Probability

Question

It is known that screws produced by a certain company will be defective with probability $0.01$ independently of each other. The company sells the screws in package of $10$ and offers a replacement guarantee that at most $1$ of the $10$ screws is defective. What proportion of packages sold must the company replace?

Answer given is $0.04$

Answer 1

The probability that a screw is defective (screwed :P) is $0.01$, independently. Thus, probability of a screw not being defective is $(1-0.01) = 0.99$

The company offers a guarantee that in any pack of $10$, at most one is defective. That means, if there are more than one defective screws in your pack, the company will replace it.

The probability that there is more than one defective piece in a pack of $10$ screws can be calculated as:

$$\large \begin{align} P\left (\substack{\text{more than}\\\text{one defective}} \right ) \;\;\;&= \quad  1 - \Biggl (\; P \left ( \substack{\text{zero}\\\text{defective}} \right ) + \;P \left (\substack{\text{one}\\\text{defective}}\right )\; \Biggr )\\[2em] &=\quad  1 - \Biggl (\;\left (0.99^{10} \right ) \; + \;  \underbrace{\left ( 10 \times 0.01 \times 0.99^9 \right )}_{\substack{\text{choose 1 screw out of 10}\\[0.5em]0.01 = P(1 \text{ defective screw})\\[0.5em]0.99^9 = P(9 \text{ fine screws})}}\; \Biggr )\\[2em] &\approx \quad  1-0.99573\\[1em] P\left (\substack{\text{more than}\\\text{one defective}} \right ) &\approx \quad  0.00427 \end{align}$$

Answer 2

Packet will be returned if at-least 2 defective screws are found in the packet.

since probability of being defective does not depend on the other : This case can be solved by Binomial Distribution with $n=10$

checking a screw in the packet (check for it being defective) is a trial and we have a total of $n=10$ such trials. each trial is independent of the other. 

Let, $x$ = # of defective screws in the packet.
Then we need to calculate 

where, $p$ = probability of success(here, which is being defective, coz we have defined $x$ like that) = $0.01$

so,

and 

which imply

Answer 3

Probability that a package will have to be replaced is : P(x≥2)
 

P(x≥2) = 1 - P(x≦1)= 1- [P{X=0} + P{X=0}]

         = 1- [10C0(0.01)⁰(0.99)¹⁰ + 10C1(0.001)¹(0.99)⁹)
        = 0.004
             0.4%  packages will have to be replaced.