Variance and Game Favorable Example on Probability Distribution

In this class, We discuss Variance and Game Favorable Example on Probability Distribution.

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The reader should have prior knowledge of the variance of a probability distribution. Click Here.

Example 1:

The below table shows the probability distribution for a random variable X.

Find variance?

Solution:

E(X) = μ = all x Σ xf(x)

E(X) = -1(0.3) + 0(0.1) + 1(0.1) + 2(0.3) + 3(0.2)

E(X) = 1

Variance = : σ2 = E(X2) – (E(X))2

E(X2) = all x Σ x2f(x)

-12 0.3 + 0 (0.1) + 22 (0.3) + 1(0.1) + 32(0.2)

= 3.4

σ2 = E(X2) – (E(X))2

σ2 = 3.4 – 1

σ2 = 2.4

Example 2:

A player wins if he get 5 on the dice.

player looses if he gets 2 or 4 on the dice.

if the player wins heget 50.

If the player losses he get 10.

Otherwise, he has to pay 15.

Is the game favourable to the player?

Solution:

Random variable X = {-15, 10, 50}

P(X) = { 1/2, 1/3, 1/6}

Player will get 50 if he throws 5 on the dice.

P(X = 50 ) = 1/6

Similarly P(X = 10) = 1/3

P(X = -15) = 1/2

Find E(X) why?

If he plays the game repeatedly, The amount the player will get is the expected value.

E(X) = μ = all x Σ xf(x)

= -15(1/2) + 10(1/3) + 50(1/6)

= 25/6

= 4.2

The average amount player will get 4.2

The game is favourable to the player.