Question

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table.

Number of defective parts	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Number of days	50	32	22	18	12	12	10	10	10	8	6	6	2	2

From these days one day is chosen at random. What is the probability that on that day, the output has
(i) no defective part?
(ii) at least 1 defective part?
(iii) not more than 5 defective parts?
(iv) more than 5 but less than 8 defective parts?
(v) more than 13 defective parts?

Answer 1

Hint: We solve this question by first considering the given table and then we solve each sub-part of the question by first considering the days that satisfy the given condition in each subpart and then apply the formula, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$. Then by simplifying the values we get the required probabilities.

Complete step by step answer:
We are given that

Number of defective parts	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Number of days	50	32	22	18	12	12	10	10	10	8	6	6	2	2

We need to find the probability that if one day is selected at random from them, it has

i) no defective part?
First let us consider the formula for probability, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$.
From the given table we can see that there are no defective parts on 50 days and total number of days are equal to 200.
So, using the above formula we get that probability that there is no defective part on the selected day is
$\Rightarrow \dfrac{50}{200}=\dfrac{1}{4}$
Hence answer is $\dfrac{1}{4}$.

ii) at least 1 defective part?
First let us consider the formula for probability, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$.
We can write the probability of selecting a day with at lest 1 defective part as,
$\Rightarrow 1-P\left( No\ defective\ part \right)$
From the above part we can substitute the value of probability that the day has no defective part. Then we get
$\Rightarrow 1-\dfrac{1}{4}=\dfrac{3}{4}$
Hence answer is $\dfrac{3}{4}$.

iii) not more than 5 defective parts
First let us consider the formula for probability, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$.
As we need to find the probability that the selected day does not have more than 5 defective parts, we need to consider the days that have less than or equal to 5 defective parts.
So, using the above formula we get that probability that there are not more than 5 defective parts on the selected day is
$\begin{align}
  & \Rightarrow \dfrac{50+32+22+18+12+12}{200}=\dfrac{146}{200} \\
 & \Rightarrow \dfrac{73}{100} \\
\end{align}$
Hence answer is $\dfrac{73}{100}$.

iv) more than 5 but less than 8 defective parts.
First let us consider the formula for probability, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$.
As we need to find the probability that the selected day have more than 5 but less than 8 defective parts, we need to consider the days that have more than 5 and less than 8 defective parts, that is the days having 6 and 7 defective parts.
So, using the above formula we get that probability that there are not more than 5 defective parts on the selected day is
\[\begin{align}
  & \Rightarrow \dfrac{10+10}{200}=\dfrac{20}{200} \\
 & \Rightarrow \dfrac{1}{10} \\
\end{align}\]
Hence answer is \[\dfrac{1}{10}\].

(v) more than 13 defective parts?
First let us consider the formula for probability, the probability of choosing n objects from a set of N objects is equal to $\dfrac{n}{N}$.
As we see in the table there are 0 days with more than 13 defective parts.
So, using the above formula we get that probability that there are more than 13 defective parts on the selected day is
$\Rightarrow \dfrac{0}{200}=0$

So, the correct answer is “0”.

Note: We can also solve this question in another method.
Let us consider the number of defective parts on the selected day as X.
Let us add a row to the table by finding the probabilities of the days. Then we can write the table as,

X	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Number of days	50	32	22	18	12	12	10	10	10	8	6	6	2	2
P(X)	\[\dfrac{50}{200}\]	$\dfrac{32}{200}$	$\dfrac{22}{200}$	$\dfrac{18}{200}$	$\dfrac{12}{200}$	$\dfrac{12}{200}$	$\dfrac{10}{200}$	$\dfrac{10}{200}$	$\dfrac{10}{200}$	$\dfrac{8}{200}$	$\dfrac{6}{200}$	$\dfrac{6}{200}$	$\dfrac{2}{200}$	$\dfrac{2}{200}$

(i) no defective part?
So, we need to find the value of $P\left( X=0 \right)$. From the table we can see that its value is,
$\Rightarrow \dfrac{50}{200}=\dfrac{1}{4}$
Hence answer is $\dfrac{1}{4}$.

(ii) at least 1 defective part?
So, here we need to find the value of $P\left( X\ge 1 \right)$.
We can write it as
$\begin{align}
  & \Rightarrow P\left( X\ge 1 \right)=1-P\left( X<1 \right) \\
 & \Rightarrow P\left( X\ge 1 \right)=1-P\left( X=0 \right) \\
\end{align}$
So, using the values from the table we get,
$\Rightarrow 1-\dfrac{1}{4}=\dfrac{3}{4}$
Hence answer is $\dfrac{3}{4}$.

(iii) not more than 5 defective parts?
So, here we need to find the value of $P\left( X\le 5 \right)$.
We can write it as
$\Rightarrow P\left( X\le 5 \right)=P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)+P\left( X=3 \right)+P\left( X=4 \right)+P\left( X=5 \right)$
So, using the values from the table we get,
$\begin{align}
  & \Rightarrow \dfrac{50+32+22+18+12+12}{200}=\dfrac{146}{200} \\
 & \Rightarrow \dfrac{73}{100} \\
\end{align}$
Hence answer is $\dfrac{73}{100}$.

(iv) more than 5 but less than 8 defective parts?
So, here we need to find the value of $P\left( 5We can write it as
$\Rightarrow P\left( 5So, using the values from the table we get,
\[\begin{align}
  & \Rightarrow \dfrac{10+10}{200}=\dfrac{20}{200} \\
 & \Rightarrow \dfrac{1}{10} \\
\end{align}\]
Hence answer is \[\dfrac{1}{10}\].

(v) more than 13 defective parts?
So, here we need to find the value of $P\left( X>13 \right)$.
As we see in the table there are 0 days with more than 13 defective parts.
So, we get the probability as 0.
Hence answer is 0