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Chapter 4: Problem 12

A box of chocolates contains 7 dark chocolate pieces and 3 milk chocolate pieces (and no others). If you randomly pick 2 pieces and eat each chocolate after choosing it, what is the probability of choosing at least one dark chocolate? Write the probability in all three forms.

Short Answer

Expert verified
The probability of choosing at least one dark chocolate is \( \frac{14}{15} \), approximately 0.9333, or 93.33%.

Step by step solution

01

Determine Total Ways to Choose 2 Chocolates

First, calculate the total number of ways to choose 2 chocolates from the box. There are 10 chocolates in total (7 dark and 3 milk). The number of ways to choose 2 chocolates from 10 is given by the combination formula: \( \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \).
02

Determine Ways to Choose 2 Milk Chocolates

Next, calculate the number of ways to pick 2 milk chocolates. Since there are 3 milk chocolates, the number of ways to choose 2 is \( \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 \).
03

Calculate Probability of Picking 2 Milk Chocolates

The probability of picking 2 milk chocolates is the number of ways to select 2 milk chocolates divided by the total number of ways to select any 2 chocolates: \( \frac{3}{45} = \frac{1}{15} \).
04

Calculate Probability of At Least One Dark Chocolate

The probability of picking at least one dark chocolate is the complement of the probability of picking 2 milk chocolates. This is calculated as: \[1 - \frac{1}{15} = \frac{14}{15}\].
05

Convert Probability into Different Forms

The probability \( \frac{14}{15} \) can be expressed as a decimal: approximately 0.9333, and as a percentage: 93.33%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. In the context of our chocolate box exercise, we use combinatorial methods to determine how many ways we can select chocolates from a total. For example, when calculating how to pick 2 chocolates out of 10, we use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]where \( n \) is the total number of chocolates, and \( r \) is the number we choose at a time. In our specific exercise, that combination \( \binom{10}{2} = 45 \) represents the total possible ways to select 2 chocolates. This helps in analyzing scenarios and computing probabilities, such as determining the likelihood of choosing milk or dark chocolates.
Complementary Probability
Complementary probability involves finding the probability of the 'complement' of an event. Simply put, if we know the probability of something not happening, we can find the probability of it happening by subtracting from 1. In this exercise, after we calculated the probability of picking 2 milk chocolates as \( \frac{1}{15} \), we used complementary probability to find the probability of choosing at least one dark chocolate. This was derived by taking the complement: \[1 - \frac{1}{15} = \frac{14}{15}\]This approach is particularly useful, as directly calculating complex events can be simplified by first finding 'what you don鈥檛 want to happen.'
Probability Forms
Probability can be expressed in different forms, which helps in understanding and communicating the concept clearly. The three most common forms are fraction, decimal, and percentage.In the exercise, the probability of picking at least one dark chocolate started as a fraction, \( \frac{14}{15} \). Converting it to a decimal, we get approximately 0.9333. When you express it as a percentage, it becomes 93.33%.
  • Fraction: This is the most mathematical form and is beneficial for further calculations.
  • Decimal: Useful for continuous scale measurements and easier comparison.
  • Percentage: It conveys probability in a familiar, everyday language, often used in reports and general communications.
Using these different forms allows you to adapt to various contexts and audiences, making probability versatile and widely applicable.

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Most popular questions from this chapter

A professor gave a test to students in a science class and in a math class during the same week. The grades are summarized below. $$ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \text { Total } \\ \hline \text { Science Class } & 7 & 18 & 13 & 38 \\ \hline \text { Math Class } & 10 & 8 & 9 & 27 \\ \hline \text { Total } & 17 & 26 & 22 & 65 \\ \hline \end{array} $$ If one student was chosen at random, find each probability: a. \(\mathrm{P}(\) in the math class) b. \(\mathrm{P}\) (earned a \(\mathrm{B}\) ) c. \(\mathrm{P}(\) earned an \(\mathrm{A}\) and was in the math class) d. \(\mathrm{P}(\) earned a \(\mathrm{B}\) given the student was in the science class) e. \(\mathrm{P}(\) is in the math class given that the student earned a \(\mathrm{B}\) )

A fitness center coach kept track over the last year of whether members stretched before they exercised, and whether or not they sustained an injury. Among the 400 members, 322 stretched before they exercised, 327 did not sustain an injury, and 270 both stretched and did not sustain an injury. a. Create a contingency table for the information. b. What is the probability that a member sustained an injury? c. What is the probability that a member sustained an injury and did not stretch? d. What is the probability that a member stretched or did not sustain an injury? e. What is the probability that a member sustained an injury given they stretched? f. What is the probability that a member sustained an injury given they did not stretch? g. Does it appear that sustaining an injury depends on whether the member stretches before exercising? Or are they independent? Use probability to support your elaim.

A ball is drawn randomly from a jar containing 6 red marbles, 2 white marbles, and 5 yellow marbles. Find the probability of: a. Drawing a white marble. b. Drawing a red marble. c. Drawing a green marble. d. Drawing two yellow marbles if you draw with replacement. e. Drawing first a red marble then a white marble if marbles are drawn without replacement.

There is a \(15 \%\) chance that a shopper entering a computer store will purchase a computer, a \(25 \%\) chance they will purchase a game/software, and there is a \(10 \%\) chance they will purchase both a computer and a game/software. a. Create a contingency table for the information. b. What is the probability that a shopper will not purchase a computer and will not purchase a game/software? c. What is the probability that a shopper will purchase a computer or purchase a game/software? d. What is the probability that a shopper will purchase a game/software given they have purchased a computer? e. What is the probability that a shopper will purchase a game/software given they did not purchase a computer? f. Does it appear that purchasing a game/software depends on whether the shopper purchased a computer? Or are they independent? Use probability to support your claim.

Compute the probability of rolling five fair six-sided dice (each side has equal probability of landing face up on each roll) and getting: a. a 3 on all five dice. b. at least one of the die shows a 3.
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