Probability

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 &lt; 0 and P(\omega_5) = -0.2 &lt; 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} &gt; 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(&gt; 6) = \frac{0}{6} = 0

(v) A number less than 6 will appear:
Favourable outcomes: $\{1, 2, 3, 4, 5\}$
P(&lt; 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 = $1 \times h - 1.5 \times t = h - 1.5(4-h) = h - 6 + 1.5h = 2.5h - 6$

| Heads (h) | Tails (t) | Amount (Rs) | No. of outcomes ($\binom{4}{h}$) | Probability |
|---|---|---|---|---|
| 0 | 4 | $2.5(0)-6 = -6$ | 1 | $\frac{1}{16}$ |
| 1 | 3 | $2.5(1)-6 = -3.50$ | 4 | $\frac{4}{16} = \frac{1}{4}$ |
| 2 | 2 | $2.5(2)-6 = -1$ | 6 | $\frac{6}{16} = \frac{3}{8}$ |
| 3 | 1 | $2.5(3)-6 = 1.50$ | 4 | $\frac{4}{16} = \frac{1}{4}$ |
| 4 | 0 | $2.5(4)-6 = 4$ | 1 | $\frac{1}{16}$ |

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 &gt; P(A) = 0.5.

But $A \cap B \subseteq A$, so $P(A \cap B) \leq P(A)$ must hold.
Since 0.6 &gt; 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.

Formula used: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

(i) $P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{5}$, $P(A \cap B) = \dfrac{1}{15}$

$$P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} = \frac{5}{15} + \frac{3}{15} - \frac{1}{15} = \frac{7}{15}$$

$$\boxed{P(A \cup B) = \frac{7}{15}}$$

(ii) $P(A) = 0.35$, $P(A \cap B) = 0.25$, $P(A \cup B) = 0.6$

$$0.6 = 0.35 + P(B) - 0.25$$
$$P(B) = 0.6 - 0.35 + 0.25 = 0.5$$

$$\boxed{P(B) = 0.5}$$

(iii) $P(A) = 0.5$, $P(B) = 0.35$, $P(A \cup B) = 0.7$

$$0.7 = 0.5 + 0.35 - P(A \cap B)$$
$$P(A \cap B) = 0.85 - 0.7 = 0.15$$

$$\boxed{P(A \cap B) = 0.15}$$

Given: $P(A) = \dfrac{3}{5}$, $P(B) = \dfrac{1}{5}$, A and B are mutually exclusive.

Formula: If A and B are mutually exclusive, $P(A \cap B) = 0$, so:
$$P(A \cup B) = P(A) + P(B)$$

$$P(A \text{ or } B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$$

P(A or B) = $\dfrac{4}{5}$.

Given: $P(E) = \dfrac{1}{4}$, $P(F) = \dfrac{1}{2}$, $P(E \cap F) = \dfrac{1}{8}$

(i) P(E or F) = P(E ∪ F):
$$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$
$$= \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$$

(ii) P(not E and not F) = P(E' ∩ F'):
By De Morgan's law: $E' \cap F' = (E \cup F)'$
$$P(E' \cap F') = 1 - P(E \cup F) = 1 - \frac{5}{8} = \frac{3}{8}$$

Answers: (i) $P(E \cup F) = \dfrac{5}{8}$, (ii) $P(E' \cap F') = \dfrac{3}{8}$

Given: $P(\text{not E or not F}) = P(E' \cup F') = 0.25$

By De Morgan's law:
$$E' \cup F' = (E \cap F)'$$

So:
$$P((E \cap F)') = 0.25$$
$$1 - P(E \cap F) = 0.25$$
$$P(E \cap F) = 0.75$$

Since $P(E \cap F) = 0.75 \neq 0$, the events E and F are not mutually exclusive.

(For mutually exclusive events, $P(E \cap F)$ must equal 0.)

Given: $P(A) = 0.42$, $P(B) = 0.48$, $P(A \cap B) = 0.16$

(i) P(not A):
$$P(A') = 1 - P(A) = 1 - 0.42 = 0.58$$

(ii) P(not B):
$$P(B') = 1 - P(B) = 1 - 0.48 = 0.52$$

(iii) P(A or B):
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$= 0.42 + 0.48 - 0.16 = 0.74$$

Given:
$P(\text{First}) = 0.8$, $P(\text{Second}) = 0.7$, $P(\text{at least one}) = P(\text{First} \cup \text{Second}) = 0.95$

Using the addition formula:
$$P(\text{First} \cup \text{Second}) = P(\text{First}) + P(\text{Second}) - P(\text{First} \cap \text{Second})$$
$$0.95 = 0.8 + 0.7 - P(\text{First} \cap \text{Second})$$
$$P(\text{First} \cap \text{Second}) = 1.5 - 0.95 = 0.55$$

The probability of passing both examinations is 0.55.

