Probability

Given: $P(E) = 0.6$, $P(F) = 0.3$, $P(E \cap F) = 0.2$

Formula used: $P(E|F) = \dfrac{P(E \cap F)}{P(F)}$ and $P(F|E) = \dfrac{P(E \cap F)}{P(E)}$

Finding $P(E|F)$:
$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}$$

Finding $P(F|E)$:
$$P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$$

Answer: $P(E|F) = \dfrac{2}{3}$ and $P(F|E) = \dfrac{1}{3}$.

Given: $P(B) = 0.5$, $P(A \cap B) = 0.32$

Formula used: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

$$P(A|B) = \frac{0.32}{0.5} = \frac{32}{50} = 0.64$$

Answer: $P(A|B) = 0.64$.

Given: $P(A) = 0.8$, $P(B) = 0.5$, $P(B|A) = 0.4$

(i) Finding $P(A \cap B)$:

Using $P(B|A) = \dfrac{P(A \cap B)}{P(A)}$:
$$P(A \cap B) = P(A) \cdot P(B|A) = 0.8 \times 0.4 = 0.32$$

(ii) Finding $P(A|B)$:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$$

(iii) Finding $P(A \cup B)$:

Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$$P(A \cup B) = 0.8 + 0.5 - 0.32 = 0.98$$

Answers: (i) $0.32$, (ii) $0.64$, (iii) $0.98$.

Given: $2P(A) = P(B) = \dfrac{5}{13}$ and $P(A|B) = \dfrac{2}{5}$

From the given condition:
$$P(B) = \frac{5}{13}, \quad P(A) = \frac{5}{26}$$

Finding $P(A \cap B)$:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} \implies P(A \cap B) = P(A|B) \cdot P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$$

Finding $P(A \cup B)$:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$$
$$= \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{11}{26}$$

Answer: $P(A \cup B) = \dfrac{11}{26}$.

Given: $P(A) = \dfrac{6}{11}$, $P(B) = \dfrac{5}{11}$, $P(A \cup B) = \dfrac{7}{11}$

(i) Finding $P(A \cap B)$:
$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11} = \frac{4}{11}$$

(ii) Finding $P(A|B)$:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{4}{11}}{\frac{5}{11}} = \frac{4}{5}$$

(iii) Finding $P(B|A)$:
$$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{4}{11}}{\frac{6}{11}} = \frac{4}{6} = \frac{2}{3}$$

Answers: (i) $\dfrac{4}{11}$, (ii) $\dfrac{4}{5}$, (iii) $\dfrac{2}{3}$.

Sample space when a coin is tossed three times has $2^3 = 8$ equally likely outcomes:
$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

(i) E: head on third toss, F: heads on first two tosses

$E = \{HHH, HTH, THH, TTH\}$
$F = \{HHH, HHT\}$
$E \cap F = \{HHH\}$

$$P(F) = \frac{2}{8} = \frac{1}{4}, \quad P(E \cap F) = \frac{1}{8}$$

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2}$$

(ii) E: at least two heads, F: at most two heads

$E = \{HHH, HHT, HTH, THH\}$
$F = \{HHT, HTH, HTT, THH, THT, TTH, TTT\}$
$E \cap F = \{HHT, HTH, THH\}$

$$P(F) = \frac{7}{8}, \quad P(E \cap F) = \frac{3}{8}$$

$$P(E|F) = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}$$

(iii) E: at most two tails, F: at least one tail

$E = \{HHH, HHT, HTH, HTT, THH, THT, TTH\}$ (all except TTT)
$F = \{HHT, HTH, HTT, THH, THT, TTH, TTT\}$
$E \cap F = \{HHT, HTH, HTT, THH, THT, TTH\}$

$$P(F) = \frac{7}{8}, \quad P(E \cap F) = \frac{6}{8} = \frac{3}{4}$$

$$P(E|F) = \frac{\frac{3}{4}}{\frac{7}{8}} = \frac{3}{4} \times \frac{8}{7} = \frac{6}{7}$$

Answers: (i) $\dfrac{1}{2}$, (ii) $\dfrac{3}{7}$, (iii) $\dfrac{6}{7}$.

Sample space: $S = \{HH, HT, TH, TT\}$, each with probability $\dfrac{1}{4}$.

