Let the probability that a person agrees to complete the survey be 0.2. Therefore, the probability that a person does not agree is 0.8.

The probability that none of the first three people agree to complete the survey is given by:

\((0.8)^3 = 0.512\)

Thus, the probability that at least one person agrees is:

\(1 - (0.8)^3 = 1 - 0.512 = 0.488\)

Alternatively, using the binomial probability formula, the probability that exactly one, two, or three people agree can be calculated and summed:

\(P(1, 2, 3) = \binom{3}{1}(0.2)(0.8)^2 + \binom{3}{2}(0.2)^2(0.8) + (0.2)^3\)

\(= 3(0.2)(0.8)^2 + 3(0.2)^2(0.8) + (0.2)^3\)

\(= 0.384 + 0.096 + 0.008 = 0.488\)

(i) The probability that the die lands on 4 is \(\frac{1}{4}\). The probability of getting exactly 2 heads in 4 coin tosses is given by the binomial probability formula:

\(P(4, 2H) = \frac{1}{4} \times \binom{4}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^2\)

\(= \frac{1}{4} \times 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{2}{27}\)

(ii) The probability that the die lands on 3 is \(\frac{1}{4}\). The probability of getting exactly 3 heads in 3 coin tosses is:

\(P(3, 3H) = \frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)

(iii) We need to find the probability that the number the die lands on is the same as the number of heads. This includes the following cases:

• Die lands on 1 and 1 head: \(\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
• Die lands on 2 and 2 heads: \(\frac{1}{4} \times \binom{2}{2} \left( \frac{1}{3} \right)^2 = \frac{1}{36}\)
• Die lands on 3 and 3 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)
• Die lands on 4 and 4 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^4 = \frac{1}{324}\)

Summing these probabilities gives:

\(\frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \frac{1}{324} = \frac{10}{81}\)

(i) The two conditions required for a binomial distribution are:

1. Constant probability of success on each trial.
2. Independent trials/events.

(ii) To find the probability that Hebe completes at least 5 puzzles in a week, we use the binomial distribution with parameters:

\(= 7 (number of trials),\)

\(= 0.7 (probability of success).\)

We need to calculate:

\(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\)

\(= \binom{7}{5}(0.7)^5(0.3)^2 + \binom{7}{6}(0.7)^6(0.3)^1 + (0.7)^7\)

\(= 0.647\)

(iii) To find the probability that Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks, we first find the probability of completing 4 or fewer puzzles in a week:

\(P(X \leq 4) = 1 - P(X \geq 5) = 1 - 0.647 = 0.3529\)

Now, use the binomial distribution with parameters:

\(= 10 (weeks),\)

\(= 0.3529 (probability of completing 4 or fewer puzzles in a week).\)

We need to calculate:

\(P(Y = 3) = \binom{10}{3}(0.3529)^3(0.6471)^7\)

\(= 0.251\)

Let the probability that Khalid rides his skateboard on a given day be 0.4. We need to find the probability that he rides on his skateboard on at least 2 out of 10 days.

This is a binomial probability problem where the number of trials is 10, and the probability of success (riding the skateboard) is 0.4.

The probability of riding on the skateboard on 0 or 1 day is:

\(P(0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = (0.6)^{10}\)

\(P(1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 10 \times 0.4 \times (0.6)^9\)

Thus, the probability of riding on the skateboard on at least 2 days is:

\(1 - (P(0) + P(1)) = 1 - [(0.6)^{10} + 10 \times 0.4 \times (0.6)^9]\)

Calculating these values:

\((0.6)^{10} \approx 0.0060466176\)

\(10 \times 0.4 \times (0.6)^9 \approx 0.0403936\)

So:

\(1 - (0.0060466176 + 0.0403936) = 1 - 0.04644 \approx 0.954\)

(i) The probability that Annabel eats pasta on exactly 2 days in a week of 7 days can be modeled using the binomial distribution. The probability of eating pasta on a given day is 0.1, and not eating pasta is 0.9. We use the binomial formula:

\(P(2) = \binom{7}{2} (0.1)^2 (0.9)^5\)

Calculating this gives:

\(P(2) = 21 \times 0.01 \times 0.59049 = 0.124\)

(ii) For the second part, we need to find the probability that Annabel eats rice on 2 days, potato on 1 day, and pasta on 2 days in a period of 5 days. The probabilities are 0.75 for rice, 0.15 for potato, and 0.1 for pasta. The number of ways to arrange these days is given by:

\(\frac{5!}{2!1!2!}\)

The probability is then:

\((0.75)^2 (0.15)^1 (0.1)^2 \times \frac{5!}{2!1!2!}\)

Calculating this gives:

\(0.5625 \times 0.15 \times 0.01 \times 30 = 0.0253\)