Probability with conditional substitution

Question

An urn contains 6 balls, 1 purple, 2 blue, and 3 brown. When a ball is selected it is replaced with a green ball unless the ball drawn is green, in which case the green ball is simply returned to the urn. What is the probability of 3 green, 1 purple, 1 blue and 1 brown ball remaining in the urn after 3 selections?

Answer 1

An urn contains 6 balls, 1 purple, 2 blue, and 3 brown.

Let the balls be labelled $\color{purple}{p}, \color{blue}{b}, \color{brown}{r}, \color{green}{g}$ for purple, blue, brown, green. Let $\color{purple}{P_n}$ be the even of drawing a purple ball on the $n^{th}$ drawing, and so on for each colour.

Thus the Initial State is: $S_0 \equiv \{\color{purple}{(p, 1)}, \color{blue}{(b,2)}, \color{brown}{(r,3)}\}$

On the First Draw, we have the probabilities: $\mathbb{P}(\color{purple}{P_1}) = \frac 16, \; \mathbb{P}(\color{blue}{B_1}) = \frac 26, \; \mathbb{P}(\color{brown}{R_1}) = \frac 36$

When a ball is selected it is replaced with a green ball unless the ball drawn is green, in which case the green ball is simply returned to the urn.

After a drawing, the urn has one less of the colour drawn, and one more of green. (and so no change if green were drawn).

So $\mathbb{P}(\color{blue}{B_2}) = \mathbb{P}(\color{purple}{P_1})\times\mathbb{P}(\color{blue}{B_2}\mid \color{purple}{P_1})+\mathbb{P}(\color{blue}{B_1})\times\mathbb{P}(\color{blue}{B_2}\mid \color{blue}{B_1}) + \mathbb{P}(\color{brown}{R_1})\times\mathbb{P}(\color{blue}{B_2}\mid \color{brown}{R_1})$

The conditional probability of a colour being drawn only depends on the count of how many of that specific colour were previously drawn. The probability of drawing a colour decreases by $\frac 16$ for each previous time that colour had been drawn (until $0$).

$\mathbb{P}(\color{blue}{B_2}\mid \color{purple}{P_1}) = \mathbb{P}(\color{blue}{B_2} \mid \color{brown}{R_1}) = \frac 2 6 = \mathbb{P}(\color{blue}{B_1})$

$\mathbb{P}(\color{blue}{B_2}\mid \color{blue}{B_1}) = \frac 1 6 = \mathbb{P}(\color{blue}{B_1}) - \frac 16$

NB: Green is the exception. The probability of green being drawn increases by $\frac 16$ for each time a different coloured ball had been previously drawn.

What is the probability of 3 green, 1 purple, 1 blue and 1 brown ball remaining in the urn after 3 selections?

What we are looking for is the probability of drawing 1 blue, and 2 brown balls, in any order.

$$\mathbb{P}(S_3\equiv\{\color{green}{(g,3)},\color{purple}{(p,1)},\color{blue}{(b,1)},\color{brown}{(r,1)}\}) = \mathbb{P}((\color{blue}{B_1}\cap\color{brown}{R_2}\cap\color{brown}{R_3)} \cup (\color{brown}{R_1}\cap\color{blue}{B_2}\cap\color{brown}{R_3})\cup(\color{brown}{R_1}\cap\color{brown}{R_2}\cap\color{blue}{B_3}))$$

$\mathbb{P}(\color{blue}{B_1}\cap\color{brown}{R_2}\cap\color{brown}{R_3}) = \mathbb{P}(\color{blue}{B_1})\times\mathbb{P}(\color{brown}{R_2}\mid \color{blue}{B_1})\times\mathbb{P}(\color{brown}{R_3} \mid \color{blue}{B_1}\cap\color{brown}{R_2}) = \frac 26 \frac 36 (\frac 36-\frac 16) = \frac {12}{216}$

$\mathbb{P}(\color{brown}{R_1}\cap\color{blue}{B_2}\cap\color{brown}{R_3}) = \mathbb{P}(\color{brown}{R_1})\times\mathbb{P}(\color{blue}{B_2}\mid \color{brown}{R_1})\times\mathbb{P}(\color{brown}{R_3} \mid \color{blue}{B_2}\cap\color{brown}{R_1}) = \frac 36 \frac 26 (\frac 36 - \frac 16) = \frac {12}{216}$

$\mathbb{P}(\color{brown}{R_1}\cap\color{brown}{R_2}\cap\color{blue}{B_3}) = \mathbb{P}(\color{brown}{R_1})\times\mathbb{P}(\color{brown}{R_2}\mid\color{brown}{R_1})\times\mathbb{P}(\color{blue}{B_3} \mid \color{brown}{R_2}\cap\color{brown}{R_1}) = \frac 36 (\frac 36-\frac 16) \frac 26 = \frac {12}{216}$

NB: Note how the order of drawing is not relevant in this case.

$$\therefore \mathbb{P}(S_3\equiv\{\color{green}{(g,3)},\color{purple}{(p,1)},\color{blue}{(b,1)},\color{brown}{(r,1)}\}) = \frac{36}{216}=\frac 16$$