Question

A sample space consists of 9 elementary outcomes e₁, e₂, ..., e₉ whose probabilities are

P(e₁) = P(e₂) = .08, P(e₃) = P(e₄) = P(e₅) = .1
P(e₆) = P(e₇) = .2, P(e₈) = P(e₉) = .07
Suppose A = {e₁, e₅, e₈}, B = {e₂, e₅, e₈, e₉}

(a) Calculate P (A), P (B), and P (A ∩ B)

(b) Using the addition law of probability, calculate P (A ∪ B)

(c) List the composition of the event A ∪ B, and calculate P (A ∪ B) by adding the probabilities of the elementary outcomes.

(d) Calculate P (\(\overline B\)) from P (B), also calculate P (\(\overline B\)) directly from the elementary outcomes of \(\overline B\).

Answer 1

Given that:

S = {e₁, e₂, e₃, e₄, e_5, e₆, e₇, e₈, e₉}

A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

P(e₁) = P(e₂) = .08, P(e₃) = P(e₄) = P(e₅) = .1

P(e₆) = P(e₇) = .2, P(e₈) = P(e₉) = .07

(a) To find: P(A), P(B) and P(A ⋂ B)

A = {e₁, e₅, e₈}

P(A) = P(e₁) + P(e₅) + P(e₈)

⇒ P(A) = 0.08 + 0.1 + 0.07 [given]

⇒ P(A) = 0.25

B = {e₂, e₅, e₈, e₉}

P(B) = P(e₂) + P(e₅) + P(e₈) + P(e₉)

⇒ P(B) = 0.08 + 0.1 + 0.07 + 0.07 [given]

⇒ P(B) = 0.32

Now, we have to find P(A ⋂ B)

A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

∴ A ⋂ B = {e₅, e₈}

⇒ P(A ⋂ B) = P(e₅) + P(e₈)

= 0.1 + 0.07

= 0.17

(b) To find: P(A ⋃ B)

By General Addition Rule:

P(A ⋃ B) = P(A) + P(B) – P(A ⋂ B)

from part (a), we have

P(A) = 0.25, P(B) = 0.32 and P(A ⋂ B) = 0.17

Putting the values, we get

P(A ⋃ B) = 0.25 + 0.32 – 0.17

= 0.40

(c) A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

∴ A ⋃ B = {e₁, e₂, e₅, e₈, e₉}

⇒ P(A ⋃ B) = P(e₁) + P(e₂) + P(e₅) + P(e₈) + P(e₉)

= 0.08 +0.08 + 0.1 + 0.07 + 0.07

= 0.40

= 0.08 + 0.1 + 0.1 + 0.2 + 0.2 [given]

= 0.68