Math Problem Statement

On three rolls of a single​ die, you will lose

​$1111

if a

22

turns up at least​ once, and you will win

​$55

otherwise. What is the expected value of the​ game?

Question content area bottom

Part 1

Let X be the random variable for the amount won on a single play of this game.

​E(X)equals=enter your response here

dollars​ (Type an integer or a decimal rounded to the nearest cent as​ needed.)

Solution

To calculate the expected value of the game, we need to find the probabilities of the possible outcomes and their respective payouts.

Step 1: Define the events

  1. You lose $1111 if the number 2 turns up at least once in three rolls.
  2. You win $55 if the number 2 does not turn up at all in three rolls.

Step 2: Probabilities

The probability of rolling a specific number (e.g., 2) on a single roll is 16\frac{1}{6}, and the probability of not rolling that number is 56\frac{5}{6}.

(a) Probability of losing (P(lose)P(\text{lose})):

The probability of rolling at least one 2 in three rolls can be calculated using the complement rule. First, calculate the probability of not rolling a 2 in all three rolls: P(no 2 in all three rolls)=(56)3P(\text{no 2 in all three rolls}) = \left( \frac{5}{6} \right)^3 P(no 2 in all three rolls)=125216P(\text{no 2 in all three rolls}) = \frac{125}{216} Thus, the probability of losing is: P(lose)=1P(no 2 in all three rolls)=1125216=91216P(\text{lose}) = 1 - P(\text{no 2 in all three rolls}) = 1 - \frac{125}{216} = \frac{91}{216}

(b) Probability of winning (P(win)P(\text{win})):

The probability of winning is the complement of losing: P(win)=P(no 2 in all three rolls)=125216P(\text{win}) = P(\text{no 2 in all three rolls}) = \frac{125}{216}

Step 3: Expected value formula

The expected value is calculated as: E(X)=P(win)(winnings)+P(lose)(losses)E(X) = P(\text{win}) \cdot (\text{winnings}) + P(\text{lose}) \cdot (\text{losses}) Substitute the values: E(X)=(125216)(55)+(91216)(1111)E(X) = \left(\frac{125}{216}\right)(55) + \left(\frac{91}{216}\right)(-1111)

Step 4: Simplify the calculations

  1. Calculate the winning term: (125216)(55)=687521631.83\left(\frac{125}{216}\right)(55) = \frac{6875}{216} \approx 31.83

  2. Calculate the losing term: (91216)(1111)=101101216467.13\left(\frac{91}{216}\right)(-1111) = \frac{-101101}{216} \approx -467.13

  3. Combine the terms: E(X)=31.83467.13=435.30E(X) = 31.83 - 467.13 = -435.30

Final Answer:

E(X)=435.30dollarsE(X) = -435.30 \, \text{dollars}

This means, on average, you lose $435.30 per game.


Would you like to see a breakdown of any specific part of this solution?

Related Questions:

  1. What is the expected value if the loss was $500 instead of $1111?
  2. How does the probability of rolling a 2 change if the die was weighted?
  3. What is the expected value if you play this game 10 times?
  4. Can you determine the variance of this game?
  5. How do you calculate expected values for a weighted die?

Tip:

Always check if the probabilities add up

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Complement Rule

Formulas

Probability of event A: P(A)
Complement Rule: P(A') = 1 - P(A)
Expected Value Formula: E(X) = Σ[P(outcome) × value(outcome)]

Theorems

Basic Probability Theorems

Suitable Grade Level

Grades 11-12