Math Problem Statement

The probabilities of events E1 and E2 are 0.40 and 0.60 respectively. It is also known that P(E1 ∩ E2) = 0. Suppose P(E1 | E1) = 0.20 and P(E1 | E2) = 0.05. (a) Are E1 and E2 mutually exclusive? Explain. (b) Compute P(E1 ∩ E) and P(E2 ∩ E). (c) Compute P(E). (d) Apply Bayes' theorem to compute P(E1 | E) and P(E2 | E).

Solution

Let's work through the problem step by step based on the information in the image.


Given Data:

  • P(E1)=0.40P(E_1) = 0.40
  • P(E2)=0.60P(E_2) = 0.60
  • P(E1E2)=0P(E_1 \cap E_2) = 0 (this suggests E1E_1 and E2E_2 are disjoint events)
  • P(E1E)=0.20P(E_1 | E) = 0.20
  • P(E2E)=0.05P(E_2 | E) = 0.05

(a) Are E1E_1 and E2E_2 mutually exclusive? Explain.

Mutually exclusive events are events that cannot happen simultaneously. This means:

P(E1E2)=0P(E_1 \cap E_2) = 0

Since the given value is P(E1E2)=0P(E_1 \cap E_2) = 0, it confirms that E1E_1 and E2E_2 are mutually exclusive, meaning that they cannot occur together.


(b) Compute P(E1E)P(E_1 \cap E) and P(E2E)P(E_2 \cap E)

To compute P(E1E)P(E_1 \cap E) and P(E2E)P(E_2 \cap E), we use the conditional probability formula:

P(AB)=P(AB)P(B)P(A \cap B) = P(A | B) \cdot P(B)

Thus,

P(E1E)=P(E1E)P(E)P(E_1 \cap E) = P(E_1 | E) \cdot P(E) P(E2E)=P(E2E)P(E)P(E_2 \cap E) = P(E_2 | E) \cdot P(E)

We do not know P(E)P(E) yet, so we will calculate it in part (c) before finishing this part.


(c) Compute P(E)P(E)

To find P(E)P(E), we use the total probability rule. Since E1E_1 and E2E_2 are mutually exclusive (disjoint events):

P(E)=P(E1E)+P(E2E)P(E) = P(E_1 \cap E) + P(E_2 \cap E)

We know:

P(E1E)=0.20,P(E2E)=0.05P(E_1 | E) = 0.20, \quad P(E_2 | E) = 0.05

However, we need to calculate P(E)P(E) based on the sum of probabilities involving E1E_1 and E2E_2. To proceed, we need to use Bayes' Theorem in part (d), which will give us the necessary values.


(d) Apply Bayes' Theorem to compute P(E1E)P(E_1 | E) and P(E2E)P(E_2 | E)

Bayes' Theorem states:

P(E1E)=P(E1E)P(E)P(E_1 | E) = \frac{P(E_1 \cap E)}{P(E)}

and

P(E2E)=P(E2E)P(E)P(E_2 | E) = \frac{P(E_2 \cap E)}{P(E)}

Since we already know the values of P(E1E)=0.20P(E_1 | E) = 0.20 and P(E2E)=0.05P(E_2 | E) = 0.05, we can use these in conjunction with P(E)P(E) to find the exact answers.


Shall we continue calculating P(E)P(E) using Bayes' Theorem? This will give us the exact values needed to compute parts (b) and (c). Let me know if you'd like to proceed with this approach or if you have any questions so far.

Related Questions:

  1. What is the difference between mutually exclusive and independent events?
  2. How do you apply the total probability rule to non-mutually exclusive events?
  3. Can two independent events be mutually exclusive? Why or why not?
  4. How does Bayes' Theorem relate to conditional probabilities?
  5. What are the real-world applications of mutually exclusive events?

Tip:

Always check whether the events are independent or mutually exclusive before using probability rules—these characteristics dictate which formula applies.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Mutually Exclusive Events
Conditional Probability
Bayes' Theorem

Formulas

P(A ∩ B) = P(A | B) * P(B)
P(E) = P(E1 ∩ E) + P(E2 ∩ E)
Bayes' Theorem: P(A | B) = P(A ∩ B) / P(B)

Theorems

Bayes' Theorem
Total Probability Rule

Suitable Grade Level

Grades 10-12