Probability — NCERT Solutions

Given: A die is rolled. Sample space S = {1, 2, 3, 4, 5, 6}.

Event E = {4} (die shows 4)

Event F = {2, 4, 6} (die shows even number)

Check for mutual exclusivity: Two events are mutually exclusive if their intersection is empty, i.e., E ∩ F = φ.

$$E \cap F = \{4\} \cap \{2, 4, 6\} = \{4\} \neq \phi$$

Since E ∩ F ≠ φ, E and F are NOT mutually exclusive.

Given: A die is thrown. Sample space S = {1, 2, 3, 4, 5, 6}.

(i) A: a number less than 7
$$A = \{1, 2, 3, 4, 5, 6\} = S$$

(ii) B: a number greater than 7
No face of a die shows a number greater than 7.
$$B = \phi$$

(iii) C: a multiple of 3
$$C = \{3, 6\}$$

(iv) D: a number less than 4
$$D = \{1, 2, 3\}$$

(v) E: an even number greater than 4
$$E = \{6\}$$

(vi) F: a number not less than 3 (i.e., ≥ 3)
$$F = \{3, 4, 5, 6\}$$

Now the required set operations:

$$A \cup B = S \cup \phi = S = \{1,2,3,4,5,6\}$$

$$A \cap B = S \cap \phi = \phi$$

$$B \cup C = \phi \cup \{3,6\} = \{3,6\}$$

$$E \cap F = \{6\} \cap \{3,4,5,6\} = \{6\}$$

$$D \cap E = \{1,2,3\} \cap \{6\} = \phi$$

$$A - C = \{1,2,3,4,5,6\} - \{3,6\} = \{1,2,4,5\}$$

$$D - E = \{1,2,3\} - \{6\} = \{1,2,3\}$$

$F' = S - F = \{1,2,3,4,5,6\} - \{3,4,5,6\} = \{1,2\}$

$$E \cap F' = \{6\} \cap \{1,2\} = \phi$$

$$F' = \{1,2\}$$

Given: A pair of dice is rolled. The sample space has 36 equally likely outcomes.

Event A: sum > 8
$$A = \{(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}$$

Event B: 2 occurs on either die
$$B = \{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)\}$$

Event C: sum is at least 7 AND a multiple of 3 (i.e., sum = 9 or 12)
$$C = \{(3,6),(4,5),(5,4),(6,3),(6,6)\}$$

Checking mutual exclusivity:

$A \cap B$: For sum > 8 with a 2 on either die, the other die must show > 6, which is impossible.
$$A \cap B = \phi$$
So A and B are mutually exclusive.

$A \cap C$: $\{(3,6),(4,5),(5,4),(6,3),(6,6)\} \neq \phi$
So A and C are not mutually exclusive.

$B \cap C$: For C, the sums are 9 or 12. With a 2 on either die, maximum sum = 2+6 = 8 < 9.
$$B \cap C = \phi$$
So B and C are mutually exclusive.

Conclusion: The pairs (A, B) and (B, C) are mutually exclusive.

Given: Three coins are tossed. Sample space:
$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

The events are:

• $A = \{HHH\}$
• $B = \{HHT, HTH, THH\}$
• $C = \{TTT\}$
• $D = \{HHH, HHT, HTH, HTT\}$

(i) Mutually exclusive events:
Two events are mutually exclusive if their intersection is empty.

• $A \cap B = \phi$ ✓
• $A \cap C = \phi$ ✓
• $B \cap C = \phi$ ✓
• $A \cap D = \{HHH\} \neq \phi$ ✗
• $B \cap D = \{HHT, HTH\} \neq \phi$ ✗
• $C \cap D = \phi$ ✓

Mutually exclusive pairs: (A, B), (A, C), (B, C), (C, D).

(ii) Simple events:
A simple event has only one sample point.

• $A = \{HHH\}$ — Simple
• $C = \{TTT\}$ — Simple

(iii) Compound events:
A compound event has more than one sample point.

• $B = \{HHT, HTH, THH\}$ — Compound
• $D = \{HHH, HHT, HTH, HTT\}$ — Compound

Given: Three coins are tossed.
$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

(i) Two mutually exclusive events:
Let $E_1$ = event of getting all heads = $\{HHH\}$
Let $E_2$ = event of getting all tails = $\{TTT\}$
$E_1 \cap E_2 = \phi$, so they are mutually exclusive.

(ii) Three mutually exclusive and exhaustive events:
Let $A$ = event of getting no head = $\{TTT\}$
Let $B$ = event of getting exactly one head = $\{HTT, THT, TTH\}$
Let $C$ = event of getting at least two heads = $\{HHT, HTH, THH, HHH\}$
$A \cap B = \phi$, $B \cap C = \phi$, $A \cap C = \phi$ (mutually exclusive)
$A \cup B \cup C = S$ (exhaustive)

(iii) Two events which are NOT mutually exclusive:
Let $P$ = event of getting at least two heads = $\{HHT, HTH, THH, HHH\}$
Let $Q$ = event of getting head on first coin = $\{HHH, HHT, HTH, HTT\}$
$P \cap Q = \{HHH, HHT, HTH\} \neq \phi$, so they are not mutually exclusive.

(iv) Two mutually exclusive but NOT exhaustive events:
Let $X$ = event of getting exactly one head = $\{HTT, THT, TTH\}$
Let $Y$ = event of getting exactly two heads = $\{HHT, HTH, THH\}$
$X \cap Y = \phi$ (mutually exclusive)
$X \cup Y \neq S$ (not exhaustive, since $HHH$ and $TTT$ are not included)

(v) Three mutually exclusive but NOT exhaustive events:
Let $L$ = $\{HHH\}$, $M$ = $\{TTT\}$, $N$ = $\{HTT\}$
$L \cap M = \phi$, $M \cap N = \phi$, $L \cap N = \phi$ (mutually exclusive)
$L \cup M \cup N = \{HHH, TTT, HTT\} \neq S$ (not exhaustive)

Given: Two dice are thrown. Sample space has 36 outcomes.

$$A = \{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

$$B = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$$

$$C = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$$
(All pairs where sum ≤ 5)

(i) A' (not A) = B (getting an odd number on the first die)
$$A' = B = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$$

(ii) not B = A (getting an even number on the first die)
$$B' = A = \{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

(iii) A or B = A ∪ B = S (every outcome has either even or odd on first die)
$$A \cup B = S$$

(iv) A and B = A ∩ B = φ (first die cannot be both even and odd)
$$A \cap B = \phi$$

(v) A but not C = A − C = A ∩ C'
From A, remove elements that are also in C:
$A \cap C = \{(2,1),(2,2),(2,3),(4,1)\}$
$$A - C = A \cap C' = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

(vi) B or C = B ∪ C
$B \cup C = B \cup (C - B)$; elements in C not in B: $\{(2,1),(2,2),(2,3),(4,1)\}$
$$B \cup C = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$$

(vii) B and C = B ∩ C
Elements common to both B and C (odd first die and sum ≤ 5):
$$B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2),(5,1)\}$$

(viii) A ∩ B' ∩ C'
Since $B' = A$, we have $A \cap B' = A \cap A = A$.
So $A \cap B' \cap C' = A \cap C' = A - C$
$$A \cap B' \cap C' = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

Using the events defined in Q.6:

• A = getting even number on first die
• B = getting odd number on first die
• C = sum ≤ 5
• $A' = B$, $B' = A$

(i) A and B are mutually exclusive — TRUE
$A \cap B = \phi$ because the first die cannot show both an even and an odd number simultaneously.

(ii) A and B are mutually exclusive and exhaustive — TRUE
$A \cap B = \phi$ (mutually exclusive) and $A \cup B = S$ (every outcome has either even or odd on first die, so exhaustive).

(iii) A = B' — TRUE
$B' = S - B$ = all outcomes where first die is NOT odd = all outcomes where first die is even = A.
Hence $A = B'$.

(iv) A and C are mutually exclusive — FALSE
$A \cap C = \{(2,1),(2,2),(2,3),(4,1)\} \neq \phi$.
So A and C are not mutually exclusive.

(v) A and B' are mutually exclusive — FALSE
Since $B' = A$, we have $A \cap B' = A \cap A = A \neq \phi$.
So A and B' are not mutually exclusive.

