First, calculate the probability of a person being group O+ given they are Rhesus +. The probability of being O+ is 37%, and the probability of being Rhesus + is the sum of A+, B+, AB+, and O+, which is 35% + 8% + 3% + 37% = 83%.

Thus, the probability of being O+ given Rhesus + is:

\(P(O \text{ given } +) = \frac{0.37}{0.83} = 0.4458\)

Now, use the binomial distribution to find the probability that fewer than 3 out of 9 people are group O+.

Let \(X\) be the number of people who are group O+. Then \(X \sim \text{Binomial}(9, 0.4458)\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

\(P(X = 0) = \binom{9}{0} (0.4458)^0 (0.5542)^9\)

\(P(X = 1) = \binom{9}{1} (0.4458)^1 (0.5542)^8\)

\(P(X = 2) = \binom{9}{2} (0.4458)^2 (0.5542)^7\)

Calculating these:

\(P(X = 0) = (0.5542)^9\)

\(P(X = 1) = 9 \times (0.4458) \times (0.5542)^8\)

\(P(X = 2) = 36 \times (0.4458)^2 \times (0.5542)^7\)

Summing these probabilities gives:

\(P(X < 3) = 0.156\)

(i) The probability that a boy chooses Music is 0.15 (since 75% choose Games and 10% choose Drama, leaving 15% for Music). We use the binomial distribution with parameters: number of trials = 6, probability of success (choosing Music) = 0.15.

The probability that fewer than 3 boys choose Music is:

\(P(0, 1, 2) = (0.85)^6 + (0.15)(0.85)^5 \binom{6}{1} + (0.15)^2(0.85)^4 \binom{6}{2}\)

Calculating each term:

\((0.85)^6 = 0.377149515625\)

\((0.15)(0.85)^5 \binom{6}{1} = 0.2679075\)

\((0.15)^2(0.85)^4 \binom{6}{2} = 0.308\)

Summing these probabilities gives \(0.953\).

(ii) The probability that a student is a Drama student is:

\(P(D) = 0.6 \times 0.1 + 0.4 \times 0.55 = 0.28\)

The probability that a Drama student is a boy is:

\(P(B|D) = \frac{P(B \cap D)}{P(D)} = \frac{0.06}{0.28} = 0.2143\)

The probability that at least 1 of the 5 Drama students is a boy is:

\(P(>1) = 1 - P(0) = 1 - (0.7857)^5\)

\(P(0) = (0.7857)^5 = 0.7078\)

Thus, \(P(>1) = 1 - 0.7078 = 0.2922\)

Therefore, the probability is \(0.701\).

(ii) The mean of \(X\) is given by \(E(X) = 14 \times 0.75 = 10.5\). We evaluate the probabilities for \(X = 10\) and \(X = 11\):

\(P(10) = \binom{14}{10} (0.75)^{10} (0.25)^4 = 0.220\)

\(P(11) = \binom{14}{11} (0.75)^{11} (0.25)^3 = 0.240\)

The value of \(X\) with the highest probability is 11.

(iii) We first find the probability that Sue completes more than 11 puzzles in a month:

\(P(>11) = \binom{14}{12} (0.75)^{12} (0.25)^2 + \binom{14}{13} (0.75)^{13} (0.25)^1 + (0.75)^{14} = 0.281\)

Now, we find the probability that in exactly 3 of the next 5 months, Sue completes more than 11 puzzles:

\(P(3) = \binom{5}{3} (0.281)^3 (0.7189)^2 = 0.115\)

(i) The three conditions for a binomial distribution are:

1. Constant probability of success.
2. Independent trials.
3. Fixed number of trials with only two outcomes (success or failure).

(ii) Let the random variable \(X\) represent the number of days Julie's train is late out of 9 days. \(X\) follows a binomial distribution with parameters \(n = 9\) and \(p = 0.3\).

We need to find \(P(X > 7) + P(X < 2)\).

\(P(X > 7) = P(X = 8) + P(X = 9)\)

\(P(X = 8) = \binom{9}{8} (0.3)^8 (0.7)^1\)

\(P(X = 9) = \binom{9}{9} (0.3)^9\)

\(P(X < 2) = P(X = 0) + P(X = 1)\)

\(P(X = 0) = \binom{9}{0} (0.3)^0 (0.7)^9\)

\(P(X = 1) = \binom{9}{1} (0.3)^1 (0.7)^8\)

Calculating these probabilities:

\(P(X = 8) = 9 \times (0.3)^8 \times (0.7) = 0.000015\)

\(P(X = 9) = (0.3)^9 = 0.00000019683\)

\(P(X = 0) = (0.7)^9 = 0.0403536\)

\(P(X = 1) = 9 \times (0.3) \times (0.7)^8 = 0.121060821\)

Adding these probabilities:

\(P(X > 7) + P(X < 2) = 0.000015 + 0.00000019683 + 0.0403536 + 0.121060821 = 0.196\)

This problem can be modeled using a binomial distribution where the probability of success (Martin playing games when his friend phones) is 0.25, and the probability of failure is 0.75. We are looking for the probability of exactly 2 successes in 8 trials.

The probability mass function for a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.

Substituting the given values:

\(P(X = 2) = \binom{8}{2} (0.25)^2 (0.75)^6\)

Calculating each part:

\(\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28\)

\((0.25)^2 = 0.0625\)

\((0.75)^6 = 0.17803125\)

Therefore,

\(P(X = 2) = 28 \times 0.0625 \times 0.17803125 = 0.311\)