Question

A fair six sided die has "1" on the face ,"2" on two of its faces and "3".on the remaining three faces. The die is rolled twice. If T is the total score write down the probability distribution of T hence determine

a) the probability that T is more than 4

b)The mean and the variance of T

Answer 1

Total number of outcomes $= 6 \times 6 = 36$

$P= \frac{1}{36}$

a) P(T>4) = P(T=5) +P(T=5)

12 times T=5

9 times T=6

$P(T>4) = \frac{12}{36} + \frac{9}{36} = \frac{12+9}{36}=\frac{21}{36} = \frac{7}{12}$

b)

$P(T=2) = \frac{1}{36} \\ P(T=3) = \frac{4}{36} \\ P(T=4) = \frac{10}{36} \\ P(T=5) = \frac{12}{36} \\ P(T=6) = \frac{9}{36}$

The mean

$E(T) = \sum^6_{T=2} T \times P(T) \\ E(T) = 2 \times \frac{1}{36} + 3 \times \frac{4}{36} + 4 \times \frac{10}{36} + 5 \times \frac{12}{36} + 6 \times \frac{9}{36} \\ = \frac{2 + 12 +40 + 60 + 54}{36} = 4.66 \\ E(T^2) = T^2 \times P(T) \\ E(T^2) = 2^2 \times \frac{1}{36} + 3^2 \times \frac{4}{36} + 4^2 \times \frac{10}{36} + 5^2 \times \frac{12}{36} + 6^2 \times \frac{9}{36} \\ = \frac{4+36+160+300+324}{36} = 22.88$

The variance

$V(T) = E(T^2) -(E(T))^2 \\ V(T) = 22.88 - (4.66)^2 = 1.13$