Question

Balls are randomly removed from an urn that initially contains 20 red and 10 blue balls.

(a) What is the probability that all of the red balls are removed before all of the blue ones have been removed? Now suppose that the urn initially contains 20 red, 10 blue, and 8 green balls.

b). Now what is the probability that all of the red balls are removed before all of the blue ones have been removed?

(c) What is the probability that the colors are depleted in the order blue, red, green?

(d) What is the probability that the group of blue balls is the first of the three groups to be removed?

Short answer

Every permutation of the balls is equally likely to be the draw order.

The wanted event is equivalent to the last drawn ball being blue.

a) $p=\frac{1}{3}$

In part b), as green balls are irrelevant for the wanted event and do not change probabilities, the answer is the same:

b) $p=\frac{1}{3}$

Combine a) - for the last drawn color, and b) - for the first exhausted color

c) $\frac{4}{57}$

Sum the possibilities that the order is blue, red, green and blue, green, red. That is the calculation of c)

d) $\frac{63}{171}$

Step-by-step solution

Given Information (Part a)

$20$ red and $10$ blue balls.

Explanation (Part a)

$20$ red and $10$ blue balls.

As the drawing is random each ball is equally likely to be dawn last.

If the last ball is blue, then all the red balls have been drawn before all the blue ones.

If all the red balls have been drawn before all the blue ones then the last ball drawn is blue.

The wanted event is equivalent to the last drawn ball being blue, and as each of the balls is equally likely to be drawn last:

$P\left(\text{the last drawn ball is blue}\right)=\frac{10}{30}=\frac{1}{3}$.

Final Answer (Part a)

The probability that all of the red balls are removed before all of the blue ones have been removed is $\frac{1}{3}$.

Given Infromation (Part b)

$20$ red, $10$ blue and $8$ green balls.

Explanation (Part b)

$20$ red, $10$ blue and $8$ green balls.

Each permutation of only red and blue balls is equally likely. Because every permutation of all the balls is equally likely, and there is an equal number of such permutations for a fixed order of red and blue balls.

Since the wanted event only regards the red and the blue balls, this problem is equivalent to part a)

$P\left(\text{the last drawn from red and blue balls is blue}\right)=\frac{10}{30}=\frac{1}{3}$

Final Answer (Part b)

The probability of$P\left(\text{the last drawn from red and blue balls is blue}\right)=\frac{1}{3}$.

Given Information (Part c)

The probability that the green is the last color to disappear is equal to the probability that the last ball drawn is green:

$P(thelastdrawnballisgreen)=\frac{4}{19}$

Given that, the order is blue, red, green if and only if all the blue balls are drawn before all the red ones. And from part b), this is:

$P\left(\text{the last drawn from red and blue balls is blue}\right)=\frac{10}{30}=\frac{1}{3}$

Explanation (Part c)

Multiplying these numbers we get the probability of the intersection by formula:

$P(A\cap B)=P(A\mid B)P\left(B\right)$

Thus

$P(alltheredballsaredrawnbeforeallblueonesandthelastdrawnballisgreen)=\frac{4}{19}*\frac{1}{3}$

And that intersection is the wanted event, so the solution is:

$P\left(\text{order is blue, red, green}\right)=P\left(BRG\right)=\frac{4}{57}$.

Final Answer (Part c)

The probability of$P\left(\text{order is blue, red, green}\right)=P\left(BRG\right)=\frac{4}{57}$.

Given Information (Part d)

$P(blueballsareexhaustedfirst)=P(BRG)+P(BGR)$

$P(BRG)=\frac{4}{57}$, $P(BGR)=?$

Explanation (Part d)

$P\left(BGR\right)=P(B\text{before}G\mid R\text{is the last ball})=\frac{10}{18}·\frac{20}{38}=\frac{50}{171}$

$P\left(\text{blue balls are exhausted first}\right)=\frac{4}{57}+\frac{50}{171}=\frac{62}{171}\approx 36.26\%$.

Final Answer (Part d)

The probability of $P\left(\text{blue balls are exhausted first}\right)=\frac{62}{171}$.