Repeat Problem $6.2$when the ball selected is replaced in the urn before the next selection.

The selected ball is replaced in the urn before the next selection.

Let$X_i$equal $1$if the ith white ball is selected $0$otherwise.

Let the first and second ball chosen be white.

Such that

$$\begin{gathered} X_1=1, \\ {X}_2=2 \end{gathered}$$

If the first ball chosen is white,

The probability of the event is$5/13.$

The selected white ball is replaced in the urn.

The second chosen ball is also white and due to replacement, the probability remains the same.

Now,

Table representing joint probability for $X_1$and $X_2$:

Simplify the above table:

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

If the ith white ball is selected, set $Xi$to $1$; otherwise, set it to $0$.

All conceivable scenarios must be investigated and the combinatory argument must be used, using the same notion as in Part (a).

Table with joint probabilities for $X1$, $X2$, and $X3$:

Simplify the above table: