A die is rolled in such a way that an even number is twice likely to occur as an odd number. If the die is rolled twice, then the mean of the number of perfect squares in two tosses is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2/3
Let each odd face (1, 3, 5) have probability p. Then each even face (2, 4, 6) has probability 2p, and the six probabilities add up to 1:
\(3p + 3(2p) = 1\)
\(9p = 1\)
\( \displaystyle p = \frac{1}{9} \)
The perfect squares on a die are 1 (odd) and 4 (even):
\( \displaystyle P(\text{perfect square}) = \frac{1}{9} + \frac{2}{9} = \frac{1}{3} \)
The number X of perfect squares in two tosses is binomial with n = 2 and \(p = \frac{1}{3}\), and its mean is np:
\( \displaystyle E(X) = 2 \times \frac{1}{3} = \frac{2}{3} \)