Question

A sample space consists of 9 elementary event E₁, E₂, E₃, ……E₈, E₉ whose probabilities are P(E₁) = P(E₂) = 0.08, P(E₃) = P(E₄) = 0.1, P(E₆) = P(E₇) = 0.2, P(E₈) = P(E₉) = 0.07 

Suppose A = {E₁, E₅, E₈}, B = {E₂, E₈, E₉} 

(i) Compute P(A), P(B) and P(A∩B). 

(ii) Using the addition law of probability, find P (A∪B). 

(iii) List the composition of the event A∪B, and calcite P(A∪B) by adding the probabilities of the elementary events. 

(iv) Calculate\(P(\bar{B})\) from P(B), also calculate \(P(\bar{B})\) directly from the element of \({(\bar B)}\)

Answer 1

Clearly according to questions sample space contains 9 elementary events(events with single outcome)

Let S represents the sample space. 

∴ S = E₁∪ E₂∪ E₃ ∪ ……∪ E₈∪ E₉ 

Given A = {E₁, E₅, E₈}

Or A = E₁∪ E₅∪E₈ 

P(E₁) = P(E₂) = 0.08, P(E₃) = P(E₄) = 0.1, P(E₆) = P(E₇) = 0.2, P(E₈) = P(E₉) = 0.07 

∴ P(A) = P(E₁∪ E₅∪ E₈) = P(E₁)+P(E₅) + P(E₈) 

⇒ P(A) = 0.08 + P(E₅) + 0.07 = 0.15 + P(E₅) …(1)

P(E₅) is missing, so we need to find it. 

Given, B = {E₂, E₈, E₉} or B = E₂∪ E₈∪ E₉ 

∴ P(B) = P(E₂∪ E₈∪ E₉) = P(E₂)+P(E₈) + P(E₉)

= 0.08 + 0.07 + 0.07 = 0.21

∴ P(B) = 0.21 ….ans (i)

∴ B’ = {E₁, E₃, E₄, E₅, E₆, E₇} or B’ = E₁∪ E₃∪ E₄∪ E₅∪ E₆∪ E₇ 

∴ P(B’) = P(E₁) + P(E₃) + P(E₄) + P(E₅) + P(E₆) + P(E₇)

1 – 0.21 = 0.08 + 0.1 + 0.1 + P(E5) + 0.2 + 0.2 

⇒ 0.79 = 0.68 + P(E₅)

∴ P(E5) = 0.79 – 0.68 = 0.11 

∴ from equation 1, we get 

P(A) = 0.15 + P(E₅) = 0.15 + 0.11 = 0.26 ..(i)

Clearly A∩B = {E₈} 

∴ P(A∩B) = P(E₈) = 0.07 (i) 

Using addition law of probability we know that

P(A∪B) = P(A) + P(B) – P(A∩B) 

⇒ P(A∪B) = 0.26 + 0.21 – 0.07 

∴ P(A ∪ B) = 0.4 (ii)

As, A ∪ B = {E₁, E₅, E₈} ∪ {E₂, E₈, E₉} 

⇒ A ∪ B = {E₁, E₅, E₈, E₂, E₉} 

∴ P(A∪B) = P(E₁)+ P(E₅)+ P(E₈)+ P(E₂)+ P(E₉) 

⇒ P(A∪B) = 0.08 + 0.11 + 0.07 + 0.08 + 0.07 = 0.41 

∴ P(A∪B) = 0.41 (iii) 

As , P(B) = 0.21 

∴ P(B’) = 1 – 0.21 = 0.79 (iv)

Calculation of P(B’) using sets –

B’ = {E₁, E₃, E₄, E₅, E₆, E₇} or B’ = E₁∪ E₃∪ E₄∪ E₅∪ E₆∪E₇ 

∴ P(B’) = P(E₁) + P(E₃) + P(E₄) + P(E₅) + P(E₆) + P(E₇) 

P(B’) = 0.08 + 0.1 + 0.1 + 0.11 + 0.2 + 0.2 

= 0.79 (iv) 

Clearly through both the ways we get the same answer.