We have $10$ white and $20$ black balls in box. We take one by one out of it all together $13$ balls. Find the probability that last ball is white.

Question

We have $10$ white and $20$ black balls in box. We take one by one (and we don't put it back in box) out of it all together $13$ balls. Find the probability that last ball is white.

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My try:

Enumerate the balls with numbers from $1$ to $30$ and arrange them all in a line.

Then we have $n= 30!$ and $m = 10\cdot 29!$ So $P= {1\over 3}$

Is this correct?

Answer 1

So, applying counting:

• there are $30$ ways to extract the 1st ball
• there are $29$ ways to extract the 2nd ball
• there are $28$ ways to extract the 3rd ball
• ...
• there are $30-13+1=18$ ways to extract the 13th ball

Altogether $30\cdot29\cdot28\cdot...\cdot18$

But we want the 13th ball to be white, so we put $1$ white ball aside and

• there are $29$ ways to extract the 1st ball
• there are $28$ ways to extract the 2nd ball
• there are $27$ ways to extract the 3rd ball
• ...
• there are $29-12+1=18$ ways to extract the 12th ball

but there are $10$ white balls and each can be put aside, altogether $\color{red}{10}\cdot29\cdot28\cdot...\cdot18$.

As a result $$P=\frac{\color{red}{10}\cdot29\cdot28\cdot...\cdot18}{30\cdot29\cdot28\cdot...\cdot18}=\frac{1}{3}$$

Answer 2

All the thirty balls have equal probability of being the thirteenth ball picked. So the answer is $\frac{10}{30}=\frac13$.