Labeling fruits as A1, O1, O2 for basket A and B1, M1 for basket B gives 6 equally likely outcomes. The sample space is {(A1, B1), (A1, M1), (O1, B1), (O1, M1), (O2, B1), (O2, M1)}. P(Apple, Banana) = 1/6.
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?
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Basket A contains 3 fruits: 1 Apple (A), 2 Oranges (O1, O2). Basket B contains 2 fruits: 1 Banana (B), 1 Mango (M).
(i) Tree Diagram:
Start -> A -> B => (A, B)
Start -> A -> M => (A, M)
Start -> O1 -> B => (O1, B)
Start -> O1 -> M => (O1, M)
Start -> O2 -> B => (O2, B)
Start -> O2 -> M => (O2, M)
(ii) Sample Space:
S = {(A, B), (A, M), (O1, B), (O1, M), (O2, B), (O2, M)}, so n(S) = 6.
(iii) Favourable outcome = {(A, B)}, which is 1.
P(Apple and Banana) = 1 / 6 (or about 0.167).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/