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 that 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)

The probability that the first board selected is green is given by the ratio of green boards to the total number of boards. Hence, \(P(A) = \frac{2000}{10000} = 0.2\).

02

Calculate P(B|A)

If event \(A\) occurs, the first board is green, leaving 1999 green boards out of a total of 9999 remaining boards. Therefore, \(P(B|A) = \frac{1999}{9999}\).

03

Calculate P(B)

Even if event \(A\) does not occur, initially, \(P(B)\) is approximately the same as \(P(A)\), so \(P(B) \approx 0.2\), assuming a large sample size.

04

Calculate P(A ∩ B)

Using the conditional probability, \(P(A \cap B) = P(A) \cdot P(B|A) = 0.2 \cdot \frac{1999}{9999}\). Since this is an approximation, compute it to get the accurate probability.

05

Check Independence of A and B

For two events \(A\) and \(B\) to be independent, \(P(A \cap B) = P(A) \cdot P(B)\). Calculate \(P(A) \cdot P(B) = 0.2 \cdot 0.2 = 0.04\). If \(P(A \cap B)\) from prior calculations does not equal \(0.04\), then the events are not independent.

06

Calculate P(A ∩ B) with Independence Assumption

Assuming \(A\) and \(B\) are independent and given \(P(A) = P(B) = 0.2\), \(P(A \cap B) = P(A) \cdot P(B) = 0.04\).

07

Analyze Approximation Validity for Part a

Compare \(P(A \cap B)\) calculated with independence assumption to \(P(A \cap B)\) calculated without. Determine if the approximation is valid.

08

Consider Smaller Sample Scenario

With a lot of ten boards (two green), calculate \(P(A \cap B)\) both with and without the independence assumption. In small samples, events are more dependent, thus differing results are expected.

09

Identify when Independence Assumption is Valid

Independence is plausible when the total number of boards is much larger than the number of boards selected, minimizing the effect of selection on remaining board probabilities and providing approximately correct results for \(P(A \cap B)\).

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Key Concepts

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

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. In this exercise, when the board is selected, we want to find the probability of the second board being green, given that the first one was green. This is expressed as \( P(B|A) \). The formula for conditional probability is:\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \] This means we consider the smaller sample space where the first board is already known to be green. From the exercise data, after selecting one green board, 1999 remain, making the sample size 9999. Hence, \( P(B|A) = \frac{1999}{9999} \).
Understanding conditional probability helps us see how events are affected by the occurrence of another one. It is useful in predicting the likelihood of dependent events.

Independence of Events

Two events are independent if the occurrence of one does not affect the probability of the other. Mathematically, if \( A \) and \( B \) are independent, then \( P(A \cap B) = P(A) \cdot P(B) \). In this problem, \( A \) and \( B \) are not independent if \( P(A \cap B) \) calculated is not equal to \( 0.04 \), because \( P(A) = P(B) = 0.2 \). Steps show that if events are independent, the simple multiplication rule can apply, simplifying probability calculations.
For example, flipping a coin and rolling a die are independent events, as one does not influence the other. But if the event influences the chances, like drawing cards without replacement, they are dependent, not independent.

Sample Size Effect

Sample size significantly affects probability outcomes, especially concerning independence assumptions. In the exercise, with 10,000 total boards, selecting one does not drastically change the subsequent probabilities, allowing a near independence assumption. But with only 10 boards, selecting affects the remainder significantly.

• Large samples approximate independence, mimicking the infinite case.
• Small samples introduce dependency, as seen in extracting from a small pool.

This is because, in larger samples, the removal of one does not notably skew the proportion of green vs. non-green. In smaller samples, each choice greatly impacts the remaining mix.

Event Dependency

Event dependency occurs when the outcome or occurrence of the first event affects the probability of the second. In the context of the board selection problem, when a green board is chosen first, it leaves fewer green boards to be chosen second, altering the probability directly.
Event dependency is vital in practical situations, altering predictions and expectations. Recognizing when events depend on each other allows for more accurate assessments of chances, unlike hypothetical scenarios where only independent events are considered, sometimes in mathematical exercises.
Understanding how actions influence subsequent ones aids in the strategic decision-making where probability plays a role, such as in risk management, logistics, and strategic planning.