(i) E: tail appears on one coin, F: one coin shows head

$E = \{HT, TH\}$ (exactly one tail)
$F = \{HT, TH\}$ (exactly one head)
$E \cap F = \{HT, TH\}$

$$P(F) = \frac{2}{4} = \frac{1}{2}, \quad P(E \cap F) = \frac{2}{4} = \frac{1}{2}$$

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$$

(ii) E: no tail appears, F: no head appears

$E = \{HH\}$ (no tail)
$F = \{TT\}$ (no head)
$E \cap F = \phi$

$$P(F) = \frac{1}{4}, \quad P(E \cap F) = 0$$

$$P(E|F) = \frac{0}{\frac{1}{4}} = 0$$

Answers: (i) $1$, (ii) $0$.

Given: A die is thrown three times. Total outcomes = $6^3 = 216$.

$F$: 6 on first toss and 5 on second toss.
$F = \{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$, so $|F| = 6$.

$E \cap F$: 6 on first, 5 on second, 4 on third = $\{(6,5,4)\}$, so $|E \cap F| = 1$.

$$P(F) = \frac{6}{216} = \frac{1}{36}, \quad P(E \cap F) = \frac{1}{216}$$

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{216}}{\frac{1}{36}} = \frac{1}{216} \times 36 = \frac{1}{6}$$

Answer: $P(E|F) = \dfrac{1}{6}$.

Given: Three people — Mother (M), Father (F), Son (S) — line up at random.

Total arrangements = $3! = 6$:
$$\{MFS, MSF, FMS, FSM, SMF, SFM\}$$

$F$ (father in middle): $\{MFS, SFM\}$, so $P(F) = \dfrac{2}{6} = \dfrac{1}{3}$.

$E$ (son on one end): $\{SMF, SFM, MFS, FMS\}$... Let me list carefully:
- Son on left end: $\{SMF, SFM\}$
- Son on right end: $\{MFS, FMS\}$
So $E = \{SMF, SFM, MFS, FMS\}$.

$E \cap F$ (father in middle AND son on one end): $\{SFM, MFS\}$, so $P(E \cap F) = \dfrac{2}{6} = \dfrac{1}{3}$.

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{3}}{\frac{1}{3}} = 1$$

Answer: $P(E|F) = 1$. (When father is in the middle, son must be on one end.)

Sample space: Each die has 6 faces, so total outcomes = $6 \times 6 = 36$.

(a) Let A = sum > 9, B = black die shows 5.

$B = \{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$, $P(B) = \dfrac{6}{36} = \dfrac{1}{6}$.

For sum > 9 with black die = 5: red die must be > 4, i.e., 5 or 6.
$A \cap B = \{(5,5),(5,6)\}$, $P(A \cap B) = \dfrac{2}{36} = \dfrac{1}{18}$.

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{18} \times 6 = \frac{2}{9}$$

(b) Let C = sum is 8, D = red die shows a number less than 4 (i.e., 1, 2, or 3).

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

Here (black, red) pairs where red < 4: $P(D) = \dfrac{18}{36} = \dfrac{1}{2}$.

For sum = 8 with red die < 4: pairs (black, red) with black + red = 8 and red ∈ {1,2,3}:
- red = 1: black = 7 (impossible)
- red = 2: black = 6 → $(6,2)$ ✓
- red = 3: black = 5 → $(5,3)$ ✓

$C \cap D = \{(6,2),(5,3)\}$, $P(C \cap D) = \dfrac{2}{36} = \dfrac{1}{18}$.

$$P(C|D) = \frac{P(C \cap D)}{P(D)} = \frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{9}$$

Answers: (a) $\dfrac{2}{9}$, (b) $\dfrac{1}{9}$.

Given: A fair die is rolled. $S = \{1,2,3,4,5,6\}$, each with probability $\dfrac{1}{6}$.

$E = \{1,3,5\}$, $F = \{2,3\}$, $G = \{2,3,4,5\}$

$$P(E) = \frac{3}{6} = \frac{1}{2}, \quad P(F) = \frac{2}{6} = \frac{1}{3}, \quad P(G) = \frac{4}{6} = \frac{2}{3}$$

$E \cap F = \{3\}$, $E \cap G = \{3,5\}$, $E \cup F = \{1,2,3,5\}$, $(E \cup F) \cap G = \{2,3,5\}$, $(E \cap F) \cap G = \{3\}$

(i) $P(E|F)$ and $P(F|E)$:
$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$$
$$P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$

(ii) $P(E|G)$ and $P(G|E)$:
$$P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{\frac{2}{6}}{\frac{2}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$
$$P(G|E) = \frac{P(E \cap G)}{P(E)} = \frac{\frac{2}{6}}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$

(iii) $P((E \cup F)|G)$ and $P((E \cap F)|G)$:

$(E \cup F) \cap G = \{2,3,5\}$, so $P((E \cup F) \cap G) = \dfrac{3}{6} = \dfrac{1}{2}$.
$$P((E \cup F)|G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}$$

$(E \cap F) \cap G = \{3\}$, so $P((E \cap F) \cap G) = \dfrac{1}{6}$.
$$P((E \cap F)|G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{4}$$

Answers: (i) $P(E|F) = \dfrac{1}{2}$, $P(F|E) = \dfrac{1}{3}$; (ii) $P(E|G) = \dfrac{1}{2}$, $P(G|E) = \dfrac{2}{3}$; (iii) $P((E\cup F)|G) = \dfrac{3}{4}$, $P((E\cap F)|G) = \dfrac{1}{4}$.

Sample space: $S = \{BB, BG, GB, GG\}$ (first child, second child), each equally likely with probability $\dfrac{1}{4}$.

Let A = both children are girls = $\{GG\}$.

(i) Given: youngest (second child) is a girl.

Let $F_1$ = youngest is a girl = $\{BG, GG\}$, $P(F_1) = \dfrac{2}{4} = \dfrac{1}{2}$.

$A \cap F_1 = \{GG\}$, $P(A \cap F_1) = \dfrac{1}{4}$.

$$P(A|F_1) = \frac{P(A \cap F_1)}{P(F_1)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

(ii) Given: at least one is a girl.

Let $F_2$ = at least one girl = $\{BG, GB, GG\}$, $P(F_2) = \dfrac{3}{4}$.

$A \cap F_2 = \{GG\}$, $P(A \cap F_2) = \dfrac{1}{4}$.

$$P(A|F_2) = \frac{P(A \cap F_2)}{P(F_2)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$$

Answers: (i) $\dfrac{1}{2}$, (ii) $\dfrac{1}{3}$.

Given:
- Easy True/False: 300
- Difficult True/False: 200
- Easy Multiple Choice: 500
- Difficult Multiple Choice: 400
- Total questions: $300 + 200 + 500 + 400 = 1400$

Let E = easy question, M = multiple choice question.

$$P(M) = \frac{500 + 400}{1400} = \frac{900}{1400} = \frac{9}{14}$$

$$P(E \cap M) = \frac{500}{1400} = \frac{5}{14}$$

$$P(E|M) = \frac{P(E \cap M)}{P(M)} = \frac{\frac{5}{14}}{\frac{9}{14}} = \frac{5}{9}$$

Answer: The required probability is $\dfrac{5}{9}$.

Sample space: Total outcomes when two dice are thrown = 36.

Let F = numbers on the two dice are different.
Number of outcomes where both dice show the same number = 6 (i.e., (1,1),(2,2),...,(6,6)).
So $|F| = 36 - 6 = 30$, $P(F) = \dfrac{30}{36} = \dfrac{5}{6}$.

Let E = sum of numbers is 4.
Outcomes with sum 4: $\{(1,3),(2,2),(3,1)\}$.
But we need different numbers, so exclude $(2,2)$.
$E \cap F = \{(1,3),(3,1)\}$, $P(E \cap F) = \dfrac{2}{36} = \dfrac{1}{18}$.

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{18}}{\frac{5}{6}} = \frac{1}{18} \times \frac{6}{5} = \frac{1}{15}$$

Answer: $P(E|F) = \dfrac{1}{15}$.

Sample space construction:
- If die shows 3 or 6 (multiples of 3): throw die again → outcomes: $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$ — 12 outcomes.
- If die shows 1, 2, 4, or 5: toss a coin → outcomes: $(1,H),(1,T),(2,H),(2,T),(4,H),(4,T),(5,H),(5,T)$ — 8 outcomes.

