Priya draws counters at random, without replacement, from a bag of 4 red and 6 blue counters. The diagram shows the outcomes of two draws. 4/10 6/10 3/9 6/9...

This tree diagram tracks two draws without replacement from a bag of 4 red and 6 blue counters (10 counters in total); the second-stage branches use a denominator of 9 because one counter has already been removed, and conditional probability (part (c)) is found by dividing the probability of the joint event by the probability of the condition.

1. (a) [2]. \(P(\text{both red}) = \dfrac{4}{10}\times\dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\).
2. (b) [2]. \(P(\text{both blue}) = \dfrac{6}{10}\times\dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).
3. (c) [3]. \(P(\text{same colour}) = P(\text{both red}) + P(\text{both blue}) = \dfrac{12}{90}+\dfrac{30}{90} = \dfrac{42}{90}\). Conditional probability restricts attention to this "same colour" outcome and asks what share of it is "both blue": \[P(\text{both blue}\mid\text{same colour}) = \frac{P(\text{both blue})}{P(\text{same colour})} = \frac{30/90}{42/90} = \frac{30}{42} = \frac{5}{7}\]
4. (d) [2]. Adding one yellow counter increases the bag to 11 counters, but the number of red counters stays at 4, so \(P(\text{first red})\) becomes \(\dfrac{4}{11}\), which is smaller than the original \(\dfrac{4}{10}\). Since the first-draw probability of red falls and the second-draw probability of red (given a red first) also falls slightly for the same reason, \(P(\text{both red})\) would be lower than before.