Question

Let B_i(i = 1, 2, 3) be three independent events in a sample space. The probability that only B₁ occur is α, only B₂ occurs is β and only B₃ occurs is γ. Let p be the probability that none of the events B_i occurs and these 4 probabilities satisfy the equations (α – 2β)p = αβ and (β – 3γ) = 2βy (All the probabilities are assumed to lie in the interval (0, 1)). Then `("P"("B"_1))/("P"("B"_3))` is equal to ______.

पर्याय

• 3

• 4

• 5

• 6

Answer 1

Let B_i(i = 1, 2, 3) be three independent events in a sample space. The probability that only B₁ occur is α, only B₂ occurs is β and only B₃ occurs is γ. Let p be the probability that none of the events B_i occurs and these 4 probabilities satisfy the equations (α – 2β)p = αβ and (β – 3γ) = 2βy (All the probabilities are assumed to lie in the interval (0, 1)). Then `("P"("B"_1))/("P"("B"_3))` is equal to 6.

Explanation:

Given: B₁, B₂, B₃ are three independent events in a sample space.

P(only B₁ occurs) = α

P(only B₂ occurs) = β

P(only B₃ occurs) = γ

P(none of events B_i occurs) = p

(α – 2β)p = αβ

Let x, y, z be probability of B₁, B₂, B₃ respectively.

⇒ x(1 – y)(1 – z) = α  ...(i)

y(1 – x)(1 – z) = β  ...(ii)

z(1 – x)(1 – y) = γ  ...(iii)

(1 – x)(1 – y)(1 – z) = p  ...(iv)

Since, (α – 2β)p = αβ

⇒ x(1 – y)(1 – z) – 2y(1 – x)(1 – z)[(1 – x)(1 – y)(1 – z)] = xy(1 – x)(1 – y)(1 – z)²

⇒ x – xy – 2y + 2xy = xy

⇒ x = 2y  ...(v)

Similarly, (β – 3r)p = 2βr

⇒ y = 3z  ...(vi)

From (v) and (vi)

x = 6z

⇒ `x/z` = 6

⇒ `("P"("B"_1))/("P"("B"_3))` = 6