Given:
$P(E \cap H) = 0.5$ (passing both)
$P(E' \cap H') = 0.1$ (passing neither)
$P(E) = 0.75$

Step 1: Find $P(E \cup H)$.
$$P(E' \cap H') = P((E \cup H)') = 1 - P(E \cup H)$$
$$0.1 = 1 - P(E \cup H)$$
$$P(E \cup H) = 0.9$$

Step 2: Use the addition formula.
$$P(E \cup H) = P(E) + P(H) - P(E \cap H)$$
$$0.9 = 0.75 + P(H) - 0.5$$
$$P(H) = 0.9 - 0.75 + 0.5 = 0.65$$

The probability of passing the Hindi examination is 0.65.

Given: Total students = 60, $n(\text{NCC}) = 30$, $n(\text{NSS}) = 32$, $n(\text{NCC} \cap \text{NSS}) = 24$

$$P(\text{NCC}) = \frac{30}{60} = \frac{1}{2}, \quad P(\text{NSS}) = \frac{32}{60} = \frac{8}{15}, \quad P(\text{NCC} \cap \text{NSS}) = \frac{24}{60} = \frac{2}{5}$$

(i) P(NCC or NSS):
$$P(\text{NCC} \cup \text{NSS}) = P(\text{NCC}) + P(\text{NSS}) - P(\text{NCC} \cap \text{NSS})$$
$$= \frac{30}{60} + \frac{32}{60} - \frac{24}{60} = \frac{38}{60} = \frac{19}{30}$$

(ii) P(neither NCC nor NSS):
$$P(\text{NCC}' \cap \text{NSS}') = 1 - P(\text{NCC} \cup \text{NSS}) = 1 - \frac{19}{30} = \frac{11}{30}$$

(iii) P(NSS but not NCC):
$$n(\text{NSS only}) = n(\text{NSS}) - n(\text{NCC} \cap \text{NSS}) = 32 - 24 = 8$$
$$P(\text{NSS but not NCC}) = \frac{8}{60} = \frac{2}{15}$$

Given: 10 red, 20 blue, 30 green marbles. Total = 60 marbles. 5 are drawn.

Total ways to draw 5 marbles from 60:
$$n(S) = \binom{60}{5}$$

(i) All 5 will be blue:
All 5 must be chosen from 20 blue marbles.
$$n(E) = \binom{20}{5}$$

$$P(\text{all blue}) = \frac{\binom{20}{5}}{\binom{60}{5}} = \frac{\frac{20!}{5! \cdot 15!}}{\frac{60!}{5! \cdot 55!}} = \frac{20 \times 19 \times 18 \times 17 \times 16}{60 \times 59 \times 58 \times 57 \times 56}$$

$$= \frac{1860480}{655381440} = \frac{34}{11316} = \frac{17}{5658} \approx \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16}{60 \cdot 59 \cdot 58 \cdot 57 \cdot 56}$$

Let us keep it in the form:
$$P(\text{all blue}) = \frac{\binom{20}{5}}{\binom{60}{5}}$$

(ii) At least one will be green:
Using complementary counting:
$$P(\text{at least one green}) = 1 - P(\text{no green})$$

P(no green) = all 5 drawn from non-green (10 red + 20 blue = 30 marbles):
$$P(\text{no green}) = \frac{\binom{30}{5}}{\binom{60}{5}}$$

$$P(\text{at least one green}) = 1 - \frac{\binom{30}{5}}{\binom{60}{5}}$$

Calculating:
$$\binom{30}{5} = \frac{30 \times 29 \times 28 \times 27 \times 26}{120} = 142506$$
$$\binom{60}{5} = \frac{60 \times 59 \times 58 \times 57 \times 56}{120} = 5461512$$

$$P(\text{at least one green}) = 1 - \frac{142506}{5461512} = 1 - \frac{12}{455} = \frac{443}{455}$$

Given: 4 cards drawn from 52 cards. We need 3 diamonds and 1 spade.

In a deck: 13 diamonds, 13 spades.

