Math Problem Statement

Complete the following probability distribution table and then calculate the stated probabilities. Outcome a b c d e Probability 0.2 0.08 0.5 0.02 (a) P({a, c, e}) P({a, c, e}) = (b) P(E ∪ F), where E = {a, c, e} and F = {b, c, e} P(E ∪ F) = (c) P(E ′), where E is as in part (b) P(E ′) = (d) P(E ∩ F ), where E and F are as in part (b) P(E ∩ F) =

Solution

Let's start by completing the probability distribution table.

Step 1: Completing the Probability Distribution Table

The given probabilities are for the outcomes a,b,c,d,ea, b, c, d, e, but we only have probabilities for a,b,c,a, b, c, and dd. We can find the missing probability for ee by using the fact that the sum of all probabilities must equal 1.

Given:

  • Probability of a=0.2a = 0.2
  • Probability of b=0.08b = 0.08
  • Probability of c=0.5c = 0.5
  • Probability of d=0.02d = 0.02

The sum of these probabilities is: 0.2+0.08+0.5+0.02=0.80.2 + 0.08 + 0.5 + 0.02 = 0.8

Therefore, the probability for ee must be: P(e)=10.8=0.2P(e) = 1 - 0.8 = 0.2

So, the completed probability distribution table is:

Outcomeabcde
Probability0.20.080.50.020.2

Step 2: Calculating the Stated Probabilities

(a) P({a,c,e})P(\{a, c, e\})

This is the probability of the union of the outcomes aa, cc, and ee. The probability of this event is simply the sum of the individual probabilities:

P({a,c,e})=P(a)+P(c)+P(e)=0.2+0.5+0.2=0.9P(\{a, c, e\}) = P(a) + P(c) + P(e) = 0.2 + 0.5 + 0.2 = 0.9

So, P({a,c,e})=0.9P(\{a, c, e\}) = 0.9

(b) P(EF)P(E \cup F), where E={a,c,e}E = \{a, c, e\} and F={b,c,e}F = \{b, c, e\}

The union EFE \cup F includes all the outcomes in either EE or FF, or in both. So: EF={a,b,c,e}E \cup F = \{a, b, c, e\} Thus, the probability of EFE \cup F is the sum of the probabilities of these outcomes: P(EF)=P(a)+P(b)+P(c)+P(e)=0.2+0.08+0.5+0.2=0.98P(E \cup F) = P(a) + P(b) + P(c) + P(e) = 0.2 + 0.08 + 0.5 + 0.2 = 0.98

So, P(EF)=0.98P(E \cup F) = 0.98

(c) P(E)P(E'), where E={a,c,e}E = \{a, c, e\}

The complement EE' consists of the outcomes not in EE, so: E={b,d}E' = \{b, d\} Thus, the probability of EE' is the sum of the probabilities of bb and dd: P(E)=P(b)+P(d)=0.08+0.02=0.1P(E') = P(b) + P(d) = 0.08 + 0.02 = 0.1

So, P(E)=0.1P(E') = 0.1

(d) P(EF)P(E \cap F), where E={a,c,e}E = \{a, c, e\} and F={b,c,e}F = \{b, c, e\}

The intersection EFE \cap F includes the outcomes that are in both EE and FF, so: EF={c,e}E \cap F = \{c, e\} Thus, the probability of EFE \cap F is the sum of the probabilities of cc and ee: P(EF)=P(c)+P(e)=0.5+0.2=0.7P(E \cap F) = P(c) + P(e) = 0.5 + 0.2 = 0.7

So, P(EF)=0.7P(E \cap F) = 0.7

Summary of Answers:

(a) P({a,c,e})=0.9P(\{a, c, e\}) = 0.9
(b) P(EF)=0.98P(E \cup F) = 0.98
(c) P(E)=0.1P(E') = 0.1
(d) P(EF)=0.7P(E \cap F) = 0.7


Would you like further explanations or details on any of these steps? Here are some related questions for practice:

  1. What happens if the sum of probabilities exceeds 1 in a probability distribution?
  2. How do we calculate the probability of the intersection of two disjoint events?
  3. Can two events have a union that has a probability greater than 1? Why or why not?
  4. How would the probability distribution change if the probability of outcome dd was increased to 0.1?
  5. What does the complement of an event represent in probability theory?

Tip: Always check that the sum of all probabilities in a probability distribution equals 1! This is a fundamental property.

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Math Problem Analysis

Mathematical Concepts

Probability
Union and Intersection of Sets
Complementary Events

Formulas

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A′) = 1 - P(A)
P(A ∩ B) = sum of probabilities of outcomes common to A and B

Theorems

Addition Rule of Probability
Complement Rule
Intersection of Events

Suitable Grade Level

Grades 7-10