Let there be three independent events E₁, E₂ and E₃. The probability that only E₁ occurs is α, only E₂ occurs is β and only E₃ occurs is γ. Let 'p' denote the probability of none of events occurs that satisfies the equations
(α - 2β)p = αβ and (β - 3γ)p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1).
Then, (Probability of occurrence of E₁)/(Probability of occurrence of E₃) is equal to _____________.

Solution

Maths sir · 5 steps

Write every probability as p times the odds of one event, and each equation shrinks to a one-liner.

1. Name the probabilities. Let P(E₁) = x, P(E₂) = y, P(E₃) = z. By independence, p = (1-x)(1-y)(1-z) and α = x(1-y)(1-z). Likewise β = (1-x)y(1-z) and γ = (1-x)(1-y)z.
2. Odds form. Compared with p, each 'only' probability swaps one factor (1-x) for x. So α = ap, β = bp, γ = cp, where a = x/(1-x), b = y/(1-y), c = z/(1-z).
3. First equation. (ap - 2bp)p = ab p² gives a - 2b = ab. Divide by ab: 1/b - 2/a = 1. Since 1/a = 1/x - 1 and 1/b = 1/y - 1, this becomes 1/y = 2/x, so y = x/2.
4. Second equation. (bp - 3cp)p = 2bc p² gives b - 3c = 2bc. Divide by bc: 1/c - 3/b = 2. In terms of y and z this is 1/z = 3/y, so z = y/3.

   z = y/3 = x/6

5. So. P(E₁)/P(E₃) = x/z = 6.

The common slip: 0.17. Writing the ratio upside down, as P(E₃)/P(E₁), gives 1/6 ≈ 0.17 instead of 6.

The idea to keep. With independent events, 'only Eᵢ' equals p × P(Eᵢ)/(1 - P(Eᵢ)). Dividing such equations by the product turns them into simple relations between 1/P(Eᵢ).

Answer from the NTA final answer key. Solution by PrepWiser.