91Ó°ÊÓ

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 ).

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

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

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

localid="1647615456092" $Outofthesefavourablecaseswhen{1}^{st}and{3}^{rd}iswhite=(W,W,W,B)(W,B,W,W)$

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

$=\frac{1}{2}$

Without replacement,

$=0.1131313×4$

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

role="math" localid="1647615865767" $=\frac{1}{2}$

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