;
$\Rightarrow$\[P\left( {X = 2} \right) = \dfrac{{\dfrac{{5!}}{{3!2!}} \times \dfrac{{20!}}{{18!2!}}}}{{\dfrac{{25!}}{{21!4!}}}}\]
Further expanding the expression into simple terms, we get;
$\Rightarrow$\[P\left( {X = 2} \right) = \dfrac{{\dfrac{{5 \times 4}}{2} \times \dfrac{{20 \times 19}}{2}}}{{\dfrac{{25 \times 24 \times 23 \times 22}}{{4 \times 3 \times 2}}}}\]
Multiplying the above expression, we get;
$\Rightarrow$\[P\left( {X = 2} \right) = \dfrac{{38}}{{253}}\]
Moving on,
\[P\left( {X = 3} \right)\], which means selecting \[3\] defective bolts and \[1\] good bolts.
\[P\left( {X = 3} \right) = \dfrac{{{}^5{C_3} \times {}^{20}{C_1}}}{{{}^{25}{C_4}}}\]
Expanding the expression, we get;
$\Rightarrow$\[P\left( {X = 3} \right) = \dfrac{{\dfrac{{5!}}{{3!2!}} \times \dfrac{{20!}}{{19!1!}}}}{{\dfrac{{25!}}{{21!4!}}}}\]
Further expanding the expression into simple terms, we get;
$\Rightarrow$\[P\left( {X = 3} \right) = \dfrac{{\dfrac{{5 \times 4}}{2} \times 20}}{{\dfrac{{25 \times 24 \times 23 \times 22}}{{4 \times 3 \times 2}}}}\]
Multiplying the above expression, we get;
$\Rightarrow$\[P\left( {X = 3} \right) = \dfrac{4}{{253}}\]
Finally,
\[P\left( {X = 4} \right)\], which means selecting \[4\] defective bolts and no good bolt.
\[P\left( {X = 4} \right) = \dfrac{{{}^5{C_4}}}{{{}^{25}{C_4}}}\]
Expanding the expression, we get;
$\Rightarrow$\[P\left( {X = 4} \right)\]\[ = \dfrac{{5 \times 4 \times 3 \times 2}}{{25 \times 24 \times 23 \times 22}}\]
Multiplying the above expression, we get;
$\Rightarrow$\[P\left( {X = 4} \right) = \dfrac{1}{{2530}}\]

$\therefore $ Probability distribution of the number of defective bolts is \[\dfrac{1}{{2530}}\]

Note: We have to remember that, the expansion of the combination formula is: \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] the students should notice and separately derive each and every frequency distribution using the binomial theorem also called as the combination expansion.