Problem 52 - A First Course in Probability

Question

A high school student is anxiously waiting to receive mail telling her whether she has been accepted to a certain college. She estimates that the conditional probabilities of receiving notification on each day of next week, given that she is accepted and that she is rejected, are as follows: $$\begin{array}{lcc} \hline \text { Day } & P(\text { mail } | \text { accepted }) & P(\text { mail } | \text { rejected }) \\ \hline \text { Monday } & .15 & .05 \\ \text { Tuesday } & .20 & .10 \\ \text { Wednesday } & .25 & .10 \\ \text { Thursday } & .15 & .15 \\ \text { Friday } & .10 & .20 \\ \hline \end{array}$$ She estimates that her probability of being accepted is .6. (a) What is the probability that she receives mail on Monday? (b) What is the conditional probability that she received mail on Tuesday given that she does not receive mail on Monday? (c) If there is no mail through Wednesday, what is the conditional probability that she will be accepted? (d) What is the conditional probability that she will be accepted if mail comes on Thursday? (e) What is the conditional probability that she will be accepted if no mail arrives that week?

Step-by-Step Solution

Answer

The short answer is as follows: (a) The probability that she receives mail on Monday is 0.11. (b) The conditional probability that she received mail on Tuesday given that she does not receive mail on Monday is approximately 0.18. (c) The conditional probability that she will be accepted if there is no mail through Wednesday is approximately 0.506. (d) The conditional probability that she will be accepted if mail comes on Thursday is 0.6. (e) The conditional probability that she will be accepted if no mail arrives that week is approximately 0.601.

Step 1: Determine probabilities

Using the given conditional probabilities, we need to find the probability of her receiving mail on Monday regardless of acceptance. Let A be the event that she is accepted, and M1 be the event that she receives mail on Monday. We are asked to find \(P(M1)\). By the rules of probability, we have \(P(A \cap M1) = P(M1|A) \times P(A)\) and \(P(M1 \cap A^c) = P(M1|A^c) \times P(A^c)\), where \(A^c\) is the event of being rejected. Using the given probabilities, we can calculate these: \(P(M1 \cap A) = 0.15 \times 0.6 = 0.09\) \(P(A^c) = 1 - 0.6 = 0.4\) \(P(M1 \cap A^c) = 0.05 \times 0.4 = 0.02\)

Step 2: Calculate the probability of receiving mail on Monday

To find \(P(M1)\), we can simply use the rule of probability: \(P(M1) = P(M1 \cap A) + P(M1 \cap A^c) = 0.09 + 0.02 = 0.11\) So, the probability that she receives mail on Monday is 0.11. (b) What is the conditional probability that she received mail on Tuesday given that she does not receive mail on Monday?

Step 1: Calculate probability of receiving mail on Tuesday

We need to calculate the probability of receiving mail on Tuesday, denoted as M2, given that mail was not received on Monday, represented as \(M1^c\). Let's find \(P(A \cap M2) = P(M2|A) \times P(A)\) and \(P(M2 \cap A^c) = P(M2|A^c) \times P(A^c)\): \(P(A \cap M2) = 0.20 \times 0.6 = 0.12\) \(P(M2 \cap A^c) = 0.10 \times 0.4 = 0.04\) Now we can find \(P(M2)\): \(P(M2) = P(M2 \cap A) + P(M2 \cap A^c) = 0.12 + 0.04 = 0.16\)

Step 2: Calculate conditional probability

Now we can calculate the conditional probability \(P(M2 | M1^c)\) using the rule of probability: \(P(M2 | M1^c) = \frac{P(M2 \cap M1^c)}{P(M1^c)}\) Note that \(M1\) and \(M2\) are mutually exclusive; if mail was received on Monday, it cannot be received again on Tuesday and vice versa. Thus, \(P(M2 \cap M1^c) = P(M2)\). So, \(P(M2 | M1^c) = \frac{P(M2)}{P(M1^c)} = \frac{0.16}{1 - 0.11} = \frac{0.16}{0.89} \approx 0.18\). The conditional probability that she received mail on Tuesday given that she does not receive mail on Monday is approximately 0.18. (c) If there is no mail through Wednesday, what is the conditional probability that she will be accepted?

