Question

A nefarious gambler has developed a weighted six-sided die, with sides marked 1,2,3,4,5, and 6. The probability that a 6 is rolled is 0.4, and all the other sides are equally likely. Define the random variable X = the value on the die when it is rolled once. (a) Write out the pmf of X. (b) Compute the probability that an even number is rolled. (c) Compute the expectation and variance of X.

Answer 1

Answer:

b) 0.64

c)

$E(x) = 4.2\\\sigma^2 = 3.36$

Step-by-step explanation:

We are given the following in the question:

x : 1, 2, 3, 4, 5, 6

$P(6) = 0.4$

The probability of rolling a 1, 2, 3,4 and 5 is same.

Now, we know that

$\displaystyle\sum P(x_i) = 1\\\\\Rightarrow P(1)+P(2)+P(3)+P(3)+P(3)+P(6)=1\\\Rightarrow 5P(1) + 0.4 = 1\\\Rightarrow 5P(1)=0.6\\\Rightarrow P(1) = 0.12\\\Rightarrow P(1) =P(2) =P(3) =P(4) =P(5) = 0.12$

a) pmf of X

x:       1         2         3        4        5       6

P(x) 0.12    0.12    0.12    0.12   0.12    0.4

b) probability that an even number is rolled

$P(\text{Even number}) \\=P(2) + P(4)+P(6)\\=0.12 + 0.12 + 0.4\\=0.64$

c) expectation and variance of X

$E(x) = \displaystyle\sum x_iP(x_i)\\\\E(x) = 1(0.12) + 2(0.12)+3(0.12)+4(0.12)+5(0.12)+6(0.4)\\E(x) = 4.2\\E(x^2) = 1^2(0.12) + 2^2(0.12)+3^2(0.12)+4^2(0.12)+5^2(0.12)+6^2(0.4)\\E(x^2) = 21\\\sigma^2 = E(x^2) - (E(x))^2 = 21 - (4.2)^2 = 3.36$