Let there be three independent events and . The probability that only occurs is , only occurs is and only occurs is . Let denote the probability of none of events occurs that satisfies the equation and . All the given probabilities are assumed to l

Question

Let there be 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 $' p'$ denote the probability of none of events occurs that satisfies the equation $(α - 2 β) p = α β$ and $(β - 3 γ) p = 2 β γ$ . All the given probabilities are assumed to lie in the interval (0, 1).

Then $\frac{probability of occurrence of E_{1}}{probability of occurrence of E_{3}}$ is equal to

Smart Achievers · Accepted Answer

(4)

$Let p (E_{1}) = x, p (E_{2}) = y and p (E_{3}) = z$

$α = p (E_{1} \cap \bar{E_{2}} \cap \bar{E_{3}}) = p (E_{1}) \cdot p (\bar{E_{2}}) \cdot p (\bar{E_{3}})$

$\Rightarrow α = x (1 - y) (1 - z)$ $\dots \dots (i)$

$Similarly,$

$β = (1 - x) y (1 - z)$ $\dots \dots (i i)$

$γ = (1 - x) (1 - y) z$ $\dots \dots (i i i)$

$p = (1 - x) (1 - y) (1 - z)$ $\dots \dots (i v)$ $and solve equation.$

Solution

1 9

2 3

3 7

4 6

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Solution

1 9

2 3

3 7

4 6

Ans. (4) Let p(E1)=x, p(E2)=y and p(E3)=z α=p(E1∩E2¯∩E3¯)=p(E1)·p(E2¯)·p(E3¯) ⇒α=x(1-y)(1-z) ……(i) Similarly, β=(1-x)y(1-z) ……(ii) γ=(1-x)(1-y)z ……(iii) p=(1-x)(1-y)(1-z) ……(iv) and solve equation.

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