Step 1: Calculate probability of no mail through Wednesday

Let W denote no mail through Wednesday. We need to find \(P(A | W)\), which can be found by dividing the probability \(P(A \cap W)\) by \(P(W)\). First, let's calculate the probability of receiving mail on Wednesday, denoted as M3, similarly to previous steps: \(P(A \cap M3) = P(M3|A) \times P(A) = 0.25 \times 0.6 = 0.15\) \(P(M3 \cap A^c) = P(M3|A^c) \times P(A^c) = 0.10 \times 0.4 = 0.04\) \(P(M3) = P(M3 \cap A) + P(M3 \cap A^c) = 0.15 + 0.04 = 0.19\) Now, let's find the probabilities of no mail on Monday, Tuesday, and Wednesday, denoted as \(M1^c\), \(M2^c\), and \(M3^c\), respectively: \(P(M1^c) = 1 - P(M1) = 1 - 0.11 = 0.89\) \(P(M2^c) = 1 - P(M2) = 1 - 0.16 = 0.84\) \(P(M3^c) = 1 - P(M3) = 1 - 0.19 = 0.81\) Since receiving mail on different days is mutually exclusive, we can find the probability of no mail through Wednesday: \(P(W) = P(M1^c) \times P(M2^c) \times P(M3^c) = 0.89 \times 0.84 \times 0.81 \approx 0.605\)

Step 2: Calculate conditional probability

We can now calculate \(P(A | W)\) using the rule of probability: \(P(A | W) = \frac{P(A \cap W)}{P(W)}\) We know that \(P(W) \approx 0.605\). Now, let's find \(P(A \cap W)\): \(P(A \cap W) = P(A) \times P(W|A)\) We can find \(P(W|A)\) similarly as \(P(M1^c) \times P(M2^c) \times P(M3^c)\) but using conditional probabilities instead of marginals: \(P(W|A) = (1 - 0.15) \times (1 - 0.20) \times (1 - 0.25) = 0.85 \times 0.80 \times 0.75 = 0.51\) Thus, \(P(A \cap W) = 0.6 \times 0.51 = 0.306\) Finally, we get: \(P(A | W) = \frac{0.306}{0.605} \approx 0.506\) So, the conditional probability that she will be accepted if there is no mail through Wednesday is approximately 0.506. (d) What is the conditional probability that she will be accepted if mail comes on Thursday?

Step 1: Calculate probability of mail on Thursday

Let's denote receiving mail on Thursday as M4 and calculate the probability of M4 the same way as before: \(P(A \cap M4) = P(M4|A) \times P(A) = 0.15 \times 0.6 = 0.09\) \(P(M4 \cap A^c) = P(M4|A^c) \times P(A^c) = 0.15 \times 0.4 = 0.06\) \(P(M4) = P(M4 \cap A) + P(M4 \cap A^c) = 0.09 + 0.06 = 0.15\)

Step 2: Calculate conditional probability

Now we need to find \(P(A | M4)\), which can be calculated using the rule of probability: \(P(A | M4) = \frac{P(A \cap M4)}{P(M4)} = \frac{0.09}{0.15} = 0.6\) The conditional probability that she will be accepted if mail comes on Thursday is 0.6. (e) What is the conditional probability that she will be accepted if no mail arrives that week?

Step 1: Calculate probability of no mail all week

Let N denote the event that no mail arrives all week. We need to find \(P(A | N)\), which can be found by dividing the probability \(P(A \cap N)\) by \(P(N)\). First, let's calculate the probability of receiving mail on Friday, denoted as M5: \(P(A \cap M5) = P(M5|A) \times P(A) = 0.10 \times 0.6 = 0.06\) \(P(M5 \cap A^c) = P(M5|A^c) \times P(A^c) = 0.20 \times 0.4 = 0.08\) \(P(M5) = P(M5 \cap A) + P(M5 \cap A^c) = 0.06 + 0.08 = 0.14\) Now, let's find the probability of no mail on Friday, denoted as \(M5^c\): \(P(M5^c) = 1 - P(M5) = 1 - 0.14 = 0.86\) For the probability of no mail all week, we multiply the probabilities of no mail on each day: \(P(N) = P(M1^c) \times P(M2^c) \times P(M3^c) \times P(M4^c) \times P(M5^c) = 0.89 \times 0.84 \times 0.81 \times 0.85 \times 0.86 \approx 0.432\)