Each face of the die has probability $\dfrac{1}{6}$. For multiples of 3, the second die has probability $\dfrac{1}{6}$ each; for others, coin has probability $\dfrac{1}{2}$ each.

Each outcome in the first group has probability $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$.
Each outcome in the second group has probability $\dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.

Let A = coin shows tail = $\{(1,T),(2,T),(4,T),(5,T)\}$.
Let B = at least one die shows 3.

$B$ includes outcomes where first die = 3: $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$ and where second die = 3: $(3,3),(6,3)$.
So $B = \{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}$.

$$P(B) = 6 \times \frac{1}{36} + 1 \times \frac{1}{36} = \frac{6}{36} + \frac{1}{36} = \frac{7}{36}$$

$A \cap B = \phi$ (A involves coin outcomes, B involves only die outcomes — they are disjoint events).

$$P(A \cap B) = 0$$

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{\frac{7}{36}} = 0$$

Answer: $P(A|B) = 0$.

Correct Answer: (C) not defined.

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$. Since $P(B) = 0$, the conditional probability $P(A|B)$ is not defined (division by zero is undefined).

Correct Answer: (D) $P(A) = P(B)$.

$P(A|B) = P(B|A)$ implies:
$$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$$

Provided $P(A \cap B) \neq 0$, we can cancel $P(A \cap B)$ from both sides to get $P(A) = P(B)$.

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

For independent events: $P(A \cap B) = P(A) \cdot P(B)$

$$P(A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$$

Answer: $P(A \cap B) = \dfrac{3}{25}$.

Given: A pack of 52 cards has 26 black cards.

Let $E_1$ = first card drawn is black, $E_2$ = second card drawn is black.

$$P(E_1) = \frac{26}{52} = \frac{1}{2}$$

After drawing one black card, 25 black cards remain out of 51 total:
$$P(E_2|E_1) = \frac{25}{51}$$

By multiplication theorem:
$$P(E_1 \cap E_2) = P(E_1) \cdot P(E_2|E_1) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$$

Answer: The probability that both cards are black is $\dfrac{25}{102}$.

Given: 15 oranges total: 12 good, 3 bad. Three are drawn without replacement.

The box is approved if all three selected oranges are good.

Let $E_1, E_2, E_3$ be the events that the 1st, 2nd, 3rd orange drawn is good.

$$P(E_1) = \frac{12}{15} = \frac{4}{5}$$
$$P(E_2|E_1) = \frac{11}{14}$$
$$P(E_3|E_1 \cap E_2) = \frac{10}{13}$$

By multiplication theorem:
$$P(E_1 \cap E_2 \cap E_3) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{1320}{2730} = \frac{44}{91}$$

Answer: The probability that the box is approved for sale is $\dfrac{44}{91}$.

Given: A fair coin and an unbiased die are tossed.

$$P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{6}$$

$$P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

The event $A \cap B$ = head on coin AND 3 on die = $\{(H,3)\}$.

Total outcomes = $2 \times 6 = 12$, so:
$$P(A \cap B) = \frac{1}{12}$$

Since $P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{12}$, A and B are independent events.

Given: Die faces: 1(red), 2(red), 3(red), 4(green), 5(green), 6(green).

$A$ = even number = $\{2, 4, 6\}$, $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$.

$B$ = red number = $\{1, 2, 3\}$, $P(B) = \dfrac{3}{6} = \dfrac{1}{2}$.

$A \cap B$ = even AND red = $\{2\}$, $P(A \cap B) = \dfrac{1}{6}$.

Now, $P(A) \cdot P(B) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.

Since $P(A \cap B) = \dfrac{1}{6} \neq \dfrac{1}{4} = P(A) \cdot P(B)$, A and B are NOT independent.

Given: $P(E) = \dfrac{3}{5}$, $P(F) = \dfrac{3}{10}$, $P(E \cap F) = \dfrac{1}{5}$.

Check: $P(E) \cdot P(F) = \dfrac{3}{5} \times \dfrac{3}{10} = \dfrac{9}{50}$.

