Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Tossing dice Three dice are tossed. (a) Find the probability that all dice show the same number of dots. (b) Find the probability that the numbers of dots on the dice are all different. (c) Work parts (a) and (b) for \(n\) dice.

Step-by-Step Solution

Answer

(a) Probability is \(\frac{1}{36}\); (b) Probability is \(\frac{5}{9}\); (c) General probabilities: \(\frac{1}{6^{n-1}}\) for same, \(\frac{_6P_n}{6^n}\) for different.

Step 1: Determine Total Possible Outcomes for Three Dice

When three dice are tossed, each die has 6 possible outcomes. Therefore, the total number of possible outcomes is the product of the individual outcomes, calculated as follows: \(6 \times 6 \times 6 = 216\). There are 216 possible combinations of dots that can appear when three dice are rolled.

Step 2: Probability of All Dice Showing the Same Number (Part a)

For all three dice to show the same number, they can either all be 1s, all 2s, ... up to all 6s. This gives 6 possible outcomes where each die shows the same number (e.g., (1,1,1), (2,2,2), ... (6,6,6)).The probability is calculated as the number of favorable outcomes divided by the total outcomes: \(\frac{6}{216} = \frac{1}{36}\).

Step 3: Probability of All Dice Showing Different Numbers (Part b)

To find the probability that all dice show different numbers, we first determine the number of favorable outcomes. Choose a number for the first die (6 options), a different number for the second die (5 options), and a different number for the third die (4 options).The number of favorable outcomes is \(6 \times 5 \times 4 = 120\). The probability is then \(\frac{120}{216} = \frac{5}{9}\).

Step 4: Generalizing for n Dice (Part c for Same Numbers)

For \(n\) dice, all showing the same number, we need all dice to show any number from 1 to 6. Thus, there are 6 favorable outcomes. The total number of outcomes is \(6^n\).The probability is given by \(\frac{6}{6^n} = \frac{1}{6^{n-1}}\).

Step 5: Generalizing for n Dice (Part c for Different Numbers)

For \(n\) dice showing different numbers, start with \(6\) options for the first die, \(5\) for the second, and so on until \((6 - n + 1)\) for the nth die, provided \(n \leq 6\). Hence, the number of favorable outcomes is given by the permutation \(_6P_n\). The probability is \(\frac{_6P_n}{6^n}\), where \(_6P_n = \frac{6!}{(6 - n)!}\).

Key Concepts

Same Number Probability

Understanding the probability of all dice showing the same number starts with recognizing what it means for three dice to display identical numbers. In this situation, each die has to show either all 1s, all 2s, up to all 6s.

This results in 6 specific forthcoming outcomes, one for each number on a die. With 6 possible outcomes of similarity out of 216, the probability calculation is straightforward:

• Favorable outcomes: 6 (e.g., all dice show 1, 2, etc.)
• Total possible outcomes: 216
• Probability: \(\frac{6}{216} = \frac{1}{36}\)

This means that there's quite a slim chance of about 2.78% that all dice will turn up with the same number on a single roll.

Different Numbers Probability

Finding the probability for dice to show different numbers can seem trickier, but it's interesting! Here, each die must display a distinct number of dots. Start with 6 options for the first die. After choosing this number, the next die has 5 remaining numbers to choose from, and the last die 4.

The process can be broken down into these steps:

• First die: 6 options (any number from 1 to 6)
• Second die: 5 options (must be different from the first)
• Third die: 4 options (distinct from both the first and second)

Thus, the number of favorable outcomes is calculated like this: \(6 \times 5 \times 4 = 120\). Therefore, the probability that all three dice will display different numbers is:

• Favorable outcomes: 120
• Total possible outcomes: 216
• Probability: \(\frac{120}{216} = \frac{5}{9}\)

This tells us that there's about a 55.56% likelihood that each die will show a different number when rolled.

Permutation

Permutations are crucial when exploring the probability of events involving a specific order. In the context of rolling dice, permutations help determine how many ways objects—in this case, numbers—can be arranged.

When showing different numbers, each die pick influences the next. Choosing numbers for dice resembles placing distinct objects in specific spots without replacement.

The formula for permutations, expressed as \(_nP_r = \frac{n!}{(n-r)!}\), guides in computing the number of favorable outcomes. Here:

• \(_6P_3\) applies since we're dealing with the permutations of 3 dice picked from 6 numbers.
• Calculate: \(_6P_3 = \frac{6!}{(6-3)!} = \frac{720}{6} = 120\).

Using permutations in probability, especially with dice, reveals all possible sequences for each different setup. This explains how 120 favorable outcomes arise for dice showing different numbers.

Combinatorial Probability

Combinatorial probability combines counting principles to assess possible outcomes' number and success in an event. When addressing dice probabilities, understanding combinations—how items can be arranged without order—becomes pivotal. However, for dice, we frequently use permutations, considering distinct order matters.

For example, in calculating the probability of different numbers on dice, permutations play a major role, as the order of select numbers is critical. With combinations, only distinct groupings matter.

• Formula for combinations: \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)
• While not used for variable dice order, understanding helps differentiate it from permutations.

Using combinatorial principles lays the foundation for deeper probability concepts beyond just dice, enriching perspectives on how events can unfold according to defined constraints and rules.