Answered: a. A₁ U A₂ b. An A [Hint: (A, U. A₁ U…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

Could you double-check if my answers are correct for this question 2.13? If not, please point out the error and present the correction.

Question. 2.13:

a. A1 ∪ A2: This event represents the project being awarded to either project 1 (A1) or project 2 (A2), or both. In words, it's the event of winning project 1 or project 2.

Probability of A1 ∪ A2:

P(A1 ∪ A2) = P(A1) + P(A2) - P(A1 ∩ A2)

= 0.22 + 0.25 - 0.11

= 0.36

b. A1' ∩ A2': This event represents not winning project 1 (A1') and not winning project 2 (A2'). In words, it's the event of not being awarded either project 1 or project 2.

Probability of A1' ∩ A2':

P(A1' ∩ A2') = P(A1' ∩ A2)

= P(A2) - P(A1 ∩ A2)

= 0.25 - 0.11

= 0.14

c. A1 ∪ A2 ∪ A3: This event represents the project being awarded to either project 1 (A1), project 2 (A2), or project 3 (A3), or any combination of them. In words, it's the event of winning any of the three projects.

Probability of A1 ∪ A2 ∪ A3:

P(A1 ∪ A2 ∪ A3) = P(A1) + P(A2) + P(A3) - P(A1 ∩ A2) - P(A1 ∩ A3) - P(A2 ∩ A3) + P(A1 ∩ A2 ∩ A3)

= 0.22 + 0.25 + 0.28 - 0.11 - 0.05 - 0.07 + 0.01

= 0.53

d. A1' ∩ A2' ∩ A3': This event represents not winning project 1 (A1'), not winning project 2 (A2'), and not winning project 3 (A3'). In words, it's the event of not being awarded any of the three projects.

Probability of A1' ∩ A2' ∩ A3':

P(A1' ∩ A2' ∩ A3') = P(A1') × P(A2') × P(A3')

= (1 - P(A1)) × (1 - P(A2)) × (1 - P(A3))

= (1 - 0.22) × (1 - 0.25) × (1 - 0.28)

= 0.58 × 0.75 × 0.72

≈ 0.315

e. A1' ∩ A2' ∩ A3: This event represents not winning project 1 (A1'), not winning project 2 (A2'), but winning project 3 (A3). In words, it's the event of not being awarded project 1 and project 2, but being awarded project 3.

Probability of A1' ∩ A2' ∩ A3:

P(A1' ∩ A2' ∩ A3) = P(A3) - P(A1 ∩ A3) - P(A2 ∩ A3) + P(A1 ∩ A2 ∩ A3)

= 0.28 - 0.05 - 0.07 + 0.01

= 0.17

f. (A1' ∩ A2') ∪ A3: This event represents not winning project 1 (A1') and not winning project 2 (A2'), or winning project 3 (A3), or both. In words, it's the event of not being awarded project 1 and project 2, or being awarded project 3, or both.

Probability of (A1' ∩ A2') ∪ A3:

P((A1' ∩ A2') ∪ A3) = P(A1' ∩ A2') + P(A3) - P((A1' ∩ A2') ∩ A3)

= 0.14 + 0.28 - 0.01

= 0.41