Math Problem Statement

  1. Suppose that P(E) = 0.73, P(F) =0.57 , and P(E ∩ F) = 0.41. Find the following: P (E ∪ F)^c options = a. 0.850 b. 0.330 c. 0.640 d. 0.110 e. 0.140 2. . Suppose that P(E) = 0.82, P(F) = 0.63, and ( ) c P E ∩ F = 0.12 . Find the following: P (E^c ∩ F^c) options= a. 0.800 b. 0.940 c. 0.794 d. 0.410 e. 0.060

Solution

Let's solve each problem step-by-step.

Problem 1

Given:

  • P(E)=0.73P(E) = 0.73
  • P(F)=0.57P(F) = 0.57
  • P(EF)=0.41P(E \cap F) = 0.41

We are asked to find P((EF)c)P((E \cup F)^c), which is the complement of the union of EE and FF.

Using the formula for the union of two events: P(EF)=P(E)+P(F)P(EF)P(E \cup F) = P(E) + P(F) - P(E \cap F) we substitute the given values: P(EF)=0.73+0.570.41=0.89P(E \cup F) = 0.73 + 0.57 - 0.41 = 0.89

Since P((EF)c)=1P(EF)P((E \cup F)^c) = 1 - P(E \cup F), we find: P((EF)c)=10.89=0.11P((E \cup F)^c) = 1 - 0.89 = 0.11

So, the correct answer is: Option (d) 0.110


Problem 2

Given:

  • P(E)=0.82P(E) = 0.82
  • P(F)=0.63P(F) = 0.63
  • P((EF)c)=0.12P((E \cap F)^c) = 0.12

We are asked to find P(EcFc)P(E^c \cap F^c), which represents the probability that neither EE nor FF occurs.

To find this, we use the formula: P(EcFc)=1P(EF)P(E^c \cap F^c) = 1 - P(E \cup F)

First, calculate P(EF)P(E \cup F) using the complement of P((EF)c)P((E \cap F)^c): P(EF)=1P((EF)c)=10.12=0.88P(E \cap F) = 1 - P((E \cap F)^c) = 1 - 0.12 = 0.88

Now, substitute into the union formula: P(EF)=P(E)+P(F)P(EF)P(E \cup F) = P(E) + P(F) - P(E \cap F) P(EF)=0.82+0.630.88=0.57P(E \cup F) = 0.82 + 0.63 - 0.88 = 0.57

Thus: P(EcFc)=10.57=0.43P(E^c \cap F^c) = 1 - 0.57 = 0.43

**Answer not listed among the options, so there may be a miscalculation

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Math Problem Analysis

Mathematical Concepts

Probability
Complement of Events
Union and Intersection of Events

Formulas

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)
P((E ∪ F)^c) = 1 - P(E ∪ F)
P(E^c ∩ F^c) = 1 - P(E ∪ F)

Theorems

Complement Rule
Addition Rule for Probabilities

Suitable Grade Level

Grade 11-12