JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 16)
Let there be three independent events E1, E2 and E3. The probability that only E1 occurs is $$\alpha$$, only E2 occurs is $$\beta$$ and only E3 occurs is $$\gamma$$. Let 'p' denote the probability of none of events occurs that satisfies the equations
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ and ($$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $$\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $$ is equal to _____________.
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ and ($$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $$\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $$ is equal to _____________.
Answer
6
Explanation
Let P(E1) = x, P(E2) = y and P(E3) = z
$$\alpha $$ = P$$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$$ = $$P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$$
$$ \Rightarrow $$ $$\alpha $$ = x(1 $$-$$ y) (1 $$-$$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$$\gamma $$ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
$${x \over {1 - x}} = {\alpha \over p}$$
$$ \Rightarrow $$ x = $${\alpha \over {\alpha + p}}$$
From (iii) and (iv)
$${z \over {1 - z}} = {\gamma \over p}$$
$$ \Rightarrow $$ z = $${\gamma \over {\gamma + p}}$$
$$ \therefore $$ $${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$$ $$ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$$ ..(v)
Also given,
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ $$ \Rightarrow $$ $$\alpha $$p = ($$\alpha $$ + 2p)$$\beta $$ ....(vi)
$$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$ $$ \Rightarrow $$ 3$$\gamma $$p = (p - 2$$\gamma $$)$$\beta $$ .....(vii)
From (vi) and (vii),
$${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$$
$$ \Rightarrow $$ p$$\alpha $$ - 6p$$\gamma $$ = 5$$\gamma $$$$\alpha $$
$$ \Rightarrow $$ $${p \over \gamma } - {{6p} \over \alpha } = 5$$
$$ \Rightarrow $$ $${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$$ ....(viii)
Now from (v) and (viii),
$${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$$ = 6
$$\alpha $$ = P$$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$$ = $$P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$$
$$ \Rightarrow $$ $$\alpha $$ = x(1 $$-$$ y) (1 $$-$$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$$\gamma $$ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
$${x \over {1 - x}} = {\alpha \over p}$$
$$ \Rightarrow $$ x = $${\alpha \over {\alpha + p}}$$
From (iii) and (iv)
$${z \over {1 - z}} = {\gamma \over p}$$
$$ \Rightarrow $$ z = $${\gamma \over {\gamma + p}}$$
$$ \therefore $$ $${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$$ $$ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$$ ..(v)
Also given,
($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ $$ \Rightarrow $$ $$\alpha $$p = ($$\alpha $$ + 2p)$$\beta $$ ....(vi)
$$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$ $$ \Rightarrow $$ 3$$\gamma $$p = (p - 2$$\gamma $$)$$\beta $$ .....(vii)
From (vi) and (vii),
$${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$$
$$ \Rightarrow $$ p$$\alpha $$ - 6p$$\gamma $$ = 5$$\gamma $$$$\alpha $$
$$ \Rightarrow $$ $${p \over \gamma } - {{6p} \over \alpha } = 5$$
$$ \Rightarrow $$ $${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$$ ....(viii)
Now from (v) and (viii),
$${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$$ = 6
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