(i) To find the probability that Sam gets a cup of coffee, we use the law of total probability:

\(P( ext{cup of coffee}) = P( ext{on time}) \times P( ext{coffee | on time}) + P( ext{not on time}) \times P( ext{coffee | not on time})\)

\(= 0.6 \times 0.9 + 0.4 \times 0.3\)

\(= 0.54 + 0.12 = 0.66\)

(ii) To find the probability that the bus is not on time given that Sam does not get a cup of coffee, we use Bayes' theorem:

\(P( ext{not on time | no cup}) = \frac{P( ext{not on time} \cap ext{no cup})}{P( ext{no cup})}\)

First, calculate \(P( ext{not on time} \cap ext{no cup})\):

\(P( ext{not on time} \cap ext{no cup}) = P( ext{not on time}) \times P( ext{no cup | not on time})\)

\(= 0.4 \times 0.7 = 0.28\)

Next, calculate \(P( ext{no cup})\):

\(P( ext{no cup}) = 1 - P( ext{cup of coffee}) = 1 - 0.66 = 0.34\)

Finally, substitute these into Bayes' theorem:

\(P( ext{not on time | no cup}) = \frac{0.28}{0.34} \approx 0.824\)

Let:

• \(A\) be the event that a person goes abroad.
• \(C\) be the event that a person goes camping.

We need to find \(P(A \mid C)\), the probability that a person goes abroad given that they go camping.

Using the formula for conditional probability:

\(P(A \mid C) = \frac{P(A \cap C)}{P(C)}\)

Calculate \(P(A \cap C)\):

\(P(A \cap C) = P(A) \times P(C \mid A) = 0.35 \times 0.15 = 0.0525\)

Calculate \(P(C)\):

\(P(C) = P(A \cap C) + P(H \cap C)\)

Where \(H\) is the event of taking a holiday in their own country.

\(P(H \cap C) = P(H) \times P(C \mid H) = 0.65 \times 0.4 = 0.26\)

Thus,

\(P(C) = 0.0525 + 0.26 = 0.3125\)

Finally, calculate \(P(A \mid C)\):

\(P(A \mid C) = \frac{0.0525}{0.3125} = 0.168\)

(i) To complete the table, we use the given probabilities and the fact that the total probability is 1.

The probability that the meal is served on time is \(\frac{1}{5}\), so the probability that it is not served on time is \(1 - \frac{1}{5} = \frac{4}{5}\).

The probability that the kitchen is left in a mess is \(\frac{3}{5}\), so the probability that it is not left in a mess is \(1 - \frac{3}{5} = \frac{2}{5}\).

We know that the probability that the meal is not served on time and the kitchen is not left in a mess is \(\frac{3}{10}\).

Using these, we can fill in the table:

(ii) To find the probability that the meal is not served on time given that the kitchen is left in a mess, we use conditional probability:

\(P(\text{Not on time} | \text{Kitchen mess}) = \frac{P(\text{Not on time and Kitchen mess})}{P(\text{Kitchen mess})}\)

From the table, \(P(\text{Not on time and Kitchen mess}) = \frac{1}{2}\) and \(P(\text{Kitchen mess}) = \frac{3}{5}\).

Thus, \(P(\text{Not on time} | \text{Kitchen mess}) = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}\)

To solve this problem, we need to find the conditional probability \(P(X > 0 \mid X \neq 2)\).

First, calculate the total number of outcomes when two spinners are spun. Each spinner has 5 sides, so there are \(5 \times 5 = 25\) possible outcomes.

Next, determine the number of outcomes where the score is not equal to 2. The score is 2 when the numbers are (1,3), (3,1), (2,4), (4,2), (3,5), and (5,3). There are 6 such outcomes.

Thus, the number of outcomes where the score is not 2 is \(25 - 6 = 19\).

Now, find the number of outcomes where the score is greater than 0 and not equal to 2. This includes all outcomes except those where the numbers are equal (score = 0) and those where the score is 2. There are 5 outcomes where the numbers are equal: (1,1), (2,2), (3,3), (4,4), (5,5).

Therefore, the number of outcomes where the score is greater than 0 and not equal to 2 is \(25 - 5 - 6 = 14\).

The conditional probability is then given by:

\(P(X > 0 \mid X \neq 2) = \frac{\text{Number of outcomes } (X > 0 \cap X \neq 2)}{\text{Number of outcomes } X \neq 2} = \frac{14}{19}\)

Thus, the probability is \(\frac{14}{19}\) or approximately 0.737.

(i) To find the probability \(x\), we use the total probability that Nikita’s mother likes the present:

\(0.3 \times 0.72 + 0.7 \times x = 0.783\)

Simplifying, we get:

\(0.216 + 0.7x = 0.783\)

\(0.7x = 0.783 - 0.216\)

\(0.7x = 0.567\)

\(x = \frac{0.567}{0.7} = 0.81\)

(ii) To find the probability that the present is a scarf given that Nikita’s mother does not like it, we use conditional probability:

\(P(S \mid NL) = \frac{P(S \cap NL)}{P(NL)}\)

\(P(S \cap NL) = 0.3 \times (1 - 0.72) = 0.3 \times 0.28 = 0.084\)

\(P(NL) = 1 - P(L) = 1 - 0.783 = 0.217\)

\(P(S \mid NL) = \frac{0.084}{0.217} \approx 0.387\)

Thus, the probability that the present is a scarf given that Nikita’s mother does not like it is \(0.387\) or \(\frac{12}{31}\).