There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
- (i) Draw a tree diagram showing all possible pairs of fruits.
- (ii) List the sample space.
- (iii) What is the probability of picking one apple and one banana?
Step-by-step solution
Idea: Give the two oranges different labels (Orange 1, Orange 2), so that all three fruits in basket A are equally likely. Then each of the 3 × 2 = 6 paths of the tree is equally likely.
(i) Draw a tree diagram showing all possible pairs of fruits.
- First stage (basket A): from a starting point draw 3 branches: Apple, Orange 1, Orange 2. Each fruit is equally likely, so each branch has probability 13.1 mark
- Second stage (basket B): from each of these draw 2 branches: Banana and Mango, each with probability 12. The tree has 3 × 2 = 6 complete paths (see the diagram).1 mark
(ii) List the sample space.
- Read each path from start to end:
S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}, n(S) = 6.
(If the oranges are not told apart, the pairs are {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}, but then the orange pairs are twice as likely as the apple pairs.)1 mark
(iii) What is the probability of picking one apple and one banana?
- Favourable outcome: (Apple, Banana), just 1 of the 6 equally likely outcomes. P = 16. (Along the path: 13 × 12 = 16, the same answer.)1 mark
Check: P(an orange and a banana) = 26 = 13, P(apple and mango) = 16, P(an orange and a mango) = 13. Total 16 + 13 + 16 + 13 = 1 ✓.
Answer to write in the exam
(i)
Basket A: Apple, Orange 1, Orange 2 (13 each)
Basket B from each: Banana, Mango (12 each)
∴ Tree with 3 × 2 = 6 paths (as drawn)
(ii)
∴ S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}
(iii)
Favourable outcomes = 1; total = 6
∴ P(apple and banana) = 16
Common mistakes that cost marks
- Drawing only 2 branches for basket A (Apple, Orange) and then saying each of the 4 pairs has probability 14. Basket A has two oranges, so an orange is twice as likely as the apple.
- Answering 15 by counting 5 fruits altogether. You pick from each basket separately; the outcomes are pairs, and there are 6 of them.
- Adding instead of multiplying along the path: 13 + 12 = 56 is not a probability of a single path.
How this can come in the exam
At a school fair, Box P has 2 blue tokens and 1 yellow token. Box Q has 1 red and 1 green token. A child takes one token from each box without looking.
(a) How many equally likely outcomes are there? (1)
(b) List the outcomes, labelling the blue tokens B1 and B2. (1)
(c) Find the probability of getting a blue and a green token. (2)
Show answer
(a) 3 × 2 = 6.(b) {(B1, R), (B1, G), (B2, R), (B2, G), (Y, R), (Y, G)}.
(c) Favourable: (B1, G), (B2, G). P = 26 = 13.
A coin is tossed and a fair die is rolled. The probability of getting heads and a 5 is
- 12
- 16
- 18
- 112
Show answer
(D) 112
2 × 6 = 12 equally likely outcomes; only (H, 5) is favourable: 112.
Try one yourself
Bag X has 1 red and 1 white ball; bag Y has 1 red, 1 white and 1 black ball. One ball is taken from each bag. Find the probability that both balls are red.
Show answer
2 × 3 = 6 equally likely pairs; only (red, red) is favourable. P = 16.
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