Problem 4 - Algebra 2 and Trigonometry

Question

In \(3-6,\) find exact probabilities showing all required computation. A fair die is tossed five times. What is the probability of tossing a \(6 :\) $$ \begin{array}{llll}{\text { a. exactly once? }} & {\text { b. exactly twice? }} & {\text { c. exactly three times? }} \\ {\text { d. exactly four times? }} & {\text { e. exactly five times? }} & {\text { f. zero times? }}\end{array} $$ g. Which is the most likely event(s) when tossing a die five times?

Step-by-Step Solution

Answer

The most likely event is tossing zero 6's in five tosses of a fair die.

Step 1: Understand the Probability Distribution

The problem deals with a binomial distribution, where each die toss is an independent trial with two outcomes: tossing a 6 (success) or not tossing a 6 (failure). The probability of success, tossing a 6, on one die roll is \( \frac{1}{6} \), and the probability of failure is \( \frac{5}{6} \).

Step 2: Identify the Binomial Formula

The probability of getting exactly \( k \) successes (tossing a 6 exactly \( k \) times) in \( n \) independent trials is given by the formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient, \( n = 5 \), and \( p = \frac{1}{6} \).

Step 3: Calculate Probability for Exactly One Success

For exactly one 6:\[P(X = 1) = \binom{5}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^4 = 5 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^4 = 5 \cdot \frac{1}{6} \cdot \frac{625}{1296} = \frac{625}{15552}\]

Step 4: Calculate Probability for Exactly Two Successes

For exactly two 6s:\[P(X = 2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \cdot \left(\frac{1}{36}\right) \cdot \left(\frac{125}{216}\right) = \frac{1250}{7776}\]

Step 5: Calculate Probability for Exactly Three Successes

For exactly three 6s:\[P(X = 3) = \binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2 = 10 \cdot \frac{1}{216} \cdot \frac{25}{36} = \frac{250}{7776}\]

Step 6: Calculate Probability for Exactly Four Successes

For exactly four 6s:\[P(X = 4) = \binom{5}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^1 = 5 \cdot \frac{1}{1296} \cdot \frac{5}{6} = \frac{25}{7776}\]

Step 7: Calculate Probability for Exactly Five Successes

For exactly five 6s:\[P(X = 5) = \binom{5}{5} \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^0 = 1 \cdot \frac{1}{7776}= \frac{1}{7776}\]

Step 8: Calculate Probability for Zero Successes

For zero 6s:\[P(X = 0) = \binom{5}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^5 = 1 \cdot \left(\frac{5}{6}\right)^5 = \frac{3125}{7776}\]

Step 9: Determine the Most Likely Event

Compare probabilities calculated in Steps 3-8. The largest probability is \( \frac{3125}{7776} \) for zero 6s, which is the most likely event.

Key Concepts

Probability Theory

Probability theory is an essential branch of mathematics focusing on the quantification of uncertainty. It provides a framework to model random phenomena observed in the real world. When we talk about tossing a fair die, probability theory helps us determine the likelihood of each possible outcome.
In a single die toss, there are six possible outcomes: each face numbered 1 through 6 is equally likely to occur. We express the probability of rolling a specific number, such as a 6, as a fraction: \( \frac{1}{6} \). This is because there is exactly one favorable outcome from six possibilities.

• Probability is always a number between 0 and 1, where 0 represents an impossible event, and 1 indicates certainty.
• The sum of probabilities for all possible outcomes of an experiment, like a die roll, is always 1.

When multiple dice are tossed, each toss is considered independent. This means that the outcome of one toss does not affect the result of another. Probability theory also involves using models like the binomial distribution to compute probabilities for multiple independent events.

Binomial Coefficient

The binomial coefficient is a mathematical concept used to determine the number of ways to choose a subset of items from a larger set, irrespective of the order. It is crucial for tasks involving combinations, such as in binomial probability scenarios.
The binomial coefficient, denoted by \( \binom{n}{k} \), can be calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \( n! \) (n factorial) is the product of all positive integers up to \( n \). The symbol "!" indicates factorial, a crucial operation in combinatorics.
This coefficient tells us how many different ways exactly \( k \) successes can occur out of \( n \) trials. In the context of dice, \( \binom{5}{2} \) calculates how many ways two sixes can appear when a die is rolled five times. Understanding this ensures we accurately measure probabilities in a binomial distribution situation.

Independent Trials

In probability theory, independent trials refer to experiments where the result of one trial does not influence the outcome of another. They are a critical part of understanding more complex probability distributions like the binomial distribution.
Tossing a fair die multiple times perfectly illustrates independent trials. Each roll does not impact the others; the probability of rolling a 6 remains constant at \( \frac{1}{6} \), regardless of previous outcomes.

• Independent trials are crucial for predicting compound events accurately.
• They enable the use of binomial probability formulas, involving calculations for several outcomes over a series of trials.

In the problem involving five die tosses, each toss is independent, making it possible to calculate the probability of specific numbers of 6s occurring using the binomial distribution formula. By recognizing trials as independent, we can confidently multiply probabilities as shown in calculations like \( \left( \frac{5}{6} \right)^4 \) when determining the likelihood of certain sequences of results.