NCERT Solutions for Class 9 Maths Chapter 7 — End-of-Chapter Exercises#

Statement	Blank
(i) The probability of an impossible event is ___.	0
(ii) The set of all possible outcomes of a random experiment is called the ___.	sample space
(iii) The probability of an event that is certain to happen is ___.	1
(iv) Tossing a fair coin has a probability of ___ for getting heads.	$\dfrac{1}{2}$

The count 15 is the frequency. The fraction of all students is the relative frequency:

$$\text{Relative frequency} = \dfrac{15}{50}$$

$$\text{Relative frequency} = \dfrac{3}{10} = 0.3$$

∴ The relative frequency is $\dfrac{3}{10}$ (0.3).

Not equally likely. A car in working order usually starts; whether it starts depends on its condition, fuel and battery, not on chance alone.

Equally likely. A fair coin is equally likely to show heads or tails.

Equally likely. Each face, 1 to 6, has the same chance on a fair die.

Not equally likely (if the outcomes are “red” and “blue”):

$$P(\text{red}) = \dfrac{3}{10}, \quad P(\text{blue}) = \dfrac{7}{10}$$

Blue is more likely. (Each individual marble is equally likely, but the two colours are not.)

Equally likely, for school problems: a boy and a girl are taken to be equally likely. (In real records, slightly more boys than girls are born, so it is only approximately even.)

$$S = \{HH, HT, TH, TT\}, \quad n(S) = 4$$

At least one head: $\{HH, HT, TH\}$, 3 outcomes:

$$P(\text{at least one head}) = \dfrac{3}{4}$$

$$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, \quad n(S) = 10$$

Even numbers: $\{2, 4, 6, 8, 10\}$, 5 outcomes:

$$P(\text{even}) = \dfrac{5}{10} = \dfrac{1}{2}$$

$$S = \{1, 2, 3, 4, 5, 6\}, \quad n(S) = 6$$

Greater than 4: $\{5, 6\}$, 2 outcomes:

$$P(\text{greater than } 4) = \dfrac{2}{6} = \dfrac{1}{3}$$

$$S = \{R_1, R_2, R_3, B_1, B_2, G\}, \quad n(S) = 6$$

Not red: $\{B_1, B_2, G\}$, 3 outcomes:

$$P(\text{not red}) = \dfrac{3}{6} = \dfrac{1}{2}$$

$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}, \quad n(S) = 8$$

Exactly two heads: $\{HHT, HTH, THH\}$, 3 outcomes:

$$P(\text{exactly two heads}) = \dfrac{3}{8}$$

$$S = \{\text{strawberry, lemon, mint}\}$$$$P(\text{strawberry}) = \dfrac{1}{3}$$

Shirt	Jeans	Khakis	Shorts
Red	Red + Jeans	Red + Khakis	Red + Shorts
Blue	Blue + Jeans	Blue + Khakis	Blue + Shorts

There are $2 \times 3 = $ 6 outfits.

$$P = \dfrac{20}{1000} = 0.02$$

This covers the two middle groups:

$$\text{Cases} = 210 + 325 = 535$$

$$P = \dfrac{535}{1000} = 0.535$$

$$P = \dfrac{445}{1000} = 0.445$$

(Check: $0.02 + 0.535 + 0.445 = 1$ ✓)

Favourable cards: P, E, E, C — 4 cards:

$$P(\text{P, E or C}) = \dfrac{4}{5}$$

$$P(\text{E}) = \dfrac{2}{5}$$

$$P(\text{not E}) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$$

$$P(8) = \dfrac{1}{8}$$

Odd numbers: 1, 3, 5, 7.

$$P(\text{odd}) = \dfrac{4}{8} = \dfrac{1}{2}$$

Greater than 2: 3, 4, 5, 6, 7, 8.

$$P(\text{greater than } 2) = \dfrac{6}{8} = \dfrac{3}{4}$$

Every number on the spinner is less than 9, so this is certain:

$$P(\text{less than } 9) = \dfrac{8}{8} = 1$$

Multiples of 3: 3, 6.

$$P(\text{multiple of } 3) = \dfrac{2}{8} = \dfrac{1}{4}$$

For the first ball there are 9 balls. The first ball is not put back, so for the second ball there are only 8 balls, and the numbers depend on the first colour.

Multiplying along the path Red → Blue:

$$P(\text{RB}) = \dfrac{4}{9} \times \dfrac{5}{8}$$

$$P(\text{RB}) = \dfrac{20}{72}$$

$$P(\text{RB}) = \dfrac{5}{18}$$

Multiplying along the path Blue → Blue:

$$P(\text{BB}) = \dfrac{5}{9} \times \dfrac{4}{8}$$

$$P(\text{BB}) = \dfrac{20}{72}$$

$$P(\text{BB}) = \dfrac{5}{18}$$

Probability 0: “the sum of the two numbers is 13”. The largest possible sum is $6 + 6 = 12$, so this cannot happen.

