Step-by-Step Guidance for Conditional Probability and Random Variables (Statistics)

Q3.1. Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers?

Background

Topic: Conditional Probability

This question tests your understanding of conditional probability, specifically how to compute the probability of an event (at least one die shows 6) given that another event (the dice show different numbers) has occurred.

Key Terms and Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

• Sample Space for Two Dice: There are 36 possible outcomes when rolling two dice.

• "At least one lands on 6": This means either the first die, the second die, or both show a 6.

• "Dice land on different numbers": The two dice show different values (no doubles).

Step-by-Step Guidance

1. List all possible outcomes when two dice are rolled. There are $6 \times 6 = 36$ total outcomes.

2. Determine the number of outcomes where the dice land on different numbers. (Hint: For each value on the first die, the second die can be any of the other 5 values.)

3. Find the number of outcomes where at least one die shows a 6 and the dice are different. (Think about cases where the first die is 6 and the second is not, and vice versa.)

4. Use the conditional probability formula: $P(\text{at least one 6} | \text{different numbers}) = \frac{P(\text{at least one 6 and different})}{P(\text{different numbers})}$

Try solving on your own before revealing the answer!

Final Answer: $\frac{10}{30} = \frac{1}{3}$

There are 30 outcomes where the dice are different, and 10 of these have at least one 6. So the conditional probability is $\frac{10}{30} = \frac{1}{3}$.

Q3.2. If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i? Compute for all values of i between 2 and 12.

Background

Topic: Conditional Probability, Discrete Sample Spaces

This question asks you to compute a conditional probability for each possible sum of two dice, focusing on the event that the first die is a 6.

Key Terms and Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

• Sum of Two Dice: For each sum $i$ (from 2 to 12), count the number of outcomes that produce that sum.

• First die is 6: For each sum $i$, determine if it is possible for the first die to be 6, and if so, what the second die must be.

Step-by-Step Guidance

1. For each sum $i$ from 2 to 12, list all possible pairs $(d_1, d_2)$ such that $d_1 + d_2 = i$.

2. For each $i$, check if $d_1 = 6$ is possible. If so, what must $d_2$ be?

3. Count the total number of outcomes for each sum $i$ (denominator).

4. Count the number of outcomes where the first die is 6 and the sum is $i$ (numerator).

5. Set up the conditional probability for each $i$ as $P(\text{first die is 6} | \text{sum is } i) = \frac{\text{number of outcomes with first die 6 and sum } i}{\text{number of outcomes with sum } i}$

Try solving on your own before revealing the answer!

Final Answer:

For each $i$ from 2 to 12, the conditional probability is:

• $i=7$: $\frac{1}{6}$

• $i=8$: $\frac{1}{5}$

• $i=9$: $\frac{1}{4}$

• $i=10$: $\frac{1}{3}$

• $i=11$: $\frac{1}{2}$

• $i=12$: $1$

• For $i < 7$, the probability is $0$.

For each sum, only one outcome has the first die as 6, and the denominator is the number of ways to get that sum.

Q3.5. An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

Background

Topic: Probability with Sequential Events, Hypergeometric Distribution

This question tests your ability to calculate the probability of a specific sequence of draws without replacement from an urn.

Key Terms and Formulas

• Without Replacement: The composition of the urn changes after each draw.

• Multiplication Rule: The probability of a sequence of dependent events is the product of the conditional probabilities at each step.

Step-by-Step Guidance

1. Calculate the probability that the first ball is white: $P_1 = \frac{6}{15}$

2. Given the first was white, calculate the probability that the second is white: $P_2 = \frac{5}{14}$

3. Given the first two were white, calculate the probability that the third is black: $P_3 = \frac{9}{13}$

4. Given the first two were white and the third was black, calculate the probability that the fourth is black: $P_4 = \frac{8}{12}$

5. Multiply these probabilities together to get the probability of the sequence: $P = P_1 \times P_2 \times P_3 \times P_4$

Try solving on your own before revealing the answer!

Final Answer: $\frac{6}{15} \times \frac{5}{14} \times \frac{9}{13} \times \frac{8}{12}$

Multiply these fractions to get the final probability. This gives the probability that the first two are white and the last two are black, drawn without replacement.

Q4.3. Find the expected value $E[X]$ where $X$ is the outcome when we roll a fair die.

Background

Topic: Expected Value of a Discrete Random Variable

This question tests your understanding of how to compute the expected value (mean) of a discrete random variable, specifically for a uniform distribution (fair die).

Key Terms and Formulas

• Expected Value: $E[X] = \sum_{i} x_i p(x_i)$

• Fair Die: Each outcome from 1 to 6 has probability $\frac{1}{6}$.

Step-by-Step Guidance

1. List all possible outcomes: $x = 1, 2, 3, 4, 5, 6$.

2. Assign the probability for each outcome: $p(x) = \frac{1}{6}$ for each $x$.

3. Set up the expected value formula: $E[X] = \sum_{x=1}^{6} x \cdot \frac{1}{6}$

4. Add the products for each value of $x$.

Try solving on your own before revealing the answer!

Final Answer: $E[X] = 3.5$

The expected value is the average of the numbers 1 through 6, each weighted equally.

Q4.5. Calculate $Var(X)$ if $X$ represents the outcome when a fair die is rolled.

Background

Topic: Variance of a Discrete Random Variable

This question tests your ability to compute the variance of a discrete random variable, using the definition $Var(X) = E[X^2] - (E[X])^2$.

Key Terms and Formulas

• Variance: $Var(X) = E[X^2] - (E[X])^2$

• Expected Value: $E[X] = \sum_{i} x_i p(x_i)$

• Expected Value of $X^2$: $E[X^2] = \sum_{i} x_i^2 p(x_i)$

Step-by-Step Guidance

1. Recall from the previous question that $E[X] = 3.5$.

2. Compute $E[X^2] = \sum_{x=1}^{6} x^2 \cdot \frac{1}{6}$.

3. Calculate each $x^2$ for $x = 1$ to $6$, multiply by$\frac{1}{6}$, and sum.

4. Plug $E[X]$ and $E[X^2]$ into the variance formula: $Var(X) = E[X^2] - (E[X])^2$.

Try solving on your own before revealing the answer!

Final Answer: $Var(X) = \frac{35}{12}$

After calculating $E[X^2] = \frac{91}{6}$ and $E[X] = 3.5$, plug into the formula to get the variance.