Question

Suppose you roll two 4 sided dice. Let the probabilities of the first die be represented by random variable X and those of the second die be represented by random variable Y. Let random variable Z be (X-Y). The range of Z will be from -3 to 3. 1. (2.5 points) Find E(Z). 2. (2.5 points) Find P(Z =-1|Z < 1). 3. (2.5 points) What is the probability that you have to play the game 3 times before you roll Z=2? 4. (2.5 points) What is the probability that when you play the game 10 times, you get Z=0 4 times out of the 10 rolls?

Answer 1

Answer:

Given that,

Suppose you roll two 4 sided dice.

Let the probabilities of the first die be represented by random variable X and those of the second die be represented by random variable Y Let random variable Z be (X-Y).

The range of Z will be from -3 to 3.

The values of Z are computed for different X and Y as:

	Y=1	Y=2	Y=3	Y=4
X=1	Z=0	Z=-1	Z=-2	Z=-3
X=2	Z=1	Z=0	Z=-1	Z=-2
X=3	Z=2	Z=1	Z=0	Z=-1
X=4	Z=3	Z=2	Z=1	Z=0

P(Z = -3) = 1/16
P(Z = -2) = 2/16
P(Z = -1) = 3/16
P(Z = 0) = 4/16
P(Z = 1) = 3/16
P(Z = 2) = 2/16
P(Z = 3) = 1/16

(1).

Find E(Z):

The expected value here is computed as:

E(Z) = (3/16)(-1 + 1) + (2/16)(-2 + 2) + (1/16)(3 - 3) + 0(4/16) = 0

Therefore E(Z) = 0

(2).

Find P(Z =-1|Z < 1):

P(Z = -1 | Z < 1) is computed using bayes theorem as:

P(Z = -1) / P(Z < 1)

= 1/10

= 0.1

Therefore 0.1 is the required probability here.

(3).

P(Z = 2) = 2/16

Probability that the game has to be played 4 times before we get z = 2 is computed here as:

= (14/16)^3(2/16)

= 0.0837

Therefore 0.0837 is the required probability here.

(4).

P(Z = 0) = 4/16

The required probability here is computed as:

$\small = \binom{10}{4}(4/16)^4(12/16)^6$

=210(0.039)(0.17798)

=1.4577