A six faced die is biased such that
3 × P(a prime number) =6 × P(a composite number) =2 × P(1).
Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :

Solution

Maths sir · 4 steps

Read each probability as a per-face value, make all six faces add to 1, then mean of X is just np.

1. Set a common value. Let 3P(prime) = 6P(composite) = 2P(1) = k, with each P meaning the probability of one face. Then each prime face (2, 3, 5) has probability k/3, each composite face (4, 6) has k/6, and face 1 has k/2.

   3 × k/3 + 2 × k/6 + k/2 = 11k/6 = 1 ⇒ k = 6/11

2. Face probabilities. So P(2) = P(3) = P(5) = 2/11, P(4) = P(6) = 1/11 and P(1) = 3/11. These add to (6 + 2 + 3)/11 = 1, as they should.
3. Perfect square. The perfect squares on a die are 1 and 4. So on one throw, p = P(1) + P(4) = 3/11 + 1/11 = 4/11.
4. So. X counts successes in n = 2 independent throws, so it is binomial and its mean is np = 2 × 4/11 = 8/11. The answer is (d).

The idea to keep. When a variable counts successes over repeated independent trials, it is binomial: find the single-trial success probability carefully, and the mean is np.

Answer from the NTA final answer key. Solution by PrepWiser.