JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Probability

Let

B_{i} (i = 1, 2, 3)

be three independent events in a sample space. The probability that only

B_{1}

occur is

α

, only

B_{2}

occurs is

β

and only

B_{3}

occurs is

γ

. Let

p

be the probability that none of the events

B_{i}

occurs and these 4 probabilities satisfy the equations

(α - 2 β) p = α β

and

(β - 3 γ) p = 2 β γ

(All the probabilities are assumed to lie in the interval

(0, 1)

). Then

\frac{P ( B _{1} )}{P ( B _{3} )}

is equal to ______.

Answer: 6

Solution

Let

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

for

i = 1, 2, 3

. Since events are independent:

From

(α - 2 β) p = α β

α p - 2 β p = α β

α (p - β) = 2 β p

Substituting and simplifying:

p_{1} (1 - p_{2}) (1 - p_{3}) \cdot (1 - p_{1}) (1 - p_{2}) (1 - p_{3}) - 2 (1 - p_{1}) p_{2} (1 - p_{3}) \cdot (1 - p_{1}) (1 - p_{2}) (1 - p_{3}) = p_{1} (1 - p_{2}) (1 - p_{3}) \cdot (1 - p_{1}) p_{2} (1 - p_{3})

After algebraic manipulation using both equations, we get:

\frac{p _{1}}{p _{2}} = 3

and

\frac{p _{2}}{p _{3}} = 2

Therefore:

\frac{P ( B _{1} )}{P ( B _{3} )} = \frac{p _{1}}{p _{3}} = \frac{p _{1}}{p _{2}} \cdot \frac{p _{2}}{p _{3}} = 3 \times 2 = 6

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer