JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 22)
Let Bi (i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is $$\alpha $$, only B2 occurs is $$\beta $$ and only B3 occurs is $$\gamma $$. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations $$\left( {\alpha - 2\beta } \right)p = \alpha \beta $$ and $$\left( {\beta - 3\gamma } \right)p = 2\beta \gamma $$ (All the probabilities are assumed to lie in the interval (0, 1)).
Then $${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$$ is equal to ________.
Then $${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$$ is equal to ________.
Answer
6
Explanation
Let x, y, z be probability of B1, B2, B3 respectively
$$\alpha $$ = P(B1 $$ \cap $$ $$\overline {{B_2}} \cap \overline {{B_3}} $$) = $$P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$$
$$ \Rightarrow $$ x(1 $$-$$ y)(1 $$-$$ z) = $$\alpha$$
Similarly, y(1 $$-$$ x)(1 $$-$$ z) = $$\beta$$
z(1 $$-$$ x)(1 $$-$$ y) = $$\gamma$$
and (1 $$-$$ x)(1 $$-$$ y)(1 $$-$$ z) = p
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$
(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 ($$\beta$$ $$-$$ 3$$\gamma $$)p = 2$$\beta$$$$\gamma $$
$$ \Rightarrow $$ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
$${x \over z} = 6$$
$$\alpha $$ = P(B1 $$ \cap $$ $$\overline {{B_2}} \cap \overline {{B_3}} $$) = $$P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$$
$$ \Rightarrow $$ x(1 $$-$$ y)(1 $$-$$ z) = $$\alpha$$
Similarly, y(1 $$-$$ x)(1 $$-$$ z) = $$\beta$$
z(1 $$-$$ x)(1 $$-$$ y) = $$\gamma$$
and (1 $$-$$ x)(1 $$-$$ y)(1 $$-$$ z) = p
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$
(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 ($$\beta$$ $$-$$ 3$$\gamma $$)p = 2$$\beta$$$$\gamma $$
$$ \Rightarrow $$ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
$${x \over z} = 6$$
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