Find the probability of the event that we run out of white balls first

Question

I have $a$ white balls, $b$ black balls, $c$ red balls in a basket. We take balls one by one and do not return. I have to find the probability of the event that we run out of white balls first. If I define:

$A_1$ - The event that we run out of white balls first.
$B_2$ - The event that we run out of black balls second.
$B_3$ - The event that we run out of black balls third.
$C_2$ - The event that we run out of red balls second.
$C_3$ - The event that we run out of red balls third.

I get that:
$P(A_1)=P(B_2 \cap C_3) + P(C_2\cap B_3)$
$P(B_2 \cap C_3) = P(B_2|C_3) * P(C_3)$
$P(C_2\cap B_3) = P(C_2|B_3) * P(B_3)$

I also found out that:

$P(C_3)=\frac{c}{a+b+c}$

$P(B_3)=\frac{b}{a+b+c}$

I need help finding $P(B_2|C_3)$ and $ P(C_2|B_3$

Thanks in advance!

Answer 1

I think you're on the right track. Your calculations for $\Pr(B_3)$ and $\Pr(C_3)$ are correct. I assume you did it something like this: line the balls up in the order they wee drawn. What is the probability that the last ball in the sequence is black (or red)?

The same idea will work to compute $\Pr(B_2)$ and $\Pr(C_2)$. If we run out of black balls second, then either,

• The last ball in the sequence is red and the last ball among the white and black balls is black, or

• The last ball in the sequence is white and the last ball among the red and black balls is black.

Can you finish it now?

Answer 2

Here's an unusual way using stars and bars to solve it.
Let us say the order in which the balls are exhausted is white, black and red.
Lay out all the white balls with one black and one red at the end as shown. $-w-w-w-w-w-w-w-b-r$ There will be $(a+1)$ interstices including the start of the line where black balls can be placed in $\binom{a+b-1}{b-1} = X, say$

There will now be $(a+b+1)$ interstices for the red balls to be placed in $\binom{a+b+c-1}{c-1}$ = Y,say

$X\cdot Y= F_1$ gives total arrangements where white balls, black balls and red balls get exhausted in that order.

Do a similar exercise for arrangements where white,red,black are exhausted in that order and get $F_2$

Unconstrained arrangements = $\frac{(a+b+c)!}{a!b!c!} = Z,say$

$Pr = \frac{F_1+F_2}{Z}$

FURTHER THOUGHTS:
A much simpler approach emerges using direct Pr,sparked by @Lulu remarks.

P(ends first) = (red ends last and black ends second last) + P(black ends last and red ends second last)

= $\frac{c}{a+b+c}\times \frac{b}{a+b}+ \frac{b}{a+b+c}\times \frac{c}{a+c}$