Conditional Probability with Dice Rolls

Conditional Probability Problem Description

• Two fair dice are considered, rolled independently.
• Outcomes of the dice are denoted by the random variables $X1$ and $X2$.

Problem Statement

• Calculate the conditional probability:
  $Pr(X<em>1 + X</em>2 \geq 8 \mid X<em>1 < X</em>2)$

Definitions

• Conditional Probability: The probability of an event given that another event has occurred is expressed as: $Pr(A \mid B) = \frac{Pr(A \cap B)}{Pr(B)}$ where:
  • $Pr(A \mid B)$ is the conditional probability of event A given B.
  • $Pr(A \cap B)$ is the joint probability that events A and B both occur.
  • $Pr(B)$ is the probability that event B occurs.

Events

• Event A: $A = {X1 + X2 \geq 8}$
• Event B: $B = {X1 < X2}$

Step 1: Find $Pr(B)$

• To find $Pr(B)$, we need to count how many pairs $(X1, X2)$ satisfy $X1 < X2$.
• The possible outcomes when rolling two dice range from 1 to 6 for each die.
• The outcomes satisfying $X1 < X2$ include:
  • (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
  • (2, 3), (2, 4), (2, 5), (2, 6)
  • (3, 4), (3, 5), (3, 6)
  • (4, 5), (4, 6)
  • (5, 6)
• Total favorable pairs count = 15.
• The total number of outcomes when rolling two dice = 36.
• Therefore,
  $Pr(B) = \frac{15}{36} = \frac{5}{12}$

Step 2: Find $Pr(A \cap B)$

• We need to count the outcomes where both $X1 + X2 \geq 8$ and $X1 < X2$.
• Suitable pairs before calculating are: (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6).
• Valid outcomes are:
  1. (2, 6) → $X1 + X2 = 8$
  2. (3, 5) → $X1 + X2 = 8$
  3. (3, 6) → $X1 + X2 = 9$
  4. (4, 5) → $X1 + X2 = 9$
  5. (4, 6) → $X1 + X2 = 10$
  6. (5, 6) → $X1 + X2 = 11$
• Total pairs satisfying both $A$ and $B$ = 6.
• Therefore,
  $Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}$

Step 3: Calculation of conditional probability

• Substitute the values into the conditional probability formula:
  $Pr(X<em>1 + X</em>2 \geq 8 \mid X<em>1 < X</em>2) = \frac{Pr(A \cap B)}{Pr(B)} = \frac{\frac{1}{6}}{\frac{5}{12}}$
• Simplifying this gives:
  $Pr(X<em>1 + X</em>2 \geq 8 \mid X<em>1 < X</em>2) = \frac{1}{6} \times \frac{12}{5} = \frac{2}{5}$

Final Answer

• The answer, rounded to two decimal places is:
  $0.40$

Summary

• Therefore, $Pr(X1 + X2 \geq 8 \mid X1 < X2) = 0.40.