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A study of the faculty at U.S. medical schools in 2006 revealed that \(32 \%\) of the faculty were women and \(68 \%\) were men. Of the female faculty, \(31 \%\) were full/associate professors, \(47 \%\) were assistant professors, and \(22 \%\) were instructors. Of the male faculty, \(51 \%\) were full/associate professors, \(37 \%\) were assistant professors, and \(12 \%\) were instructors. If a faculty member at a U.S. medical school selected at random holds the rank of full/associate professor, what is the probability that she is female?

Short Answer

Expert verified
The probability that a randomly chosen full/associate professor is female is approximately 0.2223 or 22.23%.

Step by step solution

01

Find the probability of the intersection of events A and B

To find the probability of the intersection of events A and B, i.e., P(A ∩ B), we must multiply the probability of event B by the conditional probability P(A|B). \(P(A \cap B) = P(A | B) \cdot P(B) \) Now, P(A|B)= 0.31 (31% female faculty are full/associate professors), P(B)= 0.32 (32% of the faculty are women) So, \(P(A \cap B) = 0.31 \cdot 0.32 = 0.0992\)
02

Find the probability of event A

Using the law of total probability, we can find P(A) as follows: \(P(A) = P(A|B) \cdot P(B) + P(A|\sim B) \cdot P (\sim B)\) Now, P(A|~B) = 0.51 (51% male faculty are full/associate professors), P(~B) = 0.68 (68% of the faculty are men) So, \(P(A) = 0.31 \cdot 0.32 + 0.51 \cdot 0.68 = 0.0992 + 0.3468 = 0.446\)
03

Find conditional probability

Now we can find the conditional probability P(B|A) using the formula: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\) So, \(P(B|A) = \frac{0.0992}{0.446} = 0.2223\) Therefore, the probability that a randomly chosen full/associate professor is female is approximately 0.2223 or 22.23%.

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Key Concepts

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

Conditional Probability
Understanding the concept of conditional probability is essential for students tackling problems in applied mathematics, especially those involving uncertainty and prediction. Simply put, conditional probability is the likelihood of an event occurring, given that another event has already happened.

Let's delve into the exercise which is a practical example. In the given scenario, we want to find out the probability that a randomly selected full/associate professor at a U.S. medical school is female, given that we already know they are a full/associate professor. This is a classic case of seeking a conditional probability, notated as P(B|A), where event B is 'being female' and event A is 'being a full/associate professor'.

To calculate this, we used the formula:
\(P(B|A) = \frac{P(A \cap B)}{P(A)}\)
where \(P(A \cap B)\) represents the intersection of events A and B, and P(A) represents the probability of event A. The solution to this computation provided us with the probability of a full/associate female professor, which is approximately 22.23%.
Law of Total Probability
Moving on to another vital topic, the law of total probability. This concept helps us break down complex probabilities into simpler parts. When we have a set of events that are exhaustive and mutually exclusive, we can use this law to find the overall probability of a particular event.

Referring back to our exercise, we used the law of total probability to determine the likelihood of a faculty member being a full/associate professor, regardless of gender. This involved adding the probability of a female professor to the probability of a male professor, which is mathematically represented by:
\(P(A) = P(A|B) \cdot P(B) + P(A|\sim B) \cdot P (\sim B)\)
Here, the probabilities on the right side represent the weighted likelihood of being a full/associate professor for both women and men, respectively. With the calculated probability of event A, we could then move on to find the conditional probability as described in the previous section.
Intersection of Events
The concept of the intersection of events plays a crucial role when evaluating the likelihood of two or more events occurring simultaneously. In probability terms, an intersection is represented by 'and'—we're interested in the odds of event A and event B happening together.

In the context of our example, we calculated \(P(A \cap B)\), which is the probability of a faculty member both being female (event B) and being a full/associate professor (event A). We find this intersection by multiplying the probability of one event by the conditional probability of the other:
\(P(A \cap B) = P(A | B) \cdot P(B)\)
After computing \(P(A \cap B) = 0.0992\), this value was instrumental for finding our final conditional probability. Understanding intersections is crucial for grasping more complex probability concepts and working through multifaceted probability exercises.

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Most popular questions from this chapter

Bob, the proprietor of Midway Lumber, bases his projections for the annual revenues of the company on the performance of the housing market. He rates the performance of the market as very strong, strong, normal, weak, or very weak. For the next year, Bob estimates that the probabilities for these outcomes are $$18, .27$$, $$.42, .10$$, and $$.03$$, respectively. He also thinks that the revenues corresponding to these outcomes are $$\$ 20, \$$ 18.8$$, $$\$ 16.2, \$$ 14$$, and $$\$ 12$$ million, respectively. What is Bob's expected revenue for next year?

A Christmas tree light has an expected life of \(200 \mathrm{hr}\) and a standard deviation of \(2 \mathrm{hr}\). a. Find a bound on the probability that one of these Christmas tree lights will last between \(190 \mathrm{hr}\) and \(210 \mathrm{hr}\). b. Suppose 150,000 of these Christmas tree lights are used by a large city as part of its Christmas decorations. Estimate the number of lights that are likely to require replacement between \(180 \mathrm{hr}\) and \(220 \mathrm{hr}\) of use.

The odds against an event \(E\) occurring are 2 to \(3 .\) What is the probability of \(E\) not occurring?

Maria sees the growth of her business for the upcoming year as being tied to the gross domestic product (GDP). She believes that her business will grow (or contract) at the rate of \(5 \%, 4.5 \%, 3 \%, 0 \%\), or \(-0.5 \%\) per year if the GDP grows (or contracts) at the rate of between 2 and \(2.5 \%\), between \(1.5\) and \(2 \%\), between 1 and \(1.5 \%\), between 0 and \(1 \%\), and between \(-1\) and \(0 \%\), respectively. Maria has decided to assign a probability of \(.12, .24, .40, .20\), and \(.04\), respectively, to each outcome. At what rate does Maria expect her business to grow next year?

a. Show that, for any number \(c\), $$E(c X)=c E(X)$$ b. Use this result to find the expected loss if a gambler bets \(\$ 300\) on \(\mathrm{red}\) in a single play in American roulette.

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