An urn contains $6$white and$9$black balls. If $4$balls are to be randomly selected without replacement, what is the probability that the first $2$selected is white and the last 2 black?

Select four balls from the urn containing $6$ white and $9$repudiates without substitution. Compute the probability that the initial two chosen are white and the second two are black.

Allow us to characterize four occasions $E_1,E_2,E_3$, and $E_4$as choosing a ball in first, second, third, and fourth choices separately.

The example space contains $15$prospects because the urn contains$15(6+9)$balls

that is,

$n\left(S\right)=15$

$\text{Urn}=\left\{\text{White 6, Black 9}\right\}$

The probability of choosing a white ball in first choice $E_1$is,

The probability of choosing a white ball from the urn containing $6$white balls out $15$the balls is $\frac{6}{15}$.

Second Selection $E_2$:

The example space contains 14 prospects, on the grounds that because the urn contains $14(5+9)$balls.

That is$n\left(S\right)=14$.

$Urn=White5,Black9$

The probability of choosing a white ball from the urn containing five white balls out of $14$balls is $\frac{5}{14}$because two of the six white balls among the $16$balls have been chosen in the initial two determinations as the choice is managed without substitution.

Third Selection $E_3$:

The example space contains$13$prospects, in light of the fact that the urn contains $13(4+9)$balls.

That is $n\left(S\right)=14$.

$\text{Urn}=\left\{\text{White 5, Black 9}\right\}$

The probability of choosing a black ball from the urn containing $9$black balls out $13$balls is$\frac{9}{13}$

since there are $9$among the $16$balls that have been chosen in the initial two determinations.

Fourth Selection$E_4$:

The example space contains $13$prospects that because in light of the fact that because the urn contains $13(5+9)$balls.

That is $n\left(S\right)=13$.

$\text{Urn}=\left\{\text{White 4, Black 9}\right\}$

The probability of choosing a white ball from the urn containing five white balls out of $13$balls is $\frac{5}{13}$in light of the fact that because one of the six white balls among the $14$ balls has been chosen in the past three determinations.

Hence, the probability that the first two selected are white and the last two selected balls are black is,

$P\left(E_1{E}_2{E}_3{E}_4\right)=\frac{6}{15}\times \frac{5}{14}\times \frac{9}{13}\times \frac{8}{12}$

$=0.4\times 0.3571\times 0.6923\times 0.6667$

$=0.0659$