Question

Suppose a machine produces metal parts that contain some defective parts with probability 0.05. . How many parts should be produced in order that the probability of at least one part being defective is $\dfrac{1}{2}$ or more? (Given that, ${\log _{10}}95 = 1.997$ and ${\log _{10}}2 = 0.3$)
A) 11
B) 12
C) 15
D) 14

Answer 1

Hint: First, find the probability of no part is defective by subtracting the probability of defective parts from 1. Then apply the formula $P\left( {X = r} \right){ = ^n}{C_r}{p^r}{q^{n - r}}$ to find the required parts to be produced without defective parts. After that subtract it from 1 to find the required parts to be produced with defective parts and equate it to $\dfrac{1}{2}$. Then solve it to get the number of parts to be produced.

Complete step-by-step answer:
Given: - The probability of some defective parts $ = 0.05$
Let $n$ be the required parts to be produced.
The probability of at least one part being defective = 1 - the probability of no part defective
\[ \Rightarrow \] Probability of no part defective $ = 1 - 0.05$
Subtract the values on the right side,
\[ \Rightarrow \]Probability of no part defective $ = 0.05$
The equation for the parts to be produced will be given by,
$P\left( {X = r} \right){ = ^n}{C_r}{p^r}{q^{n - r}}$
Substitute the values $p = 0.05$, $q = 0.95$,
$ \Rightarrow P\left( {X \geqslant 1} \right) = 1{ - ^n}{C_0}{\left( {0.05} \right)^0}{\left( {0.95} \right)^n}$
As it is given that,
$P\left( {X \geqslant 1} \right) \geqslant \dfrac{1}{2}$
Substitute the value,
$ \Rightarrow 1{ - ^n}{C_0}{\left( {0.05} \right)^0}{\left( {0.95} \right)^n} \geqslant \dfrac{1}{2}$
Simplify the terms,
$ \Rightarrow {\left( {0.95} \right)^n} \leqslant 1 - \dfrac{1}{2}$
Subtract the number on the right side,
$ \Rightarrow {\left( {0.95} \right)^n} \leqslant \dfrac{1}{2}$
Now taking log on both sides with base ten we get:
$ \Rightarrow {\log _{10}}{\left( {0.95} \right)^n} \leqslant {\log _{10}}\dfrac{1}{2}$
Convert the decimal into a fraction,
$ \Rightarrow n{\log _{10}}\left( {\dfrac{{95}}{{100}}} \right) \leqslant {\log _{10}}\dfrac{1}{2}$
As we know,
$\log \dfrac{m}{n} = \log m - \log n$
Use this formula in the equation,
\[ \Rightarrow n\left( {{{\log }_{10}}95 - {{\log }_{10}}100} \right) \leqslant {\log _{10}}1 - {\log _{10}}2\]
Substitute the value of ${\log _{10}}95$ and ${\log _{10}}2$,
$ \Rightarrow n\left( {1.977 - 2} \right) \leqslant 0 - 0.3$
Subtract the values on both sides,
\[ \Rightarrow - 0.023n \leqslant - 0.3\]
Divide both sides by $ - 0.023$,
$ \Rightarrow n \geqslant \dfrac{{ - 0.3}}{{ - 0.023}}$
Cancel out the common factors,
$\therefore n \geqslant 13.04$

Hence, option (C) and (D) is the correct answer.

Note: The student might make a mistake by not changing the sign when dividing by the negative number.
As we know that the total probability of an event is one, so firstly we found the probability of no part defective by subtracting the probability of one part being defective which is given $0.05$ from one and suppose n parts produced in order that the probability of at least one part being defective is one-half or more after that we formed the equation and simplified it and got the required result.