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mihalych1998 [28]
3 years ago
12

Consider a game in which a participant pays $2 to roll a die. The participant receives $3 if they roll a 1 (i.E. They go up by a

$1) and they receive $4 if they roll a 6 (i.E. They go up by $2). If they roll any other number they lose their $2 (i.E. They lost $2). What is the average expected monetary value of a single roll?
Mathematics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

The expected monetary value of a single roll is $1.17.

Step-by-step explanation:

The sample space of rolling a die is:

S = {1, 2, 3, 4, 5 and 6}

The probability of rolling any of the six numbers is same, i.e.

P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{6}

The expected pay for rolling the numbers are as follows:

E (X = 1) = $3

E (X = 2) = $0

E (X = 3) = $0

E (X = 4) = $0

E (X = 5) = $0

E (X = 6) = $4

The expected value of an experiment is:

E(X)=\sum x\cdot P(X=x)

Compute the expected monetary value of a single roll as follows:

E(X)=\sum x\cdot P(X=x)\\=[E(X=1)\times \frac{1}{6}]+[E(X=2)\times \frac{1}{6}]+[E(X=3)\times \frac{1}{6}]\\+[E(X=4)\times \frac{1}{6}]+[E(X=5)\times \frac{1}{6}]+[E(X=6)\times \frac{1}{6}]\\=[3\times \frac{1}{6}]+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]\\+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]+[4\times \frac{1}{6}]\\=1.17

Thus, the expected monetary value of a single roll is $1.17.

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The missing values in the inequality problem are;

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<h3>What is the explanation for the above?</h3>

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