Understanding Probability and Expectation with Dice

This question asks us to find the expectation of a random variable \(X\), which represents the number of fours obtained when two dice are thrown simultaneously. To solve this, we first need to understand the possible outcomes and their probabilities.

Sample Space and Outcomes

When a fair six-sided die is thrown, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}. The probability of getting any specific number, including a four, is \(\frac{1}{6}\).

When two fair six-sided dice are thrown simultaneously, the total number of possible outcomes is \(6 \times 6 = 36\). Each outcome is equally likely.

Defining the Random Variable X

The random variable \(X\) is defined as the number of fours obtained in the two throws. The possible values for \(X\) are 0, 1, or 2.

• \(X=0\): No fours are obtained.
• \(X=1\): Exactly one four is obtained.
• \(X=2\): Exactly two fours are obtained.

Calculating Probabilities for X

Let's calculate the probability for each possible value of \(X\).

• Probability of not getting a four on a single die: The outcomes that are not a four are {1, 2, 3, 5, 6}. There are 5 such outcomes. Probability of not getting a four = \(\frac{5}{6}\).
• Calculating \(P(X=0)\) (No fours): This means the first die is not a four AND the second die is not a four. Since the throws are independent: \(P(X=0) = P(\text{not 4 on 1st}) \times P(\text{not 4 on 2nd})\) \(P(X=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\)
• Calculating \(P(X=1)\) (Exactly one four): This can happen in two ways:
  1. First die is a four AND second die is not a four: \(P(\text{4 on 1st and not 4 on 2nd}) = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}\)
  2. First die is not a four AND second die is a four: \(P(\text{not 4 on 1st and 4 on 2nd}) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}\)
• Calculating \(P(X=2)\) (Exactly two fours): This means the first die is a four AND the second die is a four: \(P(X=2) = P(\text{4 on 1st}) \times P(\text{4 on 2nd})\) \(P(X=2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

Let's verify that the probabilities sum to 1:

\(P(X=0) + P(X=1) + P(X=2) = \frac{25}{36} + \frac{10}{36} + \frac{1}{36} = \frac{25+10+1}{36} = \frac{36}{36} = 1\)

This confirms our probability calculations are correct.

Probability Distribution Table

Calculating the Expectation of X

The expectation \(E[X]\) of a discrete random variable \(X\) is calculated using the formula:

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

We sum the product of each possible value of \(X\) and its corresponding probability:

\(E[X] = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))\)

Substitute the calculated probabilities:

\(E[X] = \left(0 \times \frac{25}{36}\right) + \left(1 \times \frac{10}{36}\right) + \left(2 \times \frac{1}{36}\right)\)

\(E[X] = 0 + \frac{10}{36} + \frac{2}{36}\)

\(E[X] = \frac{10 + 2}{36}\)

\(E[X] = \frac{12}{36}\)

Simplify the fraction:

\(E[X] = \frac{1}{3}\)

Conclusion on Expectation Calculation

The expectation of \(X\), the number of fours when two dice are thrown simultaneously, is \(\frac{1}{3}\).

This means that, on average, if you were to perform this experiment (throwing two dice and counting the fours) many times, the average number of fours you would expect to see per throw is \(\frac{1}{3}\).

Revision Table: Key Probability Concepts

Additional Information: Expectation and Its Properties

The expectation of a random variable is a fundamental concept in probability and statistics. It represents the center of the probability distribution.

• Interpretation: The expected value is not necessarily a value that the random variable will take in any single experiment. Instead, it's the long-term average value if the experiment is repeated many times. In our dice example, you can't roll one-third of a four, but the average number of fours over many pairs of rolls would approach \(\frac{1}{3}\).
• Linearity of Expectation: A very useful property is that for any random variables \(X_1, X_2, \dots, X_k\) and constants \(c_1, c_2, \dots, c_k\), the expectation of their linear combination is the linear combination of their expectations: \(E[c_1X_1 + c_2X_2 + \dots + c_kX_k] = c_1E[X_1] + c_2E[X_2] + \dots + c_kE[X_k]\). This property holds even if the random variables are not independent.
• Applying Linearity to our problem: Let \(Y_1\) be a random variable representing getting a four on the first die (1 if four, 0 otherwise). Let \(Y_2\) be a random variable representing getting a four on the second die (1 if four, 0 otherwise). \(E[Y_1] = 1 \cdot P(Y_1=1) + 0 \cdot P(Y_1=0) = 1 \cdot \frac{1}{6} + 0 \cdot \frac{5}{6} = \frac{1}{6}\). Similarly, \(E[Y_2] = \frac{1}{6}\). The total number of fours is \(X = Y_1 + Y_2\). By linearity of expectation: \(E[X] = E[Y_1 + Y_2] = E[Y_1] + E[Y_2] = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\). This method using indicator variables (\(Y_1, Y_2\)) and linearity confirms the result and is often simpler for sums of Bernoulli trials.