Multiple Dice

Solution & Explanation

Understanding the Problem

The problem asks us to find the probability of rolling at least one 4 when rolling two dice, and then extend this to n dice. To solve this, we need to understand the concept of complementary probability and independent events.

Key Concepts

1. Complementary Probability: The probability of an event happening is 1 minus the probability of it not happening.

   $P(A) = 1 - P(A')$

2. Independent Events: The outcome of one dice roll does not affect the outcome of another dice roll.

Solution Steps

1. Probability with Two Dice

• Single Dice:

  • Probability of rolling a 4: $P(4) = \frac{1}{6}$
  • Probability of not rolling a 4: $P(\text{not } 4) = 1 - \frac{1}{6} = \frac{5}{6}$

• Two Dice:

  • Probability of not rolling a 4 on both dice: $\left(\frac{5}{6}\right)^2 = \frac{25}{36}$

  • Probability of rolling at least one 4:

    $P(\text{at least one 4}) = 1 - P(\text{no 4 on both dice}) = 1 - \frac{25}{36} = \frac{11}{36}$

2. Extending to n Dice

• n Dice:

  • Probability of not rolling a 4 on a single die: $\frac{5}{6}$

  • Probability of not rolling a 4 on all n dice: $\left(\frac{5}{6}\right)^n$

  • Probability of rolling at least one 4:

    $P(\text{at least one 4}) = 1 - \left(\frac{5}{6}\right)^n$

Explanation

• Complementary Probability: By calculating the probability of the complementary event (not rolling a 4), we can easily find the probability of the event of interest (rolling at least one 4).

• Independence: Each dice roll is independent, allowing us to multiply probabilities across dice.

• Generalization: Extending from two dice to n dice involves using the same logic but applying it across more dice, resulting in the formula $1 - \left(\frac{5}{6}\right)^n$.

This approach efficiently calculates the probability of rolling at least one 4, leveraging the simplicity of complementary probability and the independence of dice rolls.