Q.3.6 - A First Course in Probability

Question

Consider an urn containing $12$ balls, which $8$are white. A sample of size$4$ is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly$3$ white balls? 

Step-by-Step Solution

Answer

Conditional probability with replacement is $\frac{1}{2}$.

Conditional probability without replacement is $\frac{1}{2}$.

Step 1:Given Information

Given that an urn containing$12$ balls, of which $8$ are white. A sample of size $4$ is to be drawn with replacement (without replacement ).

Step 2:Explanation(With replacement)

Number of manners by which $3$ white balls can arrive in an example of $4$ { Let it signify occasion A}

$(W,W,W,B)$,$(W,W,B,W)$,$(W,B,W,W)$,$(B,W,W,W)$

$P\left(A\right)=\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}+\frac{8}{12}\times \frac{4}{12}\times \frac{8}{12}\times \frac{8}{12}+\frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}\times \frac{8}{12}+\frac{4}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}$

$Out of these favourable cases when 1^{st} and 3^{rd} is white=(W,W,W,B)(W,B,W,W)$

$\mathrm{P}(\mathrm{B}\mid \mathrm{A})\text{ (with replacement })=\frac{\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}+\frac{8}{12}\times \frac{4}{12}\times \frac{8}{12}\times \frac{8}{12}}{\mathrm{P}\left(\mathrm{A}\right)}$

$=\frac{1}{2}$

Step 3:Explanation(Without Replacement)

Without replacement,

$P\left(A\right)=\frac{8}{12}\times \frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}+\frac{8}{12}\times \frac{4}{11}\times \frac{7}{10}\times \frac{6}{9}+\frac{8}{12}\times \frac{7}{11}\times \frac{4}{10}\times \frac{6}{9}+\frac{4}{12}\times \frac{8}{11}\times \frac{7}{10}\times \frac{6}{9}$

$=0.1131313\times 4$

$P(\mathrm{B}\mid \mathrm{A})=\frac{\frac{8}{12}\times \frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}+\frac{8}{12}\times \frac{4}{11}\times \frac{7}{10}\times \frac{6}{9}}{\mathrm{P}\left(\mathrm{A}\right)}$

$=\frac{1}{2}$

Step 4:Final Answer

The conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls with replacement is $\frac{1}{2}$.

The conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls without replacement is $\frac{1}{2}$.