Conditional probabilities& permutations

Question

There are 5 red balls and 5 yellow balls. What is the probability the third ball is red, given the second ball is red?

I tried to use the law of total probability:

P(3=R|2=R) = P(3=R|1,2=R)P(1=R) + P(3=R|2=R,1=Y)P(1=Y) = 3/16 + 4/16 = 7/16

Is that correct?

Answer 1

No; everything should be conditioned on $2=R$. See

https://en.wikipedia.org/wiki/Law_of_total_probability

So in place of $P(1=R)$ you need $P(1=R|2=R)$.

Answer 2

You have 5 red and 5 yellow balls for a total of 10 balls. You ask about the "third ball" and "second ball" so I presume they are linearly ordered.

I also presume you know that there are $\begin{pmatrix}10 \\ 5\end{pmatrix}= \frac{10!}{5!5!}= 252$ ways to order 10 things, five being identical and the other 5 identical. Knowing that the second ball is red, we are left to order 9 balls, 4 red and 5 yellow.

There are $\begin{pmatrix}9 \\ 5\end{pmatrix}= \frac{9!}{4!5!}= 126$ ways to do that. Of those 126 ways of ordering the balls, if we also have the third ball red then we are left with 8 balls, 3 red and 5 yellow. The number of ways to do that is $\begin{pmatrix}8 \\ 5\end{pmatrix}= \frac{8!}{3!5!}= 56$.

So the probability the third ball is red, given that the second ball is red, is $\frac{56}{126}= \frac{28}{63}= \frac{4}{9}$.