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Chapter 7: Problem 15

A pair of fair dice is rolled. Let \(E\) denote the event that the number falling uppermost on the first die is 5 , and let \(F\) denote the event that the sum of the numbers falling uppermost is 10 . a. Compute \(P(F)\). b. Compute \(P(E \cap F)\). c. Compute \(P(F \mid E)\). d. Compute \(P(E)\). e. Are \(E\) and \(F\) independent events?

Short Answer

Expert verified
In summary, the answers are: a. \(P(F) = \frac{1}{12}\). b. \(P(E \cap F) = \frac{1}{36}\). c. \(P(F|E) = \frac{1}{6}\). d. \(P(E) = \frac{1}{6}\). As for e, since \(P(E \cap F) \neq P(E) \times P(F)\), events E and F are not independent.

Step by step solution

01

a. Compute P(F)

To find the probability of event F occurring (sum equal to 10), we need to determine the total number of outcomes that result in a sum of 10 and divide by the total number of possible outcomes when rolling 2 dice. There are 6 numbers on both dice with 36 possible combinations, so our denominator is 36. The combinations that result in a sum of 10 are (4, 6), (5, 5), and (6, 4). Therefore, there are 3 combinations that yield a sum of 10, so the probability of event F is: \[P(F)=\frac{3}{36}=\frac{1}{12}\]
02

b. Compute P(E 鈭� F)

In order to compute the probability of events E and F occurring simultaneously (E鈭〧), we need to find the outcomes that satisfy both conditions: the first die is 5, and the sum is 10. The only combination that meets both criteria is (5, 5). Thus, there is one favorable outcome, and the probability is calculated as follows: \[P(E \cap F)=\frac{1}{36}\]
03

c. Compute P(F | E)

To compute the conditional probability of event F given event E occurred, we will use the conditional probability formula: \[P(F|E)=\frac{P(E \cap F)}{P(E)}\] We already have the intersection probability, so we need to find P(E) first.
04

d. Compute P(E)

To compute the probability of event E occurring (the first die showing 5), there is only one favorable outcome (namely, the first die is 5) out of six possible outcomes for the first die. Thus, the probability of E is: \[P(E)=\frac{1}{6}\] Now, we can find P(F | E) using the formula mentioned in step C. \[P(F|E)=\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6}\]
05

e. Are E and F independent events?

Two events are independent if the occurrence of one does not affect the probability of the other. Mathematically, events E and F are independent if \(P(E \cap F)=P(E) \times P(F)\). Let's check if this relationship holds true for our events: \[P(E \cap F)=\frac{1}{36}\] \[P(E) \times P(F)=\frac{1}{6} \times \frac{1}{12}=\frac{1}{72}\] Since \(P(E \cap F) \neq P(E) \times P(F)\), we can conclude that events E and F are not independent.

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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 how the probability of an event changes when we know that another event has occurred. It is usually denoted as \(P(A \mid B)\), meaning the probability of event \(A\) occurring given that \(B\) is true. In the dice example, we needed to find \(P(F \mid E)\), the probability that the sum of the dice is 10 given that the first die shows a 5. This uses the formula:
\[P(F \mid E) = \frac{P(E \cap F)}{P(E)}\]From the problem:
  • \(P(E \cap F)\) is the probability of both \(E\) and \(F\) occurring.
  • \(P(E)\) is the probability of the first die being 5.
  • Substituting known values, \(P(F \mid E) = \frac{1/36}{1/6} = \frac{1}{6}\).
This shows that knowing specific information about one die can affect the probability of the total event.
Independent Events
Events are independent if the occurrence of one does not influence the probability of occurrence of the other. This concept is central to understanding many probability questions. In mathematical terms, two events \(A\) and \(B\) are independent if and only if:
\[P(A \cap B) = P(A) \times P(B)\]In the dice rolling example:
  • \(P(E)\) is the probability that the first die shows a 5.
  • \(P(F)\) is the probability that the sum of the dice is 10.
  • We compared \(P(E \cap F)\) to \(P(E) \times P(F)\).
Since \(\frac{1}{36} eq \frac{1}{72}\), events \(E\) and \(F\) are not independent. They influence each other, meaning knowing one provides information about the other.
Discrete Probability
Discrete probability involves working with outcomes that occur in distinct and separate possibilities. When dealing with dice, each face represents a possible outcome, making it a classic discrete probability scenario.
Key points to grasp include:
  • The sample space for two dice is made up of 36 possible outcomes, \(6\) faces on the first die and \(6\) on the second, \(6 \times 6 = 36\).
  • For distinct probabilities, each combination of rolled numbers has equal chance, \(\frac{1}{36}\).
In our problem, we're determining specific events from this sample space, such as getting a total of 10, which required identifying specific favorable outcomes. Then, dividing by the total outcomes helps us find probabilities.
Combinatorics
Combinatorics is the mathematical study of counting and arrangements. It helps us determine the number of possible outcomes, which is vital in probability theory.
In the dice exercise:
  • Identify favorable outcomes, like rolling a total of 10 which includes combinations (4,6), (5,5), and (6,4).
  • These combinations are counted to assess probabilities.
Combinatorics aids in efficiently listing possible scenarios, as seen when solving \(P(F)\), by counting the specific outcomes instead of evaluating each possible roll manually. Understanding this enhances the ability to solve various probability problems methodically by considering all possible configurations.

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