Let there be three independent events $\mathrm{E}_{1}, \mathrm{E}_{2}$ and $\mathrm{E}_{3}$. The probability that only $\mathrm{E}_{1}$ occurs is $\alpha$, only $E_{2}$ occurs is $\beta$ and only $E_{3}$ occurs is $\gamma$. Let ' $p$ ' denote the probability of none of events occurs that satisfies the equations $(\alpha-2 \beta) \mathrm{p}=\alpha \beta$ and $(\beta-3 \gamma) \mathrm{p}=2 \beta \gamma$. All the given probabilities are assumed to lie in the interval $(0,1)$. Then, $\frac{\text { Pr obability of occurrence of } E_{1}}{\text { Pr obability of occurrence of } E_{3}}$ is equal to.....

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Probability

Let there be three independent events

E_{1}, E_{2}

and

E_{3}

. The probability that only

E_{1}

occurs is

α

, only

E_{2}

occurs is

β

and only

E_{3}

occurs is

γ

. Let '

p

' denote the probability of none of events occurs that satisfies the equations

(α - 2 β) p = α β

and

(β - 3 γ) p = 2 β γ

. All the given probabilities are assumed to lie in the interval

(0, 1)

.
Then,

\frac{Pr obability of occurrence of E _{1}}{Pr obability of occurrence of E _{3}}

is equal to.....

Answer: 6

Write

p_{i} = P (E_{i})

and

q_{i} = 1 - p_{i}

. Independence gives

α = p_{1} q_{2} q_{3}

β = q_{1} p_{2} q_{3}

γ = q_{1} q_{2} p_{3}

and

p = q_{1} q_{2} q_{3}

, so

\frac{α}{p} = \frac{p _{1}}{q _{1}}, \frac{β}{p} = \frac{p _{2}}{q _{2}}, \frac{γ}{p} = \frac{p _{3}}{q _{3}} .

Put

A = \frac{p _{1}}{q _{1}}

B = \frac{p _{2}}{q _{2}}

and

C = \frac{p _{3}}{q _{3}}

. Dividing

(α - 2 β) p = α β

p^{2}

gives

A - 2 B = A B

, so

A = \frac{2 B}{1 - B}

; dividing

(β - 3 γ) p = 2 β γ

p^{2}

gives

B - 3 C = 2 B C

, so

C = \frac{B}{3 + 2 B}

. Since

p_{1} = \frac{A}{1 + A}

and

p_{3} = \frac{C}{1 + C}

p_{1} = \frac{2 B}{1 + B}, p_{3} = \frac{B}{3 ( 1 + B )} .

The free parameter

B

cancels, and

\frac{p _{1}}{p _{3}} = 6

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer