A sample of size 4 is drawn with replacement be the first part of the problem and without replacement be the second part of the problem, then from an urn containing 12 balls, of which 8 are white, what is the conditional probability that the ball drawn on the third draw was white, given that the sample contains 3 white balls?
A. $ \dfrac{1}{4} $
B. $ \dfrac{1}{3} $
C. $ \dfrac{2}{3} $
D. $ \dfrac{3}{4} $

Hint: We have to draw 4 numbers of balls containing 3 numbers of white balls from total 12 numbers of balls having 8 numbers of balls so we have to apply permutation method accordingly because permutation is used in case of arranging the objects in order and here we have to find the probability that the ball drawn on the third draw is white, which is a case of order or arrangement rather than only selection.

Complete step-by-step answer:
Let A denote the event that the sample contains exactly three white balls and let B be the event that the ball picking from the urn takes the condition of with replacement.
So, P(A)= $ \dfrac{8}{{12}} \times \dfrac{8}{{12}} \times \dfrac{8}{{12}} \times \dfrac{4}{{12}} $
P(AB)= $ \dfrac{8}{{12}} \times \dfrac{8}{{12}} \times \dfrac{8}{{12}} \times \dfrac{3}{{12}} $
Therefore​, probability that the ball drawn on the third draw was white $ P(\dfrac{B}{A}) $ will be given by $ \dfrac{{P(AB)}}{{P(A)}} $ = $ \dfrac{3}{4} $
Condition 2: Sampling without replacement
Similarly let A denote the event that the sample contains exactly three white balls and let B be the event that the ball picking from the urn takes the condition of without replacement.
So, P(A)= $ \dfrac{8}{{12}} \times \dfrac{7}{{11}} \times \dfrac{6}{{10}} \times \dfrac{4}{9} $
P(AB)= $ \dfrac{8}{{12}} \times \dfrac{7}{{11}} \times \dfrac{6}{{10}} \times \dfrac{3}{9} $
Therefore​, probability that the ball drawn on the third draw was white $ P(\dfrac{B}{A}) $ will be given by $ \dfrac{{P(AB)}}{{P(A)}} $ = $ \dfrac{3}{4} $
So, the correct answer is “Option D”.

Note: As said permutation is used in case of arranging the objects in order. When the order of the objects matters then it should be considered as Permutation and when the order does not matters then it should be considered as Combination. So do not confuse using permutation, when required, instead of combination and vice-versa.