(vi) A', B', C are mutually exclusive and exhaustive — FALSE
$A' = B$ and $B' = A$.
$A' \cap B' = B \cap A = \phi$ ✓
But $A' \cap C = B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2),(5,1)\} \neq \phi$.
Also $B' \cap C = A \cap C \neq \phi$.
So they are NOT mutually exclusive, hence the statement is FALSE.

For a valid probability assignment, each $P(\omega_i)$ must satisfy:

1. $0 \leq P(\omega_i) \leq 1$ for all $i$
2. $\sum_{i=1}^{7} P(\omega_i) = 1$

(a) Values: 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6
All values are between 0 and 1. ✓
Sum $= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00$ ✓
Valid assignment.

(b) Values: $\frac{1}{7}$ each
All values are between 0 and 1. ✓
Sum $= 7 \times \frac{1}{7} = 1$ ✓
Valid assignment.

(c) Values: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7
All values are between 0 and 1. ✓
Sum $= 0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8 \neq 1$ ✗
NOT a valid assignment.

(d) Values: −0.1, 0.2, 0.3, 0.4, −0.2, 0.1, 0.3
$P(\omega_1) = -0.1 < 0$ and $P(\omega_5) = -0.2 < 0$ ✗
Probability cannot be negative.
NOT a valid assignment.

(e) Values: $\frac{1}{14}, \frac{2}{14}, \frac{3}{14}, \frac{4}{14}, \frac{5}{14}, \frac{6}{14}, \frac{15}{14}$
$\frac{15}{14} > 1$ ✗
Probability cannot exceed 1.
NOT a valid assignment.

Conclusion: Assignments (c), (d), and (e) are not valid.

Given: A coin is tossed twice.

Sample space: $S = \{HH, HT, TH, TT\}$, so $n(S) = 4$.

Let A = event that at least one tail occurs.
$$A = \{HT, TH, TT\}$$
$$n(A) = 3$$

$$P(A) = \frac{n(A)}{n(S)} = \frac{3}{4}$$

The probability that at least one tail occurs is $\dfrac{3}{4}$.

Given: A die is thrown. Sample space $S = \{1,2,3,4,5,6\}$, $n(S) = 6$.

(i) A prime number will appear:
Prime numbers on a die: $\{2, 3, 5\}$
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$

(ii) A number ≥ 3 will appear:
Favourable outcomes: $\{3, 4, 5, 6\}$
$$P(\geq 3) = \frac{4}{6} = \frac{2}{3}$$

(iii) A number ≤ 1 will appear:
Favourable outcomes: $\{1\}$
$$P(\leq 1) = \frac{1}{6}$$

(iv) A number more than 6 will appear:
No face shows a number > 6, so this is an impossible event.
$$P(> 6) = \frac{0}{6} = 0$$

(v) A number less than 6 will appear:
Favourable outcomes: $\{1, 2, 3, 4, 5\}$
$$P(< 6) = \frac{5}{6}$$

Given: A card is selected from a pack of 52 cards.

(a) The sample space consists of all 52 cards.
$$n(S) = 52$$
There are 52 points in the sample space.

(b) Probability that the card is an ace of spades:
There is only 1 ace of spades in the deck.
$$P(\text{ace of spades}) = \frac{1}{52}$$

(c)(i) Probability that the card is an ace:
There are 4 aces in a deck (one of each suit).
$$P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$$

(c)(ii) Probability that the card is a black card:
There are 26 black cards (13 spades + 13 clubs).
$$P(\text{black card}) = \frac{26}{52} = \frac{1}{2}$$

Given: A coin (faces: 1 and 6) and a fair die (faces: 1,2,3,4,5,6) are tossed.

Sample space: $S = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$n(S) = 12$

(i) Sum = 3:
Possible combinations: coin shows 1 and die shows 2, i.e., $(1, 2)$.
$$P(\text{sum} = 3) = \frac{1}{12}$$

(ii) Sum = 12:
Possible combinations: coin shows 6 and die shows 6, i.e., $(6, 6)$.
$$P(\text{sum} = 12) = \frac{1}{12}$$

Given: Total council members = 4 men + 6 women = 10.

Number of ways to select 1 woman = 6.
Total number of ways to select 1 member = 10.

$$P(\text{woman is selected}) = \frac{6}{10} = \frac{3}{5}$$

The probability that the selected member is a woman is $\dfrac{3}{5}$.

