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Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. Find P(O).

Short Answer

Expert verified
The probability of selecting an orange jelly bean is \(\frac{8}{75}\).

Step by step solution

01

Identify the Total Number of Jelly Beans

The problem states that there are a total of 150 jelly beans in the jar. This is the total number of possible outcomes when selecting a jelly bean.
02

Determine the Number of Orange Jelly Beans

First, calculate the number of orange jelly beans by subtracting the number of other colored jelly beans from the total.\\[\text{Total number of jelly beans} - (\text{Red} + \text{Yellow} + \text{Green} + \text{Purple} + \text{Blue})\] \\[150 - (22 + 38 + 20 + 28 + 26) = 150 - 134 = 16\] \There are 16 orange jelly beans.
03

Calculate the Probability of Selecting an Orange Jelly Bean (P(O))

The probability of selecting an orange jelly bean is calculated by dividing the number of orange jelly beans by the total number of jelly beans. \\[P(O) = \frac{\text{Number of Orange Jelly Beans}}{\text{Total Number of Jelly Beans}} = \frac{16}{150}\] \Simplifying the fraction, we get: \\[P(O) = \frac{8}{75}\] \Thus, the probability of selecting an orange jelly bean is \(\frac{8}{75}\).

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Key Concepts

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

Events
In probability, an **event** represents a specific outcome or a group of outcomes from a probabilistic experiment. When you think of an event, imagine you are trying to see if a particular thing happens. For instance, in the context of the jelly bean problem, events are defined as selecting a jelly bean of a certain color.
  • **Event B**: Obtaining a blue jelly bean.
  • **Event G**: Obtaining a green jelly bean.
  • **Event O**: Obtaining an orange jelly bean.
  • **Event P**: Obtaining a purple jelly bean.
  • **Event R**: Obtaining a red jelly bean.
  • **Event Y**: Obtaining a yellow jelly bean.
By defining events, we can easily calculate the probability of these events occurring. In our example, we specifically looked at the event of selecting an orange jelly bean (Event O).
Understanding how to define and work with events is crucial, as it simplifies the calculation and understanding of probabilities.
Sample Space
The **sample space** in probability refers to the complete set of all possible outcomes of an experiment or event. It is like the playground where all the action happens. For the jelly bean example, the sample space includes all colored jelly beans present in the jar.

To understand it better, imagine spreading out all 150 jelly beans and looking at each one as part of your sample space. Each jelly bean color, like red, yellow, green, blue, purple, and orange, contributes to this space. Therefore, the sample space, in this case, contains:
  • 22 Red Jelly Beans
  • 38 Yellow Jelly Beans
  • 20 Green Jelly Beans
  • 28 Purple Jelly Beans
  • 26 Blue Jelly Beans
  • 16 Orange Jelly Beans (since we calculated it from the remainder)
The total is 150 jelly beans, confirming our initial condition. Understanding sample space is vital because it provides the baseline against which probabilities are measured.
Counting Principles
**Counting principles** involve techniques used to count different groups or arrangements of outcomes in an event. They help in determining how many outcomes exist for a particular scenario, facilitating the calculation of probabilities.

In our exercise with the jelly beans, counting principles help us determine the quantity of each jelly bean color. By adding the number of red, yellow, green, purple, and blue jelly beans, we use a basic counting principle to find the number of orange jelly beans. This is done by subtracting the sum of known jelly beans from the total:\[150 - (22 + 38 + 20 + 28 + 26) = 16\]Counting principles are not just for subtraction; they include other rules like permutations and combinations used in more complex probability scenarios. These principles simplify the process of handling larger datasets or more intricate cases.
Fraction Simplification
In probability, once you calculate the probability expressed as a fraction, it’s often necessary to **simplify** that fraction. Simplifying fractions not only makes them easier to understand but also clearer in scientific communication.

For our jelly bean example, after determining there were 16 orange jelly beans out of a total of 150, we expressed the probability as:\[P(O) = \frac{16}{150}\]To simplify, you find the greatest common divisor (GCD) of the numerator and the denominator. For 16 and 150, the GCD is 2:\[\frac{16}{150} = \frac{16 \div 2}{150 \div 2} = \frac{8}{75}\]Thus, the probability of picking an orange jelly bean simplifies to \(\frac{8}{75}\). Always check if a further reduction is possible to ensure accuracy and neatness in your results.

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Most popular questions from this chapter

Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact. a. Find P(Person is female). b. Find P(Person obtained the disease through heterosexual contact). c. Find P(Person is female GIVEN person got the disease from heterosexual contact) d. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.

Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter Write the symbols for the probability that of all the outfielders, a player is not a great hitter.

When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin. a. Based on the given data, find P(H) and P(T). b. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice. c. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin. d. Use the tree to find the probability of obtaining at least one head.

Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. Suppose that 10,000 U.S. licensed drivers are randomly selected. a. How many would you expect to be male? b. Using the table or tree diagram, construct a contingency table of gender versus age group. c. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.

The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. $$\begin{array}{|l|l|l|l|l|}\hline \text { Name } & {\text { single }} & {\text { Double }} & {\text { Triple }} & {\text { Home Run }} & {\text { Total Hits }} \\ \hline \text { Babe Ruth } & {1,517} & {506} & {136} & {714} & {2,873} \\ \hline \text { Jackie Robinson } & {1,054} & {273} & {54} & {137} & {1,518} \\ \hline \text { Ty Cobb } & {3,603} & {174} & {295} & {114} & {4,189} \\ \hline \text { Hank Aaron } & {2,294} & {624} & {98} & {755} & {3,771} \\ \hline\end{array}$$ Are "the hit being made by Hank Aaron" and "the hit being a double" independent events? a. Yes, because P(hit by Hank Aaron|hit is a double) = P(hit by Hank Aaron) b. No, because P(hit by Hank Aaron|hit is a double) ? P(hit is a double) c. No, because P(hit is by Hank Aaron|hit is a double) ? P(hit by Hank Aaron) d. Yes, because P(hit is by Hank Aaron|hit is a double) = P(hit is a double)

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