Answer:
The probability of selecting a yellow pencil from a bag containing five green and four yellow pencils can be solved by using probability tree diagrams.
i) Probability tree diagram:
Here, the first event is the selection of the first pencil, which can either be yellow or green. The second event is the selection of the second pencil, which can also be either yellow or green. The diagram can be drawn as follows:
```
G Y
/ \ / \
G Y G Y
/ \ / \ / \ / \
G Y G Y G Y G Y
```
The probability of selecting a yellow pencil is represented by the branches leading to the Y node, and the probability of selecting a green pencil is represented by the branches leading to the G node.
ii) Probability of getting both counters chosen as yellow:
The probability of getting both counters chosen as yellow is the probability of selecting a yellow pencil on the first draw and a yellow pencil on the second draw. The probability of selecting a yellow pencil on the first draw is 4/9, and the probability of selecting a yellow pencil on the second draw is 3/8 (since there are now only 3 yellow pencils left in the bag). The probability of both events occurring is:
(4/9) x (3/8) = 1/6
Therefore, the probability of getting both counters chosen as yellow is 1/6.
iii) Probability of getting one green counter and one yellow counter are chosen:
The probability of getting one green counter and one yellow counter can be found by adding the probabilities of two possible outcomes:
1. The first pencil is green and the second pencil is yellow.
2. The first pencil is yellow and the second pencil is green.
The probability of the first outcome is (5/9) x (4/8) = 5/18, and the probability of the second outcome is (4/9) x (5/8) = 5/18.
Adding these probabilities, we get:
5/18 + 5/18 = 10/18 = 5/9
Therefore, the probability of getting one green counter and one yellow counter are chosen is 5/9.
Explanation: