/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Twelve percent of people in West... [FREE SOLUTION] | 91Ó°ÊÓ

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Twelve percent of people in Western countries consider themselves lucky. If 3 people are selected at random, what is the probability that at least one will consider himself lucky?

Short Answer

Expert verified
The probability that at least one of the 3 people considers themselves lucky is approximately 0.3185.

Step by step solution

01

Define the Probability of a Single Event

First, identify the probability that one person considers themselves lucky. Given, this probability is 12%, or \( P(L) = 0.12 \). Similarly, the probability that one person does not consider themselves lucky is \( P( ext{not } L) = 1 - 0.12 = 0.88 \).
02

Calculate the Probability of None Being Lucky

To find the probability that none of the 3 people selected consider themselves lucky, use the probability that one person is not lucky, raised to the power of 3 (since we have 3 independent events). This probability is \( P( ext{none lucky}) = 0.88^3 \).
03

Compute the Probability

Now, calculate \( 0.88^3 = 0.681472 \). Thus, the probability that none of the 3 people consider themselves lucky is approximately 0.6815.
04

Determine the Probability of At Least One Being Lucky

Finally, the probability that at least one person considers themselves lucky is 1 minus the probability that none consider themselves lucky. Thus, \( P( ext{at least one is lucky}) = 1 - 0.6815 = 0.3185 \).

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Key Concepts

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

Independent Events
In probability, independent events are those events where the outcome of one event does not affect the outcome of another. This is an important concept to grasp when calculating probabilities for multiple scenarios.
Imagine you have a bag of marbles. If you pull out a marble, note its color, then return it before drawing another, each draw is independent. The result of the first draw doesn't change the chances for the second one. This is similar to our exercise, where selecting the first person who thinks they are lucky doesn't change the odds for selecting the next person.
For our given problem, selecting 3 people with the probability that each considers themselves lucky is an independent event. This means each person's belief in luck doesn't linger or influence the next. The independence is crucial when calculating the combined probabilities, like finding out the chance that none are considered lucky.
  • Probability for one person being lucky = 0.12
  • Probability for one person not being lucky = 0.88
Complement Rule
The complement rule is a useful tool in probability that helps calculate the likelihood of an event not occurring. In simple terms, if you know the probability of an event happening, the complement rule helps you find the probability it doesn't happen.
The rule works as follows: If an event has a probability of occurring, its complement is 1 minus that probability. This makes sense because probability must add up to 1 (or 100%). Thus, if the probability of event A occurring is \( P(A) \), then the probability of it not occurring is \( P(\text{not }A) = 1 - P(A) \).
In our exercise, we wanted to know the probability of at least one person believing they are lucky. Understanding the complement rule allowed us to first calculate the probability of none of the 3 people considering themselves lucky and then use this result to find the probability that at least one does. By subtracting from one, we focus on the broader scenario and simplify the complex calculations.
Power of Probability
Raising a probability to a power is a common technique used in probability calculations when dealing with multiple independent events. This is often seen as: \( (\text{probability of an event})^n \), where \( n \) is the number of times the event occurs independently.
In our scenario, we have the probability of an individual thinking themselves lucky as \( 0.12 \), and likewise, not considering themselves lucky as \( 0.88 \). If you want to calculate the probability of more than one person being unlucky in a group, you raise the probability of a single event happening to the power of the number of people in the group.
This can look daunting, but it can be simplified by:
  • Probability of none being lucky = \( 0.88^3 \)
  • Simplifies complex event calculations
By using this method, you are breaking down a complex situation into smaller, more manageable parts. Calculating \( 0.88^3 \) gave us the probability of none being lucky, making it possible to then use the complement rule for finding out when at least one person considers themselves lucky.

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Most popular questions from this chapter

State which events are independent and which are dependent. a. Having a large shoe size and having a high IQ b. A father being left-handed and a daughter being left-handed c. Smoking excessively and having lung cancer d. Eating an excessive amount of ice cream and smoking an excessive amount of cigarettes

In how many ways can you choose 3 kinds of ice cream and 2 toppings from a dessert buffet with 10 kinds of ice cream and 6 kinds of toppings?

How many different ID cards can be made if there are 6 digits on a card and no digit can be used more than once?

Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a \(6,\) then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet \(\$ 1\) and won, you could win \(\$ 5 .\) On average, you would win \(\$ 5\) once in 6 rolls and lose \(\$ 1\) on the other 5 rolls - hence the term fair game. In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1 the house might offer 15 to 1 in order to make a profit. Odds can be expressed as a fraction or as a ratio, such as \(\frac{5}{1}, 5: 1,\) or 5 to \(1 .\) Odds are computed in favor of the event or against the event. The formulas for odds are $$ \begin{array}{l} \text { Odds in favor }=\frac{P(E)}{1-P(E)} \\ \text { Odds against }=\frac{P(\bar{E})}{1-P(\bar{E})} \end{array} $$ In the die example, $$ \begin{aligned} &\text { Odds in favor of a } 6=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5} \text { or } 1: 5\\\ &\text { Odds against a } 6=\frac{\frac{5}{6}}{\frac{1}{6}}=\frac{5}{1} \text { or } 5: \end{aligned} $$ Find the odds in favor of and against each event. a. Rolling a die and getting a 2 b. Rolling a die and getting an even number c. Drawing a card from a deck and getting a spade d. Drawing a card and getting a red card e. Drawing a card and getting a queen f. Tossing two coins and getting two tails g. Tossing two coins and getting exactly one tail

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