Problem 33 - A First Course in Probability

Question

On rainy days, Joe is late to work with probability \(.3 ;\) on nonrainy days, he is late with probability . \(1 .\) With probability.7, it will rain tomorrow. (a) Find the probability that Joe is early tomorrow. (b) Given that Joe was early, what is the conditional probability that it rained?

Step-by-Step Solution

Answer

(a) The probability that Joe is early tomorrow is \(0.76\). (b) Given that Joe was early, the conditional probability that it rained is approximately \(0.645\).

Step 1: Identify the Events

We are given probabilities for two events: 1. Event R: It will rain tomorrow (with probability 0.7) 2. Event NR: It will not rain tomorrow (with probability 0.3) 3. Event L: Joe is late (with probability 0.3, when it rains; and 0.1, when it doesn't rain) We are tasked to find the probability of Joe being early (Event E) tomorrow and the conditional probability of it raining given that Joe is early (Event R | E).

Step 2: Calculate the Probability of Joe Being Early (Event E)

To find P(E), we need to consider both situations where it rains and it doesn't rain. We can use the law of total probability. P(E) = P(E | R)P(R) + P(E | NR)P(NR) We have P(R) = 0.7, P(NR) = 0.3, and we know that on rainy days Joe is late with probability 0.3, so P(E | R) = 1 - P(L | R) = 1 - 0.3 = 0.7. Likewise, on non-rainy days, Joe is late with probability 0.1, thus P(E | NR) = 1 - P(L | NR) = 1 - 0.1 = 0.9. Now, we can plug in the values: P(E) = (0.7)(0.7) + (0.9)(0.3) = 0.49 + 0.27 = 0.76

Step 3: Calculate the Conditional Probability of It Raining Given Joe is Early (Event R | E)

Now we need to find P(R | E), which can be calculated using the conditional probability formula: P(R | E) = P(R and E) / P(E) We already found P(E) in step 2, which is 0.76. To calculate P(R and E), we use: P(R and E) = P(E | R)P(R) = (0.7)(0.7) = 0.49 Plug the values into the formula: P(R | E) = 0.49 / 0.76 = 0.64473684 Therefore, the conditional probability of it raining given that Joe is early is approximately 0.645. #Summary# (a) The probability that Joe is early tomorrow is 0.76. (b) Given that Joe was early, the conditional probability that it rained is approximately 0.645.

Key Concepts

Law of Total Probability

The Law of Total Probability is a fundamental concept in probability theory that allows us to calculate the probability of an event by considering all possible scenarios that could lead to it. In simple terms, it breaks down a complex probability into simpler parts by using known probabilities. This is particularly helpful when dealing with events that depend on multiple conditions.

In the exercise, we needed to find the probability that Joe is early (Event E) by considering two possible scenarios: whether it rains (Event R) or not (Event NR). When using the law, you consider the probability of Event E happening in each scenario and then sum these probabilities weighted by the likelihood of each scenario occurring:

\[P(E) = P(E | R)P(R) + P(E | NR)P(NR)\]

Here:

• The probability of it raining tomorrow, \(P(R)\), is 0.7. Meanwhile, the probability of it not raining, \(P(NR)\), is 0.3.
• If it rains, Joe is early with probability \(P(E | R) = 0.7\).
• If it does not rain, Joe is early with probability \(P(E | NR) = 0.9\).

Plugging these into the equation provides \(P(E) = 0.49 + 0.27 = 0.76\), meaning Joe has a 76% chance of being early tomorrow.

Bayes' Theorem

Bayes' Theorem allows us to update our predictions or hypotheses based on new evidence. It is particularly effective for calculating reverse probabilities, that is, the probability of a cause given an effect.

Bayes' Theorem can be expressed as:

\[P(A | B) = \frac{P(B | A)P(A)}{P(B)}\]

In our exercise, we want to know the probability that it rained, given that Joe was early (i.e., \(P(R | E)\)). We already calculated \(P(E)\) as 0.76.

Let's break it down:

• \(P(R)\) is the probability that it will rain, which is 0.7.
• \(P(E | R)\) is the probability that Joe is early given that it rained, calculated as 0.7.
• \(P(R \text{ and } E)\) is calculated as \(P(E | R)P(R) = 0.49\).

Thus, the conditional probability that it rained given Joe was early is:

\[P(R | E) = \frac{0.49}{0.76} \approx 0.645\]

This result, around 64.5%, reflects how likely rain was the cause of Joe being early, showing how Bayes' Theorem helps find these probabilities in backward scenarios.

Probability of Complementary Events

The probability of complementary events offers a simple yet powerful insight into probability. When thinking about complementary events, consider two mutually exclusive outcomes that span all possibilities for a particular scenario.

For example, if you know the probability of a particular event occurring, the probability of it not occurring is simply 1 minus the probability of the event. If Joe has a probability of being late (Event L), the probability of him being early (Event E) is the complement.

In our exercise:

• The probability Joe is late on a rainy day is 0.3. Thus, the probability he is early that day, \(P(E | R)\), is \(1 - 0.3 = 0.7\).
• Similarly, the probability Joe is late on a non-rainy day is 0.1, so the probability he is early then, \(P(E | NR)\), is \(1 - 0.1 = 0.9\).

This notion of complementary events not only simplifies calculations but also strengthens our understanding of probability distributions, emphasizing how every event outcome and its complement complete the whole set of possibilities (adding up to 1). Understanding this concept strengthens one's ability to make intuitive sense of complex probability scenarios.