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:
=210(0.039)(0.17798)
=1.4577
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