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Chapter 5: Problem 21

A candy dish contains five brown and three red M&Ms. A child selects three M&Ms without checking the colors. Use this information to answer the questions in Exercises \(18-21 .\) Write down \(p(x)\), the probability distribution for \(x\), the number of red M&Ms in the selection for \(x=0,1,2,3\)

Short Answer

Expert verified
Answer: The probability distribution for the number of red M&Ms (x) in the selection is as follows: - p(x=0) = 5/28 - p(x=1) = 15/28 - p(x=2) = 15/56 - p(x=3) = 1/56

Step by step solution

01

Understand the problem and identify the given information

In this problem, there are a total of 8 M&Ms: 5 brown and 3 red. A child selects 3 M&Ms without checking the colors. We need to find the probability distribution for the number of red M&Ms in the selection for \(x=0,1,2,3\).
02

Calculate the probabilities for each value of \(x\)

We will use the combinations formula to calculate the probabilities for each value of \(x\) separately. The combinations formula is given by: \(\binom{n}{k} =\frac{n!}{k!(n-k)!}\), where \(n!\) is the factorial of \(n\) and \(k!\) is the factorial of \(k\). Probability that \(x=0\) (no red M&Ms in the selection): In this case, all 3 M&Ms are brown. The number of ways to choose 3 brown M&Ms out of 5 is \(\binom{5}{3}=\frac{5!}{3!(5-3)!}=10\). The total number of ways to choose any 3 M&Ms out of the 8 is \(\binom{8}{3}=\frac{8!}{3!(8-3)!}=56\). So, \(p(x=0)=\frac{10}{56}=\frac{5}{28}\). Probability that \(x=1\) (1 red M&M in the selection): In this case, there are 2 brown and 1 red M&M. The number of ways to choose 2 brown M&Ms out of 5 is \(\binom{5}{2}=\frac{5!}{2!(5-2)!}=10\). The number of ways to choose 1 red M&M out of 3 is \(\binom{3}{1}=\frac{3!}{1!(3-1)!}=3\). So, the total number of ways to have 1 red M&M and 2 brown M&Ms in the selection is \(10\times3=30\), and \(p(x=1)=\frac{30}{56}=\frac{15}{28}\). Probability that \(x=2\) (2 red M&Ms in the selection): In this case, there is 1 brown and 2 red M&Ms. The number of ways to choose 1 brown M&M out of 5 is \(\binom{5}{1}=\frac{5!}{1!(5-1)!}=5\). The number of ways to choose 2 red M&Ms out of 3 is \(\binom{3}{2}=\frac{3!}{2!(3-2)!}=3\). So, the total number of ways to have 2 red M&Ms and 1 brown M&M in the selection is \(5\times3=15\), and \(p(x=2)=\frac{15}{56}\). Probability that \(x=3\) (all red M&Ms in the selection): In this case, all 3 M&Ms are red. The number of ways to choose 3 red M&Ms out of 3 is \(\binom{3}{3}=\frac{3!}{3!(3-3)!}=1\). So, \(p(x=3)=\frac{1}{56}\).
03

Write down the probability distribution for \(x\)

Based on our calculations in step 2, the probability distribution \(p(x)\) for \(x=0,1,2,3\) is as follows: \(p(x=0)=\frac{5}{28}\) \(p(x=1)=\frac{15}{28}\) \(p(x=2)=\frac{15}{56}\) \(p(x=3)=\frac{1}{56}\)

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Key Concepts

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

Combinations Formula
In probability and statistics, the combinations formula is a critical tool used to compute the number of different ways to select items from a group, where the order of selection does not matter. This concept plays a pivotal role when dealing with situations where we need to calculate probabilities involving various combinations of outcomes.

The formula for finding the number of combinations of selecting \(k\) items from a set of \(n\) items is given by the binomial coefficient, denoted as \( \binom{n}{k} \), and calculated using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Here, the notation \(n!\)鈥攔ead as 'n factorial'鈥攔efers to the product of all positive integers up to \(n\) (with \(0!\) defined as 1 for mathematical consistency).

To provide a tangible example, let's consider choosing 3 M&M's from a selection of 8. The formula indicates there are \( \binom{8}{3} \) combinations possible, calculated as \( \frac{8!}{3!(8-3)!} \), which simplifies to 56 different combinations. It's important to note that when using the combinations formula, the outcome set '3 M&M's' is the same regardless of the order in which we choose them. The formula elegantly accounts for this by eliminating order as a factor entirely, thus simplifying our probability calculations.
Factorials in Probability
The concept of factorials is intimately linked with probability calculations, especially when dealing with permutations and combinations. A factorial, denoted as \(n!\), represents the product of all positive integers up to a number \(n\). For example, \(5!\) is equal to \(5 \times 4 \times 3 \times 2 \times 1\) or 120. In probability, factorials determine the total number of ways in which elements can be arranged (permutations) or selected (combinations).