Given: Coin tossed 4 times. Win Re 1 per head, lose Rs 1.50 per tail.

Sample space has $2^4 = 16$ equally likely outcomes.

Let $h$ = number of heads, $t$ = number of tails, where $h + t = 4$.

Amount = $1 \times h - 1.5 \times t = h - 1.5(4-h) = h - 6 + 1.5h = 2.5h - 6$

There are 5 different amounts possible:

• $P(\text{amount} = -\text{Rs }6) = \dfrac{1}{16}$
• $P(\text{amount} = -\text{Rs }3.50) = \dfrac{1}{4}$
• $P(\text{amount} = -\text{Re }1) = \dfrac{3}{8}$
• $P(\text{amount} = \text{Rs }1.50) = \dfrac{1}{4}$
• $P(\text{amount} = \text{Rs }4) = \dfrac{1}{16}$

Given: Three coins are tossed.
$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}, \quad n(S) = 8$$

(i) 3 heads: $\{HHH\}$
$$P = \frac{1}{8}$$

(ii) 2 heads: $\{HHT, HTH, THH\}$
$$P = \frac{3}{8}$$

(iii) At least 2 heads: $\{HHT, HTH, THH, HHH\}$
$$P = \frac{4}{8} = \frac{1}{2}$$

(iv) At most 2 heads: All outcomes except HHH: $\{HHT, HTH, THH, HTT, THT, TTH, TTT\}$
$$P = \frac{7}{8}$$

(v) No head: $\{TTT\}$
$$P = \frac{1}{8}$$

(vi) 3 tails: $\{TTT\}$
$$P = \frac{1}{8}$$

(vii) Exactly two tails: $\{HTT, THT, TTH\}$
$$P = \frac{3}{8}$$

(viii) No tail: $\{HHH\}$
$$P = \frac{1}{8}$$

(ix) At most two tails: All outcomes except TTT: $\{HHH, HHT, HTH, THH, HTT, THT, TTH\}$
$$P = \frac{7}{8}$$

Given: $P(A) = \dfrac{2}{11}$

Formula: $P(\text{not } A) = 1 - P(A)$

$$P(A') = 1 - \frac{2}{11} = \frac{11-2}{11} = \frac{9}{11}$$

The probability of event 'not A' is $\dfrac{9}{11}$.

Given: The word is ASSASSINATION.

Let us count the letters:
A-S-S-A-S-S-I-N-A-T-I-O-N

Total letters = 13

Vowels: A, A, A, I, I, O → 6 vowels
Consonants: S, S, S, S, N, T, N → 7 consonants

(i) Probability that the letter is a vowel:
$$P(\text{vowel}) = \frac{6}{13}$$

(ii) Probability that the letter is a consonant:
$$P(\text{consonant}) = \frac{7}{13}$$

Given: A person chooses 6 different numbers from 1 to 20.

The order of numbers does not matter.

Total ways to choose 6 numbers from 20:
$$n(S) = \binom{20}{6} = \frac{20!}{6! \cdot 14!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38760$$

There is only 1 winning combination.

$$P(\text{winning}) = \frac{1}{38760}$$

The probability of winning the prize is $\dfrac{1}{38760}$.

For probabilities to be consistently defined, we need:

• $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$
• $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
• All probabilities must be between 0 and 1.

(i) $P(A) = 0.5$, $P(B) = 0.7$, $P(A \cap B) = 0.6$

Here $P(A \cap B) = 0.6 > P(A) = 0.5$.

But $A \cap B \subseteq A$, so $P(A \cap B) \leq P(A)$ must hold.
Since $0.6 > 0.5$, this is a contradiction.

Not consistently defined.

(ii) $P(A) = 0.5$, $P(B) = 0.4$, $P(A \cup B) = 0.8$

Using the formula:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$0.8 = 0.5 + 0.4 - P(A \cap B)$$
$$P(A \cap B) = 0.9 - 0.8 = 0.1$$

Check: $P(A \cap B) = 0.1 \leq P(A) = 0.5$ ✓ and $P(A \cap B) = 0.1 \leq P(B) = 0.4$ ✓

Consistently defined.