Write the sample space and calculate the probability based on the given information.

(A) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

(B) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

(C) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

(D) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

(E) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

(A) Sample space has 6 × 6 = 36 (equally likely outcomes or ordered pairs).

Prime number greater than 5 upto 12: 7, 11.

Sum = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 outcomes.

Sum = 11: (5, 6), (6, 5) → 2 outcomes.

Total favourable outcomes = 6 + 2 = 8.

$$P(\text{prime greater than } 5) = \frac{8}{36} = \frac{2}{9}$$

(B) Total balls = 9.

Total ways to draw 2 balls (ordered) = $9 \times 8 = 72$.

$$P(\text{same colour}) = P(RR) + P(GG) + P(BB)$$

$$= \left(\frac{4}{9}\right)\left(\frac{3}{8}\right) + \left(\frac{3}{9}\right)\left(\frac{2}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{1}{8}\right)$$

$$= \frac{12}{72} + \frac{6}{72} + \frac{2}{72}$$

$$=\frac{20}{72}=\frac{5}{18}$$

$$P(\text{different colours}) = 1 - \frac{5}{18} = \frac{13}{18}$$

(C) Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, $n(S) = 8$ .

Favourable outcomes: $\mathrm{HHT}$, $\mathrm{HTH}$. (Both have first coin $\mathrm{H}$ and 2 total heads.)

$$P = \frac{2}{8} = \frac{1}{4} = 0.25$$

(D) Total four-digit arrangements = 4! = 24.

Number is even iff last digit is even, i.e., 2 or 4.

Favourable: last digit = 2 or 4.

For each of the 2 choices of last digit, the remaining 3 digits can be arranged in 3! = 6 ways.

Total favourable = $2 \times 6 = 12$ .

$$P(\text{even}) = \frac{12}{24} = \frac{1}{2}$$

Each question: $P(\text{correct}) = \frac{1}{4}$, $P(\text{wrong}) = \frac{3}{4}$.

Since the three questions are independent, 'exactly two correct' means two questions are right and one is wrong.

The single wrong answer can occur in any of the three questions, giving three equally likely cases.

$$\text{For each such case: } P = \left(\frac{1}{4}\right)^2 \times \left(\frac{3}{4}\right) = \frac{3}{64}.$$

$$P(\text{exactly 2 correct}) = 3 \times \frac{3}{64} = \frac{9}{64}.$$