Total ways to draw 4 cards from 52:
$$n(S) = \binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$$

Favourable outcomes (3 diamonds from 13 AND 1 spade from 13):
$$n(E) = \binom{13}{3} \times \binom{13}{1} = 286 \times 13 = 3718$$

$$P(\text{3 diamonds and 1 spade}) = \frac{3718}{270725} = \frac{286 \times 13}{270725}$$

$$= \frac{3718}{270725} = \frac{858}{62475} = \frac{286}{20825}$$

$$\boxed{P = \frac{286 \times 13}{\binom{52}{4}} = \frac{3718}{270725}}$$

Given: Die has 6 faces: two faces with '1', three faces with '2', one face with '3'.

Total outcomes = 6.

(i) P(2):
Number of faces showing 2 = 3
$$P(2) = \frac{3}{6} = \frac{1}{2}$$

(ii) P(1 or 3):
Faces showing 1 = 2, faces showing 3 = 1
$$P(1 \text{ or } 3) = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$$

(iii) P(not 3):
$$P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}$$

Given: Total tickets = 10,000; prizes = 10; non-prize tickets = 9,990.

(a) One ticket:
$$P(\text{not getting prize}) = \frac{9990}{10000} = \frac{999}{1000}$$

(b) Two tickets:
Favourable outcomes (both tickets are non-prize):
$$n(E) = \binom{9990}{2}$$
Total ways to choose 2 tickets from 10,000:
$$n(S) = \binom{10000}{2}$$

$$P(\text{no prize}) = \frac{\binom{9990}{2}}{\binom{10000}{2}} = \frac{9990 \times 9989}{10000 \times 9999}$$

(c) Ten tickets:
$$P(\text{no prize}) = \frac{\binom{9990}{10}}{\binom{10000}{10}}$$

$$= \frac{9990 \times 9989 \times 9988 \times \cdots \times 9981}{10000 \times 9999 \times 9998 \times \cdots \times 9991}$$

Given: 100 students; Section I has 40, Section II has 60.

Total ways to select 40 students for Section I from 100:
$$n(S) = \binom{100}{40}$$

(a) Both in the same section:

Case 1: Both in Section I (40 students).
Remaining 38 places filled from remaining 98 students:
$$n(E_1) = \binom{98}{38}$$

Case 2: Both in Section II (60 students).
Section I is filled from remaining 98 students (none of us):
$$n(E_2) = \binom{98}{40}$$

$$P(\text{same section}) = \frac{\binom{98}{38} + \binom{98}{40}}{\binom{100}{40}}$$

Now, $\binom{100}{40} = \frac{100 \times 99}{40 \times 60} \times \binom{98}{38}$... Let us use a simpler approach:

Consider just the placement of 2 specific students among 100.

Total ways to assign 2 specific students to sections = ways to choose which section each goes to.

Actually, the simplest method:

Total ways to choose 2 students' positions from 100 for Section I = $\binom{100}{2}$ (choosing 2 from 100 for any 2 spots).

Both in Section I: $\binom{40}{2}$ ways.
Both in Section II: $\binom{60}{2}$ ways.

$$P(\text{same section}) = \frac{\binom{40}{2} + \binom{60}{2}}{\binom{100}{2}} = \frac{780 + 1770}{4950} = \frac{2550}{4950} = \frac{17}{33}$$

(b) Both in different sections:
$$P(\text{different sections}) = 1 - \frac{17}{33} = \frac{16}{33}$$

Given: 3 letters and 3 envelopes. Letters inserted randomly.

Total ways to insert 3 letters into 3 envelopes = $3! = 6$

Let us list all permutations (1,2,3 represent correct positions):
- (1,2,3): All correct ✓
- (1,3,2): Letter 1 correct
- (2,1,3): Letter 3 correct
- (2,3,1): None correct
- (3,1,2): None correct
- (3,2,1): Letter 2 correct

Outcomes where at least one letter is in correct envelope: (1,2,3), (1,3,2), (2,1,3), (3,2,1) = 4 outcomes

$$P(\text{at least one correct}) = \frac{4}{6} = \frac{2}{3}$$

Given: $P(A) = 0.54$, $P(B) = 0.69$, $P(A \cap B) = 0.35$

(i) P(A∪B):
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.54 + 0.69 - 0.35 = 0.88$$

(ii) P(A'∩B') = P((A∪B)'):
By De Morgan's law: $A' \cap B' = (A \cup B)'$
$$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.88 = 0.12$$

(iii) P(A∩B') = P(A) − P(A∩B):
(Elements in A but not in B)
$$P(A \cap B') = P(A) - P(A \cap B) = 0.54 - 0.35 = 0.19$$

(iv) P(B∩A') = P(B) − P(A∩B):
(Elements in B but not in A)
$$P(B \cap A') = P(B) - P(A \cap B) = 0.69 - 0.35 = 0.34$$