But $P(E \cap F) = \dfrac{1}{5} = \dfrac{10}{50}$.

Since $\dfrac{10}{50} \neq \dfrac{9}{50}$, i.e., $P(E \cap F) \neq P(E) \cdot P(F)$, E and F are NOT independent.

Given: $P(A) = \dfrac{1}{2}$, $P(A \cup B) = \dfrac{3}{5}$, $P(B) = p$.

(i) Mutually exclusive: $P(A \cap B) = 0$.
$$P(A \cup B) = P(A) + P(B) \implies \frac{3}{5} = \frac{1}{2} + p \implies p = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$$

(ii) Independent: $P(A \cap B) = P(A) \cdot P(B) = \dfrac{p}{2}$.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2} = \frac{1}{2} + \frac{p}{2}$$
$$\frac{p}{2} = \frac{3}{5} - \frac{1}{2} = \frac{1}{10} \implies p = \frac{1}{5}$$

Answers: (i) $p = \dfrac{1}{10}$, (ii) $p = \dfrac{1}{5}$.

Given: A and B are independent, $P(A) = 0.3$, $P(B) = 0.4$.

(i) $P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.4 = 0.12$

(ii) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.12 = 0.58$

(iii) Since A and B are independent:
$$P(A|B) = P(A) = 0.3$$

(iv) Since A and B are independent:
$$P(B|A) = P(B) = 0.4$$

Answers: (i) $0.12$, (ii) $0.58$, (iii) $0.3$, (iv) $0.4$.

Given: $P(A) = \dfrac{1}{4}$, $P(B) = \dfrac{1}{2}$, $P(A \cap B) = \dfrac{1}{8}$.

P(not A and not B) = $P(A' \cap B') = P((A \cup B)')= 1 - P(A \cup B)$.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$$

$$P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8}$$

Answer: $P(\text{not A and not B}) = \dfrac{3}{8}$.

Given: $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{7}{12}$, $P(A' \cup B') = \dfrac{1}{4}$.

Using De Morgan's law: $P(A' \cup B') = 1 - P(A \cap B)$.

$$\frac{1}{4} = 1 - P(A \cap B) \implies P(A \cap B) = \frac{3}{4}$$

Now check: $P(A) \cdot P(B) = \dfrac{1}{2} \times \dfrac{7}{12} = \dfrac{7}{24}$.

Since $P(A \cap B) = \dfrac{3}{4} = \dfrac{18}{24} \neq \dfrac{7}{24} = P(A) \cdot P(B)$, A and B are NOT independent.

Given: A and B independent, $P(A) = 0.3$, $P(B) = 0.6$.

(i) P(A and B):
$$P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.6 = 0.18$$

(ii) P(A and not B):
$$P(A \cap B') = P(A) \cdot P(B') = P(A) \cdot [1 - P(B)] = 0.3 \times 0.4 = 0.12$$

(iii) P(A or B):
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.6 - 0.18 = 0.72$$

(iv) P(neither A nor B):
$$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.72 = 0.28$$

Answers: (i) $0.18$, (ii) $0.12$, (iii) $0.72$, (iv) $0.28$.

Given: A die is tossed three times.

P(odd number on one toss) = $\dfrac{3}{6} = \dfrac{1}{2}$.

P(no odd number in three tosses) = P(even on all three tosses) $= \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$.

P(at least one odd number) $= 1 - \dfrac{1}{8} = \dfrac{7}{8}$.

Answer: $\dfrac{7}{8}$.

Given: Box has 10 black and 8 red balls. Total = 18. Draws are with replacement.

$P(\text{black}) = \dfrac{10}{18} = \dfrac{5}{9}$, $P(\text{red}) = \dfrac{8}{18} = \dfrac{4}{9}$.

(i) Both balls are red:
$$P = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$$

(ii) First ball is black and second is red:
$$P = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$

(iii) One black and one red (in any order):
$$P = P(\text{black then red}) + P(\text{red then black}) = \frac{5}{9} \times \frac{4}{9} + \frac{4}{9} \times \frac{5}{9} = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$$

Answers: (i) $\dfrac{16}{81}$, (ii) $\dfrac{20}{81}$, (iii) $\dfrac{40}{81}$.