Question

A manufacturing firm produces pads of bound paper. $3\%$ of all paper pads produced are improperly bound. An inspector randomly selects two pads of paper, one at a time. Because a large number of pads are being produced during the inspection, the sampling being done, in essence, with replacement. What is the probability that the two pads selected are both improperly bound?

Answer 1

Hint : To find the probability of selecting two pads, both are improperly bound can be calculated by first calculating the probability of selecting one pad which is improperly bound then multiplying it with the probability of another one pad which is also bound improperly.

Complete step-by-step answer :
It is given that the $3\%$ of all paper pads are bound improperly.
We have to find the probability of selecting two pads which are bound improperly.
We know that probability = number favorable cases / number of total cases
Here the favorable cases is $3\%$ and number of total cases is 100%
So probability of selecting only one pad at a time which is bounded improperly = $\dfrac{3}{{100}}$
With replacement there is no change in total number of cases
So the probability of selecting second pad which is bounded improperly = $\dfrac{3}{{100}}$
Now the probability of selecting two pad which is improperly bounded = $\dfrac{3}{{100}} \times \dfrac{3}{{100}} = \dfrac{9}{{10000}}$
Hence the probability of selecting two pad which is improperly bounded = $\dfrac{9}{{10000}}$

Note : Here in this question it is clearly mentioned that after selecting one pad it is replaced means there will not be any changes in total number of cases to finding the probability of a second pad.