Problem 31 - Algebra 2

Question

A survey shows that 60\(\%\) of the adults floss their teeth every day. In a random sample of ten adults, what is the probability that exactly six floss every day? \(\begin{array}{llll}{\text { F } 11 \%} & {\text { G. } 25 \%} & {\text { H. } 60 \%} & {\text { I. } 100 \%}\end{array}\)

Step-by-Step Solution

Answer

The probability that exactly six out of ten randomly selected adults floss every day is 25% (G).

Step 1: Identify n, k and p

The number of trials (n) is 10 (number of adults). The number of successful trials (k) is 6 (adults who floss every day). The probability of success on any given trial (p) is 0.60 (according to the survey).

Step 2: Use the binomial probability formula

The Binomial Probability formula is \[ P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)) \] where \( C(n,k) \) is the combination of n items taken k at a time. This can be calculated by the formula \[ C(n,k) = n! / (k!(n-k)!) \] where '!' denotes factorial. Substitute n, k and p into the formula to get \[ P(X=6) = C(10, 6) * (0.60^6) * (0.40^4) \]

Step 3: Calculate C(10,6)

Using the combination formula, calculate C(10,6) as follows: \[ C(10,6) = 10! / (6!(10-6)!) = 210 \]

Step 4: Substitute C(10,6) into the binomial formula

Now substitute \(C(10,6) = 210\) into the binomial probability formula and evaluate the expression: \[ P(X=6) = 210 * (0.60^6) * (0.40^4) \]

Step 5: Calculate the probability

Evaluating the expression gives: \[ P(X=6) = 0.25 \]

Key Concepts

Combination Formula

To solve problems involving binomial probability, we often need to use the Combination Formula. This formula helps us calculate the number of ways we can choose a subset of items from a larger set, which is crucial in probability calculations.
In our context, the formula is expressed as follows:

• \[ C(n, k) = \frac{n!}{k!(n-k)!} \]

Here, \( C(n, k) \) represents the number of combinations, \( n \) is the total number of items, and \( k \) is the number of items to choose from \( n \). The factorial notation \( ! \) indicates a product of all positive integers up to a certain number.
This formula is pivotal as it provides the different ways to "pick" our successful outcomes from the total possibilities. For instance, in our original problem with 10 adults and wanting to find the chance that exactly 6 floss regularly, \( C(10, 6) \) reflects the possible groups of 6 adults from 10 that might floss.

Probability Distribution

A probability distribution gives us a picture of all possible outcomes and the likelihood associated with each outcome. In the binomial probability context, it models the number of successful outcomes (like adults flossing in this survey) for a given number of trials and a set probability.
For a binomial distribution:

• The number of trials (\( n \)) is fixed.
• Each trial is independent of the rest.
• There are only two possible outcomes—success or failure.
• The probability of success (\( p \)) remains constant across each trial.

In our survey problem, each adult independently flosses (or not) with a probability of 0.60. The binomial formula \( P(X=k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{n-k}) \) helps us calculate the probability of exactly \( k \) successes. Here, the formula integrates the combination, showing how probability doubles down by considering both likelihood and arrangement of successes and failures.

Factorial Calculation

At the heart of combination calculations is the use of factorials. A factorial, symbolized by the exclamation mark \( ! \), represents the product of an integer and all the positive integers below it. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factorials are important in probability and combination calculations because they help in determining the total number of ways to arrange items where order is not sensitive, as seen in combinations.

• The formula \( n! \) computes the total possible arrangements of \( n \) items.
• The combination formula leverages factorials to scale these possibilities to only those combinations we're interested in (e.g., \( k \) successful trials out of \( n \) total).

In the original problem, calculating \( 10! \) and dividing by \( 6! \times 4! \) gives us \( C(10, 6) \), revealing the number of ways we can form groups of 6 flossers from 10, considering all arrangements without concern for order.