Verified Solution • 2025–26 Session moderate

QJEEM21Ma9804: JEE Main 2021 Mathematics - Class 12 JEEM21Ma9804 Solution: Let B i (i = 1, 2, 3) be three independent events in a sample…

Let Bi (i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is α\alpha , only B2 occurs is β\beta and only B3 occurs is γ\gamma . Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α−2β)p=αβ\left( {\alpha - 2\beta } \right)p = \alpha \beta and (β−3γ)p=2βγ\left( {\beta - 3\gamma } \right)p = 2\beta \gamma (All the probabilities are assumed to lie in the interval (0, 1)).
Then P(B1)P(B3){{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}} is equal to ________.

Step-by-Step Verified Solution

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

α\alpha = P(B1 ∩\cap B2‾∩B3‾\overline {{B_2}} \cap \overline {{B_3}} ) = P(B1)P(B2‾)P(B3‾)P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)

⇒\Rightarrow x(1 −- y)(1 −- z) = α\alpha

Similarly, y(1 −- x)(1 −- z) = β\beta

z(1 −- x)(1 −- y) = γ\gamma

and (1 −- x)(1 −- y)(1 −- z) = p

( α\alpha −- 2 β\beta )p = α\alpha β\beta

(x(1 −- y)(1 −- z) −- 2y(1 −- x)(1 −- z)) (1 −- x)(1 −- y)(1 −- z) = xy(1 −- x)(1 −- y)(1 −- z)

x −- xy −- 2y + 2xy = xy

x = 2y ...... (1)

Similarly ( β\beta −- 3 γ\gamma )p = 2 β\beta γ\gamma

⇒\Rightarrow y = 3z .... (2)

From (1) & (2)

x = 6z

Now

xz=6{x \over z} = 6

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