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.
Draw a tree diagram showing all possible pairs of fruits.
The tree diagram has three initial branches for Basket A: Apple, Orange 1, and Orange 2. From each of these, there are two branches for Basket B: Banana and Mango.
List the sample space.
S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}. Number of possible outcomes = 6
What is the probability of picking one apple and one banana?
Favourable outcome: (Apple, Banana) Number of favourable outcomes = 1 Theoretical Probability (P) = Number of favourable outcomes / Number of possible outcomes = 1/6
Explanation
To solve this, we first identify the distinct fruits in each basket. Basket A has 3 fruits (1 Apple, 2 Oranges treated as distinct objects Orange 1 and Orange 2) and Basket B has 2 fruits (1 Banana, 1 Mango). The tree diagram branches from each fruit in Basket A to every fruit in Basket B, resulting in 3 x 2 = 6 possible pairs. The sample space lists these 6 pairs. The probability is found using the theoretical probability formula provided in the text: P = Number of favourable outcomes / Number of possible outcomes. Since only one pair matches the event of picking an apple and a banana, the probability is 1/6.