Question

An urn contains four tickets marked with numbers 112, 121, 122, 222 and one ticket is drawn at random. Let A_i (i = 1, 2, 3) be the event that i^th digit of the number of the ticket drawn is 1. Discuss the independence of the events A₁, A₂, and A₃.

Answer 1

One ticket can be drawn out of 4 tickets in ⁴C₁ = 4 ways.
∴ n(S) = 4
According to given information,
Let A₁ be the event that 1^st digit of the number of ticket is 1
A₂ be the event that 2^nd digit of the number of ticket is 1.
A₃ be the event that 3^rd digit of the number of ticket is 1.
∴ A₁ = {112, 121, 122}, A₂ = {112}, A3 = {121}

∴ `"P"("A"_1) = ("n"("A"_1))/("n"("S")) = 3/4`,

`"P"("A"_2) = ("n"("A"_2))/("n"("S")) = 1/4`,

`"P"("A"_3) = ("n"("A"_3))/("n"("S")) = 1/4`

`{:("P"("A"_1) "P"("A"_2) = 3/16),("P"("A"_2) "P"("A"_3) = 1/16),("P"("A"_1) "P"("A"_3) = 3/16):}}`   ...(i)

A₁ ∩ A₂ = {112}, A₂ ∩ A₃ = Φ, A₁ ∩ A₃ = {121}

`{:("P"("A"_1 ∩ "A"_2) = ("n"("A"_1 ∩ "A"_2))/("n"("S")) = 1/4),("P"("A"_2 ∩ "A"_3) = 0),("P"("A"_1 ∩ "A"_3) = 1/4):}}`   ...(ii)

∴ From (i) and (ii),
`{:("P"("A"_1)*"P"("A"_2) ≠ "P"("A"_1 ∩ "A"_2)),("P"("A"_2)*"P"("A"_3) ≠ "P"("A"_2 ∩ "A"_3)),("P"("A"_1)*"P"("A"_3) ≠ "P"("A"_1 ∩ "A"_3)):}}`   ...(iii)

∴ A₁, A₂, A₃ are not pairwise independent
For mutual independent of events A₁, A₂, A₃ We require to have
P(A₁ ∩ A₂ ∩ A₃) = P(A₁) P(A₂) P(A₃)
and P(A₁) P(A₂) = P(A₁ ∩ A₂),
P(A₂) P(A₃) = P(A₂ ∩ A₃),
P(A₁) P(A₃) = P(A₁ ∩ A₃)
∴ From (iii),
A₁, A₂, A₃ are not mutually independent.