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Chapter 3: Problem 85

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card. 鈥� G = card drawn is green 鈥� E = card drawn is even-numbered a. List the sample space. b. P(G) = _____ c. P(G|E) = _____ d. P(G AND E) = _____ e. P(G OR E) = _____ f. Are G and E mutually exclusive? Justify your answer numerically

Short Answer

Expert verified
a. Sample space: G1, G2, G3, G4, G5, Y1, Y2, Y3; b. \( P(G) = \frac{5}{8} \); c. \( P(G|E) = \frac{2}{3} \); d. \( P(G \text{ AND } E) = \frac{1}{4} \); e. \( P(G \text{ OR } E) = \frac{7}{8} \); f. No, \( G \) and \( E \) are not mutually exclusive.

Step by step solution

01

List the Sample Space

The sample space consists of all possible outcomes of drawing one card. There are eight cards: 5 green cards numbered 1, 2, 3, 4, and 5, and 3 yellow cards numbered 1, 2, and 3. Thus, the sample space can be listed as: G1, G2, G3, G4, G5, Y1, Y2, Y3.
02

Calculate P(G)

To find the probability of drawing a green card, we divide the number of green cards by the total number of cards. There are 5 green cards out of 8, so \( P(G) = \frac{5}{8} \).
03

Determine the event E and calculate P(G|E)

The event E refers to drawing an even-numbered card. The even-numbered cards are G2, G4, and Y2. Hence, there are 3 even-numbered cards. To find \( P(G|E) \), we calculate the probability of drawing a green card given that the card is even. There are 2 green even-numbered cards (G2 and G4) out of the 3 even-numbered, so \( P(G|E) = \frac{2}{3} \).
04

Find P(G AND E)

\( P(G \text{ AND } E) \) is the probability that the card drawn is both green and even-numbered. The possible outcomes here are G2 and G4, so there are 2 favorable outcomes out of 8. Thus, \( P(G \text{ AND } E) = \frac{2}{8} = \frac{1}{4} \).
05

Calculate P(G OR E)

To find \( P(G \text{ OR } E) \), we find the probability that the card is either green or even-numbered or both. This can be calculated using the formula: \( P(G \text{ OR } E) = P(G) + P(E) - P(G \text{ AND } E) \). We already have \( P(G) = \frac{5}{8} \) and \( P(G \text{ AND } E) = \frac{1}{4} \). For \( P(E) \), there are 3 even-numbered cards out of 8, so \( P(E) = \frac{3}{8} \). Substitute these values to find \( P(G \text{ OR } E) = \frac{5}{8} + \frac{3}{8} - \frac{1}{4} = \frac{7}{8} \).
06

Determine if G and E are Mutually Exclusive

Two events are mutually exclusive if they cannot happen at the same time. Since there are cards that are both green and even-numbered (G2 and G4), \( G \) and \( E \) are not mutually exclusive.

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Key Concepts

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

Sample Space
In probability, the **sample space** is a complete set of all possible outcomes of an experiment or event. For the exercise mentioned, drawing one card from a set of eight creates different possible outcomes.

There are five green cards numbered 1 through 5 and three yellow cards numbered 1 to 3, meaning each card is a unique outcome in our context. Therefore, the sample space is represented as follows:
  • G1: Green card number 1
  • G2: Green card number 2
  • G3: Green card number 3
  • G4: Green card number 4
  • G5: Green card number 5
  • Y1: Yellow card number 1
  • Y2: Yellow card number 2
  • Y3: Yellow card number 3
Listing out all possible outcomes is crucial because it helps define the scope of potential results and makes calculating probabilities much easier.
Conditional Probability
**Conditional Probability** involves finding the probability of an event occurring given that another event has already occurred. In this scenario, we want to find the probability that a card is green given we know it is even-numbered.

To calculate conditional probability, we use the formula:
\( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

Here, the probability of drawing a green card given it's even (\( P(G|E)\)) is found by evaluating:
  • There are 3 even-numbered cards (G2, G4, Y2).
  • Out of these, 2 are green (G2, G4).

Thus, the probability is \( \frac{2}{3} \). Conditional probability is indispensable in assessing likelihoods where prior knowledge about one event influences another.
Mutually Exclusive Events
**Mutually Exclusive Events** are those that cannot occur at the same time. In simpler terms, if one event happens, the other cannot.

In the card example, checking if the events G (the card is green) and E (the card is even-numbered) are mutually exclusive requires checking if any card can be both green and even-numbered.

Here, we see:
  • Two cards, G2 and G4, are both green and even.
Since these two cards satisfy both conditions simultaneously, we conclude that G and E are not mutually exclusive events. Understanding mutually exclusive events is essential because it affects how we calculate the probability of multiple events occurring.
Probability of Union
The **Probability of Union** involves determining the probability of one event occurring or another event occurring, or both. In mathematical terms, this is expressed as \( P(A \cup B) \).

For figuratively finding \( P(G \text{ OR } E) \), use the formula for the probability of the union:
\( P(G \cup E) = P(G) + P(E) - P(G \cap E) \)

Looking at the card example:
  • \( P(G) = \frac{5}{8} \), as 5 out of 8 cards are green.
  • \( P(E) = \frac{3}{8} \), because there are 3 cards that are even-numbered.
  • We subtract the overlap \( P(G \cap E) = \frac{1}{4} \) from the sum to avoid double-counting cards that are both green and even.
Thus, \( P(G \cup E) = \frac{7}{8} \), indicating a high likelihood of drawing a card that is either green, even, or both. Knowing how to calculate the probability of union expands your ability to predict outcomes when dealing with overlapping events.

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Most popular questions from this chapter

Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino. In this problem, let: 鈥� C = Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder. 鈥� L = Latino Californians Suppose that one Californian is randomly selected. Are L and C independent events? Show why or why not.

Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino. In this problem, let: 鈥� C = Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder. 鈥� L = Latino Californians Suppose that one Californian is randomly selected. Find P(C).

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 great hitters, a player is an outfielder.

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(B)

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) 鈥� Let F be the event that a student is female. 鈥� Let M be the event that a student is male. 鈥� Let S be the event that a student has short hair. 鈥� Let L be the event that a student has long hair. a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female.
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