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.

• I expect it's easier to work with probabilities. So, let $R, Y,C$ be the respective probabilities. Then $R+B+C=1$ is automatic, so you can eliminate one variable right at the start. – lulu Feb 7 at 15:16
• Got it, probability of obtaining a yellow is 1/3 i think. Thanks for the advice. – Ahmed Hussain-Shah Feb 7 at 15:35
• Inspired guesswork can also give an answer, recognizing that $13 = 3^2 + 2^2$ and $36 = 6^2.$ – David K Feb 7 at 15:41

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 $$\frac{1}{3}$$.