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91Ó°ÊÓ

A lot contains 15 castings from a local supplier and 25 castings from a supplier in the next state. Two castings are selected randomly, without replacement, from the lot of \(40 .\) Let \(A\) be the event that the first casting selected is from the local supplier, and let \(B\) denote the event that the second casting is selected from the local supplier. Determine: a. \(P(A)\) b. \(P(B \mid A)\) c. \(P(A \cap B)\) d. \(P(A \cup B)\) Suppose that 3 castings are selected at random, without replacement, from the lot of \(40 .\) In addition to the definitions of events \(A\) and \(B,\) let \(C\) denote the event that the third casting selected is from the local supplier. Determine: e. \(P(A \cap B \cap C)\) f. \(P\left(A \cap B \cap C^{\prime}\right)\)

Short Answer

Expert verified
a. 3/8, b. 14/39, c. 7/52, d. 39/52, e. 91/988, f. 175/1976.

Step by step solution

01

Calculate P(A)

The probability of the first casting being from the local supplier, denoted as P(A), is calculated by considering the number of favorable outcomes (castings from the local supplier) over the total number of castings. There are 15 local castings out of a total of 40 castings.\[ P(A) = \frac{15}{40} = \frac{3}{8} \]
02

Calculate P(B | A)

The probability of the second casting being from the local supplier, given that the first one was (event B given A), is calculated by reducing both the total number of castings and the number of local castings by one since the first drawing was a local casting.\[ P(B \mid A) = \frac{14}{39} \]
03

Calculate P(A ∩ B)

The probability of both the first and the second castings being from the local supplier (P(A ∩ B)) is the product of the probabilities calculated in the previous steps since they are independent draws.\[ P(A \cap B) = P(A) \times P(B \mid A) = \frac{3}{8} \times \frac{14}{39} = \frac{7}{52} \]
04

Calculate P(A ∪ B)

The probability of at least one casting being from the local supplier (P(A ∪ B)) can be calculated using the formula for the probability of the union of two events:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \].First, calculate P(B), the probability that the second casting is local without regard to the first:\[ P(B) = \frac{15}{40} = \frac{3}{8} \] (note: this is calculated similarly to P(A) as events A and B are about separate selections without dependency consideration.)Then use the union formula:\[ P(A \cup B) = \frac{3}{8} + \frac{3}{8} - \frac{7}{52} = \frac{39}{52} \]
05

Calculate P(A ∩ B ∩ C)

For three consecutive local castings, we multiply the conditional probabilities:- First (P(A)) casting is from local as calculated earlier: \( \frac{3}{8} \)- Second (P(B | A)) casting is from local: \( \frac{14}{39} \)- Third (P(C | A ∩ B)) with further reduced castings:\[ P(C \mid A \cap B) = \frac{13}{38} \]So, the probability of all three (A ∩ B ∩ C) is:\[ P(A \cap B \cap C) = \frac{3}{8} \times \frac{14}{39} \times \frac{13}{38} = \frac{91}{988} \]
06

Calculate P(A ∩ B ∩ C′)

For two castings being local and one not, consider failing on the third. For P(A ∩ B ∩ C′):- P(A ∩ B) from Step 3: \( \frac{7}{52} \)- The probability that the third casting is not from the local supplier: reduce non-locals out of leftover:\[ P(C^{\prime} \mid A \cap B) = \frac{25}{38} \]Calculate:\[ P(A \cap B \cap C^{\prime}) = \frac{7}{52} \times \frac{25}{38} = \frac{175}{1976} \]

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Key Concepts

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

Conditional Probability
Conditional probability helps us understand the likelihood of an event happening given that another event has already occurred. In mathematical terms, if we want to find the probability of event B occurring given that event A has already happened, we use the formula:
  • \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \)
This formula expresses how the probability of B is adjusted by the occurrence of A. For example, in selecting castings without replacement, knowing the first casting was local impacts the probability of the second being local too. The reduced total number of items means we adjust probabilities based on prior selections. By calculating \( P(B \mid A) = \frac{14}{39} \), we're recognizing that one local casting is no longer in the pool.
Independent Events
Independent events are those whose occurrence or outcome does not affect each other. In other words, the probability of one event does not change if we know the outcome of another. This is a contrast to dependent events, where one event can directly impact the other. In our exercise, drawing castings without replacement tells us we are dealing with dependent events. The probability of subsequent draws is influenced by previous draws. For example, if the first casting is local, this affects the probability of the second casting. Therefore, in scenarios like this exercise, we use conditional probabilities rather than treating events as independent. If events were independent, we'd calculate with simple multiplication of probabilities without adjusting for previous outcomes.
Probability of Union
The probability of the union of events is finding how likely it is for at least one of the events to occur. The union of events A and B, denoted as \( A \cup B \), includes all outcomes where either A, B, or both occur. This is useful because it helps identify common cases or outcomes in combined events.To determine the probability of a union, we use the formula:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
This formula accounts for double-counting that may occur when combining probabilities. For instance, when calculating \( P(A \cup B) = \frac{39}{52} \), it includes all scenarios where at least one of the drawn castings is from the local supplier. This helps in considering all possible favorable outcomes.
Combined Events
When dealing with combined events, it means considering situations where multiple criteria need to be met. This can involve intersecting events, where we require several conditions to occur together.In this exercise, we deal with combined events like \( A \cap B \cap C \), where we want all three castings to be from the local supplier. To calculate this, we multiply through the conditional probabilities of each subsequent event, adjusting for each reduction in total outcomes as castings are drawn:
  • \( P(A \cap B \cap C) = \frac{3}{8} \times \frac{14}{39} \times \frac{13}{38} \)
Similarly, for an event like \( A \cap B \cap C^{\prime} \), we look at the likelihood of the first two castings being local and the third from another supplier. Calculating combined events in this way allows us to evaluate scenarios where multiple specific conditions are required to be met in a sequence of actions.

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