NCERT Solutions Class 9 Maths Exercise 7.4 — The Mathematics of Maybe: Introduction to Probability

NCERT Class 9 Mathematics — Ganita Manjari Part I, The Mathematics of Maybe: Introduction to Probability, Exercise Set 7.4. All 2 questions solved.

A tree diagram lays a multi-stage experiment out as branches: one level of branching per stage, one complete path per outcome. Two rules make it work.

• Along a path, multiply. The probability of a complete outcome is the product of the probabilities on the branches you travelled.
• Across the leaves, add. The probability of an event is the sum over every leaf that belongs to it, and the probabilities of all the leaves together must come to $1$.

Key insight. Draw one branch per object, not one per name. Basket A holds an apple and two oranges, so its stage has three branches, not two. A tree with an “Apple” branch beside an “Orange” branch has leaves that are not equally likely, and counting them will give the wrong answer while looking perfectly reasonable.

Question 1

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.

Solution.

(i) Draw a tree diagram showing all possible pairs of fruits

Basket $A$ contributes the first stage. It holds three fruits, so it gets three branches — the apple and each orange separately, written $O_1$ and $O_2$ — each with probability $\frac13$. Basket $B$ contributes the second, with two branches of probability $\frac12$ each.

Here $A$ is the apple, $O_1$ and $O_2$ the two oranges, $B$ the banana and $M$ the mango.

Each complete path has probability $\frac13 \times \frac12 = \frac16$, and there are six paths — so the six leaf probabilities add to $1$, which is the check that the tree is complete.

(ii) List the sample space

Reading the six leaves off the tree:

$$S = \{(A, B), (A, M), (O_1, B), (O_1, M), (O_2, B), (O_2, M)\}, \qquad n(S) = 3 \times 2 = 6$$

Described by fruit type rather than by individual fruit, these six outcomes are apple–banana, apple–mango, orange–banana (twice) and orange–mango (twice).

(iii) What is the probability of picking one apple and one banana?

Exactly one leaf is favourable — the path apple, then banana:

$$P(\text{apple and banana}) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$$

Why the two oranges had to be separated. Suppose the tree had been drawn with just an “Apple” branch and an “Orange” branch. It would have four leaves — $(A,B)$, $(A,M)$, $(O,B)$, $(O,M)$ — and counting one favourable out of four would give $\frac14$. That is wrong, and the reason is that those four leaves are not equally likely: an orange is twice as likely to be drawn as the apple, so the two orange leaves each carry $\frac13$ and the two apple leaves each carry $\frac16$.

$$\tfrac16 + \tfrac16 + \tfrac13 + \tfrac13 = 1 \ \checkmark$$

Drawing the branches one per fruit avoids the whole problem, because then every leaf really does carry the same probability.

Question 2

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.

Solution. There are $3 + 4 + 2 = 9$ pens, so a single draw gives

$$P(R) = \frac{3}{9} = \frac13, \qquad P(K) = \frac{4}{9}, \qquad P(G) = \frac{2}{9}$$

writing $K$ for black. Because the pen is put back, the box is identical for the second draw, so the second stage carries exactly the same three probabilities. That is what “with replacement” buys you.

(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?

Three colours at each of two stages give $3 \times 3 = 9$ outcomes:

The second-stage branches carry the same $\frac39$, $\frac49$, $\frac29$ as the first, so they are not labelled again on the diagram. The nine outcomes are

$$S = \{RR,\ RK,\ RG,\ KR,\ KK,\ KG,\ GR,\ GK,\ GG\}$$

and the nine leaf probabilities add to $\frac{81}{81} = 1$ ✓ — note that $RK$ and $KR$ are different outcomes, since the first letter is your pen and the second your friend’s.

(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

“Same colour” is the event $\{RR, KK, GG\}$ — three of the nine leaves. Add their probabilities:

$$P(\text{same}) = \left(\frac39\right)^2 + \left(\frac49\right)^2 + \left(\frac29\right)^2 = \frac{9}{81} + \frac{16}{81} + \frac{4}{81} = \frac{29}{81}$$

which is about $0.358$.

Note that it is not $\frac39$, the answer you would get by wrongly reasoning that “three of the nine leaves are favourable, so the probability is $\frac13$”. The nine leaves are not equally likely — $KK$ alone carries nearly four times the probability of $GG$ — so they have to be added, not counted.

As a check, the complementary event (different colours) has probability $1 – \frac{29}{81} = \frac{52}{81}$, and adding the six mixed leaves directly gives $\frac{12 + 6 + 12 + 8 + 6 + 8}{81} = \frac{52}{81}$ ✓

Common mistakes

• Question 1, drawing only two branches from basket $A$. The basket holds three fruits. Two branches give four leaves that are not equally likely, and the answer $\frac14$ that follows is wrong.
• Question 1(iii), adding instead of multiplying. $\frac13 + \frac12 = \frac56$ is not a probability of anything here. Along a path you multiply; it is across leaves that you add.
• Question 2, forgetting that the pen is replaced. With replacement the second stage is $\frac39, \frac49, \frac29$ again. Without it, the second stage would be out of $8$ pens and would depend on what the first draw took — a different and harder tree.
• Question 2(ii), counting leaves instead of adding probabilities. Three of the nine leaves are favourable, but $\frac39$ is not the answer. Unequal leaves must be added.
• Treating $RK$ and $KR$ as the same outcome. They are different: one is you red and your friend black, the other is the reverse. Merging them leaves six outcomes and breaks the total.
• Not checking that the leaves add to $1$. It costs one line and catches a missing branch or a mislabelled probability straight away.

Practise next

• The End-of-Chapter Exercises, Questions 10 and 13 — tree diagrams again, but drawing without replacement, where the second stage changes.
• Exercise Set 7.3 — revise how a sample space is listed before a tree is drawn over it.