GCSE Probability question relating to trees, help please. There are only red counters, yellow counters and blue counters in a bag.
Kevin takes at random a counter from the bag.
He puts the counter back in the bag.
Lethna takes at random a counter from the bag.
She puts the counter back in the bag.
The probability that both counters are red or that both counters are yellow is 13/36.
The probability that the first counter is red and the second counter is not red is 1/4.
Seb takes at random a counter from the bag.
Work out the probability that Seb takes a yellow counter.
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So far, from the text, I've formed $2$ equations but since the number of unknowns is greater than the number of equations, I cannot solve.
$[(R^2)/(R+Y+B)^2] + [(Y^2)/(R+Y+B)^2] = 13/36$
$[(R)/(R+Y+B)]*[(Y+B)/(R+Y+B)] = 1/4$
Any help would be appreciated. Thanks.

Let $R$, $Y$ and $B$ be the events in which a red, yellow and blue counter is drawn from the bag, respectively. We have:
$$P(R) \cdot (1 - P(R)) = \frac{1}{4} \iff P(R)^2 - P(R) + \frac{1}{4} = 0 \iff P(R) = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1}{2}$$
$$P(R) \cdot P(R) + P(Y) \cdot P(Y) = \frac{13}{36} \iff P(Y)^2 = \frac{13 - 9}{36} = \frac{4}{36} = \frac{1}{9} \iff P(Y) = \frac{1}{3}$$
The probability that Seb draws a yellow counter, thus equals $\frac{1}{3}$.
