1 Answer

Shawn Esterman · Answer 1 · 2022-06-21T22:25:26+0000

Given:

A fair die is rolled.

It pays off $10 for 6, $7 for a 5, $4 for a 4 and no payoff otherwise.

To find:

The expected winning for this game.

Solution:

If a die is rolled then the possible outcomes are 1, 2, 3, 4, 5, 6.

The probability of getting a 6 is:

P(6)=(1)/(6)

The probability of getting a 5 is:

P(5)=(1)/(6)

The probability of getting a 4 is:

P(4)=(1)/(6)

The probability of getting other numbers (1,2,3) is:

$P(\text{Otherwise})=(3)/(6)$

$P(\text{Otherwise})=(1)/(2)$

We need to find the sum of product of payoff and their corresponding probabilities to find the expected winning for this game.

$E(x)=10* P(6)+7* P(5)+4* P(4)+0* P(\text{Otherwise})$

E(x)=10* (1)/(6)+7* (1)/(6)+4* (1)/(6)+0* (1)/(2)

E(x)=(10)/(6)+(7)/(6)+(4)/(6)+0

E(x)=(10+7+4)/(6)

E(x)=(21)/(6)

E(x)=3.5

Therefore, the expected winnings for this game are $3.50.

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