Repeat Problem $6.3a$when the ball selected is replaced in the urn before the next selection

In probability theory, a joint likelihood is the opportunity that two events will occur simultaneously. In different terms, joint likelihood is the likelihood of two occasions occurring simultaneously.

The white balls are numbered, so if the ith white ball is chosen, set Yi to 1 and 0 otherwise.

Before the next selection, the picked ball is replaced in the urn. If the ith white

ball is selected, set Y₁to 1; otherwise, set Y₂to 0. If it have not been picked for the first and second white balls.

After that, we must utilize conditional probability.

$P(Y_1=0,Y_2=0)P(Y_2=0,Y_1=0)P(Y_1=0)$

Because of the symmetry, all of the balls have an identical chance of being chosen or not being chosen.

Thus,

role="math" localid="1647231330635" $P(Y_1=0)={\left(\frac{12}{13}\right)}^3$

Let's pretend the first white ball was not chosen. Thus, this thing can be excluded. Then the conditional probability for the second white ball has not been chosen, $P(Y_2=0,Y_1=0)={\left(\frac{11}{12}\right)}^3$

Hence,

$P(Y_1=0,Y_2=0)={\left(\frac{11}{12}\right)}^3\times {\left(\frac{12}{13}\right)}^3=\frac{1331}{2197}$

Similarly, we have

$$\begin{gathered} P(Y_1=0,Y_2=1)=P(Y_2=1,Y_1=0)P(Y_1=0) \\ =(1-P(Y_2=0,Y_1=0)P(Y_1=0) \\ =(1-{\left(\frac{11}{12}\right)}^3){\left(\frac{12}{13}\right)}^3=\frac{397}{2197} \end{gathered}$$

And due to symmetry we have,

$P(Y_1=1,Y_2=0)=P(Y_1=0,Y_2=1)=\frac{397}{2197}$

Thus, we can calculate the remaining probability as final probability by one minus all the probabilities which are calculated above. Therefore,

$P(Y_1=1,Y_2=1)=1-\frac{1331}{2197}-\frac{397}{2197}=\frac{72}{2197}$