NCERT Solutions for Class 9th Maths Chapter 7 Exercise Set 7.4 — Tree diagrams
Book page 169 Updated on2026-09-08
Q1.
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?
Answer
(i) Basket A holds 3 fruits, so the first stage has three branches; label the two oranges separately so that all three are equally likely. Basket B holds 2 fruits, so each branch then splits in two.
Basket A has 3 fruits, Basket B has 2, so the tree ends in 3 × 2 = 6 equally likely pairs.
(ii) Reading the six leaves:
S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)} n(S) = 3 × 2 = 6, each with probability 1/6
(iii) Exactly one leaf is favourable:
P(apple and banana) = 1/6 ≈ 0.167 or, multiplying along that path: 1/3 × 1/2 = 1/6 ✓
Why it happens: the two oranges must be kept apart. If you write the sample space by type — {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)} — you get four outcomes, and dividing gives the wrong answer 1/4. Those four are not equally likely: picking an orange from Basket A has probability 2/3 while picking the apple has probability 1/3, because there are two oranges and only one apple. Labelling them Orange 1 and Orange 2 restores equal likelihood, and only then is counting leaves legitimate.
Check it yourself: add the six leaf probabilities: 6 × 1/6 = 1. Now group them by type — (Apple, ·) totals 2/6 = 1/3 and (Orange, ·) totals 4/6 = 2/3, matching the 1 apple to 2 oranges in the basket. The tree and the basket agree.
Q2.
Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same. (i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes? (ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Answer
The box holds 3 + 4 + 2 = 9 pens. Because the pen is put back, the box is identical for your friend's turn, so both stages carry the same branch probabilities: red 3/9, black 4/9, green 2/9.
(i) Three colours at each stage give 3 × 3 = 9 ordered colour pairs:
S = {RR, RB, RG, BR, BB, BG, GR, GB, GG}
Two picks with replacement. The nine path probabilities add to 81/81 = 1; the three shaded-alike paths RR, BB, GG give 29/81.
These nine pairs are not equally likely, so their probabilities must be got by multiplying along each path, not by writing 1/9 each.
(ii) "Same colour" is made of the three paths RR, BB and GG. Multiply along each, then add:
Why it happens: the branches on one path happen one after the other, so their probabilities multiply; different paths are alternatives, so their probabilities add. Replacement is what makes the second-stage fractions the same as the first — the box your friend picks from is exactly the box you picked from. Had you kept the pen, the second denominator would be 8 and black-then-black would be 4/9 × 3/8.
Check it yourself: all nine path probabilities add to 81/81 = 1, and 29/81 + 52/81 = 1, so different colours is the more likely result. Black is the single most likely match (16/81) simply because there are more black pens than any other colour.