Understanding how to manipulate factorials is essential when simplifying probability problems. For instances involving combinations, we often divide factorials to cancel out terms, making the math more manageable. This cancellation allows us to operate effectively even with seemingly large numbers, as shown in the selection of M&M's example. The arithmetic involving factorials might appear daunting at first, but with practice, learners become adept at breaking down and simplifying the factorial expressions to reveal probabilities of various outcomes with clarity and confidence.
Probability Calculations
Probability calculations are the heart of understanding random events. These calculations can range from simple to highly complex, and they are built upon the principles of factorials and combinations, among other mathematical tools. Probability is essentially a way to quantify the likelihood of a particular event occurring out of all possible outcomes, often expressed as a fraction or a percentage.

Let's take the example of finding out the probability of selecting red and brown M&M's. Here, we identified the total number of outcomes using combinations and then calculated the probability of each event (like selecting a specific number of red M&M's) by considering the fraction of favorable combinations to the total combinations possible. In mathematical terms, the probability of an event \(A\) is given by \( P(A) = \frac{\text{Number of ways event A can occur}}{\text{Total number of possible outcomes}} \). For the M&M's problem, this allows us to conclude that the chance of selecting all three red M&M's is \( \frac{1}{56} \), a valuable point of knowledge for the curious child capable of pondering probabilities while selecting candies.

Mastering probability calculations not only enhances analytical skills but also prepares students for real-world applications, such as predicting outcomes in genetics, finance, and even sports. It's a mathematical area as rich in theory as it is in pragmatic utility.

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Most popular questions from this chapter

Use Table 1 in Appendix I to evaluate the following probabilities for \(n=6\) and \(p=.8\) : a. \(P(x \geq 4)\) b. \(P(x=2)\) c. \(P(x<2)\) d. \(P(x>1)\) Verify these answers using the values of \(p(x)\) calculated in Exercise 27 .

A shipping company knows that the cost of delivering a small package within 24 hours is \(\$ 14.80 .\) The company charges \(\$ 15.50\) for shipment but guarantees to refund the charge if delivery is not made within 24 hours. If the company fails to deliver only \(2 \%\) of its packages within the 24 -hour period, what is the expected gain per package?

You can insure a \(\$ 50,000\) diamond for its total value by paying a premium of \(D\) dollars. If the probability of loss in a given year is estimated to be .01, what premium should the insurance company charge if it wants the expected gain to equal \(\$ 1000 ?\)

Let \(x\) be a binomial random variable with \(n=20\) and \(p=.1\). a. Calculate \(P(x \leq 4)\) using the binomial formula. b. Calculate \(P(x \leq 4)\) using Table 1 in Appendix I. c. Use the following Excel output to calculate \(P(x \leq 4)\). Compare the results of parts a, b, and c. d. Calculate the mean and standard deviation of the random variable \(x\). e. Use the results of part d to calculate the intervals \(\mu \pm \sigma, \mu \pm 2 \sigma,\) and \(\mu \pm 3 \sigma .\) Find the probability that an observation will fall into each of these intervals. f. Are the results of part e consistent with Tchebysheff's Theorem? With the Empirical Rule? Why or why not? Excel output for Exercise 33: Binomial with \(n=20\) and \(p=.1\) $$ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} \\ \hline 1 & \mathrm{x} & \mathrm{p}(\mathrm{x}) & \mathrm{x} & \mathrm{p}(\mathrm{x}) \\ \hline 2 & 0 & 0.1216 & 11 & 7 \mathrm{E}-07 \\ \hline 3 & 1 & 0.2702 & 12 & 5 \mathrm{E}-08 \\ \hline 4 & 2 & 0.2852 & 13 & 4 \mathrm{E}-09 \\ \hline 5 & 3 & 0.1901 & 14 & 2 \mathrm{E}-10 \\ \hline 6 & 4 & 0.0898 & 15 & 9 \mathrm{E}-12 \\ \hline 7 & 5 & 0.0319 & 16 & 3 \mathrm{E}-13 \\ \hline 8 & 6 & 0.0089 & 17 & 8 \mathrm{E}-15 \\ \hline 9 & 7 & 0.0020 & 18 & 2 \mathrm{E}-16 \\ \hline 10 & 8 & 0.0004 & 19 & 2 \mathrm{E}-18 \\ \hline 11 & 9 & 0.0001 & 20 & 1 \mathrm{E}-20 \\ \hline 12 & 10 & 0.0000 & & \\ \hline \end{array} $$

Use the probability distribution for the random variable \(x\) to answer the questions in Exercises 12-16. $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ $$ \text { Find } p(4) $$
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