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Quantitative Aptitude

Dependent Events

Of the three ...

Question

Of the 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 the probability $p$ that none of events $E_{1}, E_{2}$ or $E_{3}$ occurs satisfy 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 $\frac{{Probability of occurence of E}_{1}}{{Probability of occurence of E}_{3}}$

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Solution

$P (E_{1}) P ({¯ ¯¯ ¯ E}_{2}) P ({¯ ¯¯ ¯ E}_{3}) = α . . . . (i)$
$P ({¯ ¯¯ ¯ E}_{1}) P (E_{2}) P ({¯ ¯¯ ¯ E}_{3}) = β . . . . (i i)$
$P ({¯ ¯¯ ¯ E}_{1}) P ({¯ ¯¯ ¯ E}_{2}) P (E_{3}) = γ . . . . (i i)$
$P ({¯ ¯¯ ¯ E}_{1}) P ({¯ ¯¯ ¯ E}_{2}) P ({¯ ¯¯ ¯ E}_{3}) = ρ . . . . (i v)$
Divide (i) by (iv),
$\frac{P (E_{1})}{P (E_{1})} = \frac{α}{ρ} = \frac{α}{\frac{α β}{α - β}} = \frac{α - 2 β}{β}$

$\Rightarrow \frac{p (_{1})}{P (E_{1})} = \frac{β}{α - 2 β}$

$\Rightarrow \frac{1 - P (E_{1})}{P (E_{1})} = \frac{β}{α - 2 β}$

$\Rightarrow \frac{1}{P (E_{1})} - 1 = \frac{β}{α - 2 β}$

$\Rightarrow \frac{1}{P (E_{1})} = \frac{α - β}{α - 2 β}$

$\Rightarrow P (E_{1}) = \frac{α - 2 β}{α - β} . . . . (v)$

Now, $\frac{α β}{α - 2 β} = \frac{2 β γ}{β - 3 γ}$

$\Rightarrow \frac{α}{α - 2 β} = \frac{2 γ}{β - 3 γ}$

$\Rightarrow α β - 3 γ α = 2 γ α - 4 β γ$

$\Rightarrow α β = 5 γ α - 4 β γ$

$γ = \frac{α β}{5 α - 4 β} . . . . (v i)$

Divide (iii) by (iv)

$\frac{P (E_{3})}{P ({¯ ¯¯ ¯ E}_{3})} = \frac{γ}{ρ} = \frac{γ}{\frac{α β}{α - 2 β}} = \frac{γ (α - 2 β)}{α β}$

$\frac{P (E_{3})}{P ({¯ ¯¯ ¯ E}_{3})} = \frac{α - 2 β}{5 α - 4 β}$

$\frac{P (E_{3})}{P ({¯ ¯¯ ¯ E}_{3})} = \frac{5 α - 4 β}{α - 2 β}$

$\frac{1 - P (E_{3})}{P (E_{3})} = \frac{5 α - 4 β}{α - 2 β}$

$\frac{1}{P (E_{3})} = \frac{6 α - 6 β}{α - 2 β} = \frac{6 (α - β)}{α - 2 β}$

$P (E_{3}) = \frac{α - 2 β}{6 (α - β)}$

$\frac{P (E_{1})}{P (E_{3})} = \frac{\frac{α - 2 β}{α - β}}{\frac{α - 2 β}{6 (α - β)}} = 6$

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