Step 2: Calculate conditional probability

We can now calculate \(P(A | N)\) using the rule of probability: \(P(A | N) = \frac{P(A \cap N)}{P(N)}\) We know that \(P(N) \approx 0.432\). Now, let's find \(P(A \cap N)\): \(P(A \cap N) = P(A) \times P(N|A)\) We can find \(P(N|A)\) similarly as \(P(M1^c) \times P(M2^c) \times P(M3^c) \times P(M4^c) \times P(M5^c)\) but using conditional probabilities instead of marginals: \(P(N|A) = (1 - 0.15) \times (1 - 0.20) \times (1 - 0.25) \times (1 - 0.15) \times (1 - 0.10) = 0.85 \times 0.80 \times 0.75 \times 0.85 \times 0.90 = 0.433\) Thus, \(P(A \cap N) = 0.6 \times 0.433 = 0.2598\) Finally, we get: \(P(A | N) = \frac{0.2598}{0.432} \approx 0.601\) So, the conditional probability that she will be accepted if no mail arrives that week is approximately 0.601.

Key Concepts

Introduction to Probability Theory

At the heart of understanding random events and making predictions about uncertain outcomes is probability theory. It's a mathematical framework that allows us to calculate the likelihood of events occurring, which can range from the flip of a coin to more complex situations like a student awaiting college acceptance letters.

When talking about probability, we typically work with values between 0 and 1, where 0 indicates impossibility and 1 denotes certainty. The probability of an event E is denoted as P(E). A fundamental aspect of probability theory is that the sum of probabilities of all possible outcomes of a random experiment is equal to 1.

In problems like the one involving the high school student waiting for mail, we assign probabilities to reflect our expectations about future events. We use these probabilities to answer various questions about the outcomes, both unconditional and conditional—a concept which represents the likelihood of an event given that another event has already occurred. For instance, the probability of receiving mail on a day, given that the student is accepted, is a conditional probability.

Understanding Mutually Exclusive Events

Events are described as mutually exclusive when the occurrence of one event excludes the possibility of another. In simple terms, both can't happen at the same time. For example, if the student receives mail on Monday, this precludes the chance of also receiving a first mail on Tuesday.

In the context of our exercise, the student estimates the probability of receiving mail on different days, which are treated as mutually exclusive events because she expects to receive the notification mail on one day only. This concept is crucial when calculating probabilities over a series of days. If events are mutually exclusive, the probability of their union is the sum of their individual probabilities, which is applied in solving part (b) of our exercise.

It's important to remember that just because events are mutually exclusive does not mean they are independent. Independence, another key concept in probability, occurs when the occurrence of one event does not affect the probability of another; however, this is not explored within this specific problem.

The Role of Bayes' Theorem in Conditional Probability

Bayes' theorem is an essential principle in probability theory that provides a way to revise existing predictions or theories (prior probabilities) in light of new evidence. It's especially powerful for calculating the conditional probabilities when we have information about the reverse condition.

The theorem's equation is given by:
\( P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \)
where:

• \( P(A|B) \) is the probability of event A given that B has happened,
• \( P(B|A) \) is the probability of event B given that A has happened,
• \( P(A) \) is the probability of event A, and
• \( P(B) \) is the probability of event B.

In part (d) of our exercise, Bayes' theorem could be used to answer the question about the conditional probability of acceptance given that mail comes on Thursday. It allows us to update the probability of the student's acceptance based on the new information that mail has been received on a specific day.

Bayes' theorem is particularly powerful in iterative scenarios, such as the successive days without mail, where with each passing day without confirmation, the conditional probabilities are adjusted.