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Chapter 15: Problem 18

The soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. The box contains 4 medium, 10 large, and 6 extra-large shirts. You want a medium for you and one for your sister. Find the probability of each event described. a) The first two you grab are the wrong sizes. b) The first medium shirt you find is the third one you check. c) The first four shirts you pick are all extra-large. d) At least one of the first four shirts you check is a medium.

Short Answer

Expert verified
a) 12/19, b) 10/57, c) 1/323, d) 123/323.

Step by step solution

01

Calculate Probability of Picking 2 Wrong Sizes

To calculate the probability that the first two shirts you grab are wrong sizes (either large or extra-large), start by finding the probability of not picking a medium shirt on both draws:- Probability of not picking a medium shirt on the first draw is the probability of picking either a large or extra-large: \[ P(\text{Wrong size first}) = \frac{16}{20} \]- Probability of not picking a medium shirt on the second draw, given that the first was not medium, is: \[ P(\text{Wrong size second}) = \frac{15}{19} \]Now multiply these probabilities together to find the total probability:\[ P(\text{Two wrong sizes}) = \frac{16}{20} \times \frac{15}{19} = \frac{240}{380} = \frac{12}{19} \]
02

Calculate Probability of Finding Medium on Third Draw

To find the probability that the first medium shirt is the third one you check:- First two picks are wrong sizes (similar to Step 1): \[ \frac{16}{20} \times \frac{15}{19} \]- The probability that the third pick is a medium: \[ \frac{4}{18} \]Combining these, the total probability is:\[ P(\text{First medium on third}) = \frac{16}{20} \times \frac{15}{19} \times \frac{4}{18} = \frac{240}{1368} = \frac{20}{114} = \frac{10}{57} \]
03

Calculate Probability of Picking 4 Extra-Large Shirts

For the first four shirts all being extra-large:- Probability of first pick being extra-large: \[ \frac{6}{20} \]- Similarly, probabilities for picking extra-large on subsequent draws: \[ \frac{5}{19}, \frac{4}{18}, \text{and } \frac{3}{17} \]Multiply these probabilities:\[ P(\text{Four extra-larges}) = \frac{6}{20} \times \frac{5}{19} \times \frac{4}{18} \times \frac{3}{17} = \frac{360}{116280} = \frac{1}{323} \]
04

Calculate Probability of At Least One Medium in First Four

The probability of at least one medium shirt in the first four draws is the complement of no mediums being drawn.Calculate the probability of picking only wrong sizes (large and extra-large) for the first four shirts:- Probability for each draw: \[ \frac{16}{20}, \frac{15}{19}, \frac{14}{18}, \text{and } \frac{13}{17} \]Multiply these probabilities:\[ P(\text{No mediums in four}) = \frac{16}{20} \times \frac{15}{19} \times \frac{14}{18} \times \frac{13}{17} = \frac{4368}{58140} = \frac{104}{165} \approx \frac{200}{323} \]Thus, the probability of at least one medium is:\[ P(\text{At least one medium}) = 1 - \frac{200}{323} = \frac{123}{323} \]

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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 counting, arrangement, and combination of objects. It helps us understand how many different ways we can arrange items or choose them from a larger set. In the context of the shirt picking example, combinatorics is used to determine the number of ways shirts can be selected. The process usually involves:
  • Calculating probabilities based on potential arrangements and selections of items.
  • Using fractions to represent the likelihood of picking certain items under different conditions.
  • Reducing complex probabilities by breaking them into sequences of simpler events and multiplying the individual probabilities together.
Thus, by understanding the basic arrangements and probabilities involved, we can solve real-life problems such as predicting the chances of picking desired or undesired shirt sizes.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already happened. It's a key tool when dealing with sequences of events where each event's outcome depends on the previous ones.
In the exercise about selecting shirts, we calculated the probabilities of the second shirt being a wrong size given that the first was a wrong size. The conditional probability can be calculated by:
  • Identifying the initial condition that has already occurred i.e., the shirt already picked.
  • Determining how the initial condition affects the probability of the next selections.
  • Updating the overall probability based on the assumption of the initial condition.
This approach helps tackle complex situations by simplifying each subsequent calculation, taking into account previous outcomes. It's essential for understanding scenarios with dependent events.
Independent Events
Independent events are those where the outcome of one does not affect the outcome of another. In the context of probability, knowing whether events are independent or not can greatly influence how probabilities are calculated. However, in the exercise, no events are truly independent as each shirt selected affects subsequent choices.
For instance:
  • If two events are independent, the probability of both occurring is the product of their probabilities.
  • However, when selecting shirts from a finite group without replacement, the act of selecting one shirt reduces the available pool, thus linking the events.
Understanding the concept of independence helps in identifying the right approach to solve probability problems and anticipate possible factors affecting those probabilities.

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Most popular questions from this chapter

A junk box in your room contains a dozen old batteries, five of which are totally dead. You start picking batteries one at a time and testing them. Find the probability of each outcome. a) The first two you choose are both good. b) At least one of the first three works. c) The first four you pick all work. d) You have to pick 5 batteries to find one that works.

Police often set up sobriety checkpointsroadblocks where drivers are asked a few brief questions to allow the officer to judge whether or not the person may have been drinking. If the officer does not suspect a problem, drivers are released to go on their way. Otherwise, drivers are detained for a Breathalyzer test that will determine whether or not they will be arrested. The police say that based on the brief initial stop, trained officers can make the right decision \(80 \%\) of the time. Suppose the police operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national traffic safety experts suspect that about \(12 \%\) of drivers have been drinking. a) You are stopped at the checkpoint and, of course, have not been drinking. What's the probability that you are detained for further testing? b) What's the probability that any given driver will be detained? c) What's the probability that a driver who is detained has actually been drinking? d) What's the probability that a driver who was released had actually been drinking?

Leah is flying from Boston to Denver with a connection in Chicago. The probability her first flight leaves on time is \(0.15\). If the flight is on time, the probability that her luggage will make the connecting flight in Chicago is \(0.95\), but if the first flight is delayed, the probability that the luggage will make it is only \(0.65\). a) Are the first flight leaving on time and the luggage making the connection independent events? Explain. b) What is the probability that her luggage arrives in Denver with her?

Real estate ads suggest that \(64 \%\) of homes for sale have garages, \(21 \%\) have swimming pools, and \(17 \%\) have both features. What is the probability that a home for sale has a) a pool or a garage? b) neither a pool nor a garage? c) a pool but no garage?

Suppose that \(23 \%\) of adults smoke cigarettes. It's known that \(57 \%\) of smokers and \(13 \%\) of nonsmokers develop a certain lung condition by age 60 . a) Explain how these statistics indicate that lung condition and smoking are not independent. b) What's the probability that a randomly selected 60 -year-old has this lung condition?
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