Probability 1: “the sum of the two numbers is between 2 and 12” (or “each die shows a number less than 7”). Every one of the 36 outcomes satisfies this, so it is certain.

(No single outcome such as (3, 4) has probability 1 — each of the 36 outcomes has probability $\dfrac{1}{36}$. A certain result has to be an event that contains every outcome.)

$S$ = all 36 ordered pairs $(a, b)$ with $a, b \in \{1, \ldots, 6\}$. The sums go from 2 to 12; the primes greater than 5 are 7 and 11.

• Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) — 6 outcomes
• Sum 11: (5, 6), (6, 5) — 2 outcomes

$$P = \dfrac{6 + 2}{36}$$

$$P = \dfrac{8}{36} = \dfrac{2}{9}$$

There are 9 balls. First find the probability that both have the same colour, drawing one after the other:

$$P(\text{both red}) = \dfrac{4}{9} \times \dfrac{3}{8} = \dfrac{12}{72}$$

$$P(\text{both green}) = \dfrac{3}{9} \times \dfrac{2}{8} = \dfrac{6}{72}$$

$$P(\text{both blue}) = \dfrac{2}{9} \times \dfrac{1}{8} = \dfrac{2}{72}$$$$P(\text{same colour}) = \dfrac{12 + 6 + 2}{72} = \dfrac{20}{72} = \dfrac{5}{18}$$

So:

$$P(\text{different colours}) = 1 - \dfrac{5}{18}$$

$$P(\text{different colours}) = \dfrac{13}{18}$$

$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}, \quad n(S) = 8$$

First coin H and exactly two heads: $\{HHT, HTH\}$, 2 outcomes:

$$P = \dfrac{2}{8} = \dfrac{1}{4}$$

The sample space is all arrangements of 1, 2, 3, 4, such as 1234, 1243, …, 4321:

$$n(S) = 4 \times 3 \times 2 \times 1 = 24$$

The number is even when its last digit is 2 or 4. For each choice of last digit, the other three digits can be arranged in $3 \times 2 \times 1 = 6$ ways:

$$\text{Even numbers} = 2 \times 6 = 12$$

$$P(\text{even}) = \dfrac{12}{24} = \dfrac{1}{2}$$

Each guess is correct (✓) with probability $\dfrac{1}{4}$ and wrong (✗) with probability $\dfrac{3}{4}$. The sample space of results is

$$S = \{✓✓✓, ✓✓✗, ✓✗✓, ✗✓✓, ✓✗✗, ✗✓✗, ✗✗✓, ✗✗✗\}$$

Exactly 2 correct: ✓✓✗, ✓✗✓, ✗✓✓. Each has probability

$$\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{3}{64}$$$$P(\text{exactly 2 correct}) = 3 \times \dfrac{3}{64}$$

$$P(\text{exactly 2 correct}) = \dfrac{9}{64}$$

(Counting answers directly: $4^3 = 64$ equally likely ways to answer, and $3 \times 3 = 9$ of them have exactly two right.)

With replacement, every number 1–4 is possible on both draws.

Without replacement, the second number cannot equal the first.

$$n(S_{\text{with replacement}}) = 4 \times 4 = 16$$

$$n(S_{\text{without replacement}}) = 4 \times 3 = 12$$

$$S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}$$

$n(S) = 2 \times 6 = 12$.

With three coins, the number of heads can be 0 (TTT), 1, 2 or 3 (HHH).

List	Sample space?	Reason
(i) {1, 2, 3}	No	misses 0 — the outcome TTT is not included
(ii) {0, 1, 2}	No	misses 3 — the outcome HHH is not included
(iii) {0, 1, 2, 3, 4}	No	includes 4, which cannot happen with 3 coins
(iv) {0, 1, 2, 3}	Yes	contains every possible number of heads, each once

∴ (iv) {0, 1, 2, 3} is the sample space. (Its outcomes are not equally likely: $P(0) = \dfrac{1}{8}$, $P(1) = \dfrac{3}{8}$, $P(2) = \dfrac{3}{8}$, $P(3) = \dfrac{1}{8}$.)

When a drop lands at random, the probability of landing in a region is that region’s share of the total area.

$$\text{Rectangle} = 3 \times 2 = 6 \text{ m}^2$$

The circle has radius $\dfrac{1}{2}$ m:

$$\text{Circle} = \pi\left(\dfrac{1}{2}\right)^2 = \dfrac{\pi}{4} \text{ m}^2$$$$P(\text{inside circle}) = \dfrac{\pi/4}{6}$$

$$P(\text{inside circle}) = \dfrac{\pi}{24}$$

With $\pi \approx \dfrac{22}{7}$:

$$P(\text{inside circle}) \approx \dfrac{22}{168} = \dfrac{11}{84} \approx 0.13$$

∴ The probability is $\dfrac{\pi}{24}$ ≈ 0.13.