Question

A 6-sided die is biased. Now, the numbers one to four are equally likely to happen, but five and six is thrice as likely to land face up as each of the other numbers. If X is the number shown on the uppermost face, determine the expected value of X when 6 is shown on the uppermost face.

(a) \(\frac{13}{4}\)

(b) \(\frac{3}{5}\)

(c) \(\frac{2}{7}\)

(d) \(\frac{21}{87}\)

I got this question in homework.

The query is from Discrete Probability in chapter Discrete Probability of Discrete Mathematics

Answer 1

Right choice is (a) \(\frac{13}{4}\)

The best I can explain: Let P(1) = P(2) = P(3) = P(4) = p; P(5) = P(6) = 2p. We know that the sum of all probabilities must be 1 ⇒ p + p + p + p + 2p + 2p = 1

 ⇒ 8p = 1 ⇒ p = \(\frac{1}{8}\)

 Expected Value:

μ = E(X) = ∑x * P(x) =  \(1 * \frac{1}{8} + 2 * \frac{1}{8} + 3 * \frac{1}{8} + 4 * \frac{1}{8} + 5 * \frac{2}{8} +  6 * \frac{2}{8} = \frac{13}{4}\).