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. Eleven
B. Twelve
C. Fifteen
D. Fourteen

Answer 1

- Hint:- In this question the probability of at least one part being defective is equals to one minus probability of no defective part.

Complete step-by-step solution -

Given that probability of some defective parts $ = 0.05$
Since probability of at least one part being defective $ = 1 - $ probability of no part defective.
Therefore probability of no part defective $ = 1 - 0.05 = 0.95$
Let $n$ be the required parts to be produced then equation will be:
$ = 1 - {\left( {0.95} \right)^n} \geqslant \dfrac{1}{2}$
So ${\left( {0.95} \right)^n} \leqslant \dfrac{1}{2}$
Now taking log on both sides with base ten we get:
$n\left[ {{{\log }_{1o}}95 - 2} \right] \leqslant {\log _{10}}\dfrac{1}{2}$
$ \Rightarrow n \leqslant \dfrac{{300}}{{23}} \leqslant 13$
So $n = 11,12$
Hence the correct option is A & B.

Note:- As we know that total probability of an event is one, so firstly we found the probability of no part defective by subtracting 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.