91Ó°ÊÓ

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

Step by step solution

01

Calculate P(A)

We are given that there are 10,000 boards, of which 2,000 are green. The probability that the first board selected is green (event A) is the number of green boards divided by the total number of boards: \( P(A) = \frac{2000}{10000} = 0.2 \).

02

Calculate P(B|A)

After selecting one green board, there are 1,999 green boards left out of the remaining 9,999 boards. Hence, the probability that the second board selected is green, given the first one was green, is \( P(B|A) = \frac{1999}{9999} \approx 0.2 \).

03

Calculate P(A ∩ B) without Independence

Since both boards are selected without replacement, we use conditional probability: \( P(A \cap B) = P(A) \times P(B|A) = 0.2 \times \frac{1999}{9999} \approx 0.1998 \).

04

Determine Independence of A and B

For two events to be independent, \( P(B) = P(B|A) \) should hold. Here, \( P(B) = \frac{2000}{10000} = 0.2 \), but \( P(B|A) \approx 0.2 \) as well, so the events are nearly independent for such a large sample.

05

Calculate P(A ∩ B) with Independence Assumed

Under the independence assumption and given \( P(A) = P(B) = 0.2 \), we have \( P(A \cap B) = P(A) \times P(B) = 0.2 \times 0.2 = 0.04 \).

06

Compare P(A ∩ B) from Part (a) and Assumed Independence

The difference is \( 0.002 \), suggesting independence does not yield an accurate \( P(A \cap B) \) for a small sample.

07

Evaluate Part (c) with Smaller Sample

For a sample of 10 boards with 2 green, \( P(A) = \frac{2}{10} = 0.2 \), and \( P(B|A) = \frac{1}{9} \approx 0.111 \), which leads to \( P(A \cap B) = 0.2 \times 0.111 \approx 0.0222 \). This starkly differs from \( 0.04 \), so independence is less valid.

08

Conclusion on Independence Assumption

In large populations like part (a), independence can be assumed to simplify calculations, but with smaller samples, exact methods are preferred, while independence fails due to a larger effect of each selection.

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Key Concepts

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

Independent Events

In probability, events are said to be independent when the occurrence of one event does not affect the probability of another event occurring. This is a significant concept as it helps us understand how different occurrences may or may not influence each other.
For instance, when flipping a coin twice, the result of the first flip doesn't affect the result of the second; each flip stands on its own. This means the events are independent. Conversely, if we're drawing cards from a deck without replacement, the outcome of one draw affects the next, indicating dependency.
In our original exercise, we initially questioned if selecting a green board affected the chances of the next being green. To confirm independence, we check if the probability of the second event occurring is unchanged by the first. This requires that \( P(B|A) = P(B) \).

• If true, the events are independent.
• If false, they are dependent.

Deciding independence is crucial as it determines the method we use for calculating probabilities.

Conditional Probability Formula

Conditional probability is used to find the probability of an event occurring given that another event has already occurred. The formula is written as \( P(B|A) = \frac{P(A \cap B)}{P(A)} \), and is pivotal in understanding situations where events are related.
In simpler terms, it means the probability of event B happening given that A has occurred. This concept shows us that the occurrence of event A can reshape the probability landscape for B.
Applying this to the board selection activity: Since we have two outcomes dependent on each other, we calculate \( P(B|A) \). First, \( P(A) \) is determined by what fraction of total boards are green. Then conditional probability helps find \( P(A \cap B) \) by considering how what's left of the green affects the subsequent choices.

• This formula plays a vital role when calculating probabilities for events that don't occur in isolation.
• For dependent events like this—where the sample size influences outcomes—it offers precision.

Understanding and using conditional probability clarifies interconnected event scenarios.

Probability without Replacement

Under scenarios where items are selected without being returned, probability calculations must consider the reduced sample size. This is what is described as 'probability without replacement.', a key aspect of the original exercise.
Whenever you select an item and do not replace it, the total number of items available for the next selection decreases. This sequential change affects all subsequent probabilities.
Look at the board selection scenario: initially, selecting a green board results in fewer green boards for the next pick. This leads to different probabilities than if the first were replaced. Thus, we use the conditional probability formula here to adjust for the changing sample.

• Because the situation involves no replacements, choices must reflect a dynamic sample pool change, impacting each subsequent probability.
• Across smaller samples, like selecting from 10 to 9, the effect drastically alters probability outcomes.

Understanding probability without replacement is essential, as it leads to more accurate real-world predictions when choices affect available options.