Two dice are thrown simultaneously. If \( X \) denotes the number of fours, then the expectation of \( X \) will be:
\(\frac{1}{3}\)
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:
- First die is a four AND second die is not a four: \(P(\text{4 on 1st and not 4 on 2nd}) = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}\)
- 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
| \(x\) (Number of Fours) | \(P(X=x)\) |
|---|---|
| 0 | \(\frac{25}{36}\) |
| 1 | \(\frac{10}{36}\) |
| 2 | \(\frac{1}{36}\) |
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
| Concept | Description | Formula/Example |
|---|---|---|
| Sample Space (\(\Omega\)) | The set of all possible outcomes of a random experiment. | Two dice: \(\{(1,1), (1,2), ..., (6,6)\}\), \(|\Omega|=36\) |
| Event | A subset of the sample space. | Getting a four on the first die: \(\{(4,1), (4,2), ..., (4,6)\}\) |
| Probability of an Event (A) | The likelihood of event A occurring. | For equally likely outcomes: \(P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}}\) |
| Random Variable (X) | A variable whose value is a numerical outcome of a random phenomenon. | Number of fours when rolling two dice. |
| Expectation \(E[X]\) | The average value of a random variable over many repetitions of the experiment. Also called the expected value or mean. | \(E[X] = \sum x \cdot P(X=x)\) for a discrete random variable. |
| Binomial Distribution | Describes the number of successes in a fixed number of independent Bernoulli trials. | Our problem is a Binomial case: \(n=2\) trials (dice), success = getting a four (\(p=1/6\)). \(X \sim B(2, 1/6)\). For \(X \sim B(n,p)\), \(E[X] = np\). Here, \(E[X] = 2 \times \frac{1}{6} = \frac{1}{3}\). |
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.
Important Questions from Probability
A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throws, and a number less than 4 in the third throw?
If the random variable \( X \) has the following distribution:
X 0 1 2 otherwise P(X) k 2k 3k 0 Match List-I with List-II:
List-I List-II (A) k (I) \(\frac{5}{6}\) (B) P(X < 2) (II) \(\frac{4}{3}\) (C) E(X) (III) \(\frac{1}{2}\) (D) P(1 ≤ X ≤ 2) (IV) \(\frac{1}{6}\) Choose the correct answer from the options given below:
Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form, where c is some constant:
\[ P(X = x) = \begin{cases} 0.1, & \text{if } x = 0 \\ cx, & \text{if } x = 1 \text{ or } x = 2 \\ c(5 - x), & \text{if } x = 3 \text{ or } x = 4 \\ 0, & \text{otherwise} \end{cases} \]
Match List-I with List-II:
List-I List-II (A) c (I) 0.75 (B) P(X ≤ 2) (II) 0.3 (C) P(X = 2) (III) 0.55 (D) P(X ≥ 2) (IV) 0.15 Choose the correct answer from the options given below:
For the differential equation \( (x \log_e x) dy = (\log_e x - y) dx \):
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.
(D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where A is an arbitrary constant.
Choose the correct answer from the options given below:
Which of the following is the probability of \( x \) successes in a binomial distribution with number of trials \( n \) and probability of success as \( \theta \) ( \( 0 < \theta < 1 \) ) in each trial?
