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Chapter 2: Problem 191

A researcher receives 100 containers of oxygen. Of those containers, 20 have oxygen that is not ionized, and the rest are ionized. Two samples are randomly selected, without replacement, from the lot. (a) What is the probability that the first one selected is not ionized? (b) What is the probability that the second one selected is not ionized given that the first one was ionized? (c) What is the probability that both are ionized? (d) How does the answer in part (b) change if samples selected were replaced prior to the next selection?

Short Answer

Expert verified
(a) 0.2, (b) 20/99, (c) 316/495, (d) 0.2 with replacement.

Step by step solution

01

Total Containers

You are given that there are 100 containers, with 20 non-ionized and 80 ionized containers. This helps in calculating probabilities as you know how many containers are in each category.
02

Part (a): First Selection Probability

To find the probability that the first container selected is not ionized, use the fact that there are 20 not ionized out of 100 total containers.\[ P( ext{not ionized first}) = rac{20}{100} = rac{1}{5} = 0.2 \]
03

Part (b): Second Without Replacement

If the first selected was ionized (80 possibilities), the total number of containers decreases by 1 to 99, and non-ionized containers remain 20. Calculate the probability that the second one is not ionized given the first was ionized:\[ P( ext{not ionized second | ionized first}) = rac{20}{99} \]
04

Part (c): Both Ionized Probability

The probability of both containers being ionized is the product of the probability that each pick results in an ionized container.First pick ionized: \[ P( ext{ionized first}) = rac{80}{100} = rac{4}{5} \]Second pick ionized, given first was ionized:\[ P( ext{ionized second | ionized first}) = rac{79}{99} \]The combined probability is:\[ P( ext{both ionized}) = rac{4}{5} imes rac{79}{99} = rac{316}{495} \]
05

Part (d): Second With Replacement

If the selection is with replacement, after the first pick, the total number of containers and their distribution remains the same. For the second pick probability being not ionized:\[ P( ext{not ionized}) = rac{20}{100} = rac{1}{5} = 0.2 \]

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in probability theory. It refers to the probability of an event occurring given that another event has already occurred. This is essential when the outcome of one event affects the probability of another.
For example, in the exercise, finding the probability that the second container is not ionized, given the first was ionized, involves conditional probability. The probability is determined by recalculating it over a smaller sample size, as the first event changes the conditions for the second event.
  • Initial condition: First container ionized, meaning 80 containers are ionized.
  • Now, only 99 containers remain, as one has already been removed.
  • The non-ionized containers remain at 20, so the conditional probability is computed as \( \frac{20}{99} \).
This means our previous knowledge of the first container being ionized changes the likelihood of what is drawn next.
Combinatorics
Combinatorics involves counting, arranging, and selecting objects within a set, a cornerstone in probability calculations. It helps us determine how many possible ways an event can occur, especially useful in sampling questions.
In the given problem, understanding how to quantify possible selections is crucial. You need to figure out how these selections affect probabilities when picking containers:
  • Two samples are taken from 100 containers, which involves counting combinations of taking one container at a time.
  • For part (c), to find the probability both containers are ionized, we calculate the chance of picking one ionized following another ionized container.
The beauty of combinatorics in probability is that it simplifies complex problems into manageable calculations. By knowing the number of possible configurations, we gain insights into probabilities across scenarios.
Sampling Without Replacement
Sampling without replacement is a method of selection where an item is not returned to the pool after being chosen. This approach significantly affects probabilities because the pool size reduces with each selection.
In our exercise, since two containers are selected without replacement, the second selection depends on the outcome of the first selection:
  • The total number of containers decreases from 100 to 99 after the first draw.
  • This reduction changes the probabilities of subsequent selections.
  • For instance, if we first choose an ionized container, only 79 ionized ones remain out of 99 for the next draw.
With this change, sampling without replacement introduces dependencies between events, influencing the probability calculations as seen in contrasts between parts (b) and (d) of the exercise.

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Most popular questions from this chapter

\(+\) If \(A, B,\) and \(C\) are mutually exclusive events with \(P(A)=0.2, P(B)=0.3,\) and \(P(C)=0.4,\) determine the follow. ing probabilities: (a) \(P(A \cup B \cup C)\) (b) \(P(A \cap B \cap C)\) (c) \(P(A \cap B)\) (d) \(P[(A \cup B) \cap C]\) (e) \(P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)\)

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A computer system uses passwords that are exactly seven characters and each character is one of the 26 letters \((a-z)\) or 10 integers \((0-9)\). You maintain a password for this computer system. Let \(A\) denote the subset of passwords that begin with a vowel (either \(a, e, i, o,\) or \(u\) ) and let \(B\) denote the subset of passwords that end with an even number (either \(0,2,4,6,\) or 8 ). (a) Suppose a hacker selects a password at random. What is the probability that your password is selected? (b) Suppose a hacker knows that your password is in event \(A\) and selects a password at random from this subset. What is the probability that your password is selected? (c) Suppose a hacker knows that your password is in \(A\) and \(B\) and selects a password at random from this subset. What is the probability that your password is selected?

Use the axioms of probability to show the following: (a) For any event \(E, P\left(E^{\prime}\right)=1-P(E)\). (b) \(P(\varnothing)=0\) (c) If \(A\) is contained in \(B,\) then \(P(A) \leq P(B)\).
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