Question

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. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

Let A denote the event that the sample contains exactly three white balls and let B be the event that the ball taking the condition of with replacement.
Now $p\left(A\right)=\left({}^4{C}_3\right)\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}$

$P\left(AB\right)=\left({}^3{C}_2\right)\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}$
$\therefore P\left[B/A\right]=\frac{P\left[AB\right]}{P\left(A\right)}=\frac{{}^3{C}_2}{{}^4{C}_3}=\frac{3}{4}$
In the case of sampling without replacement,
$P\left(A\right)=\left({}^4{C}_3\right)\frac{8}{12}\times \frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$

$P\left(AB\right)=\left({}^3{C}_2\right)\frac{8}{12}\times \frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$
$\therefore P\left[B/A\right]=\frac{{}^3{C}_2}{{}^4{C}_3}=\frac{3}{4}$