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STD 12 - 13. probability — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 13. probability4 Marks
MCQ
Let there be three independent events $E _{1}, E _{2}$ and $E _{3}$. The probability that only $E _{1}$ occurs is $\alpha$, only $E _{2}$ occurs is $\beta$ and only $E _{3}$ 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, $\frac{\text { Probability of occurrence of } E _{1}}{\text { Probability of occurrence of } E _{3}}$ is equal to ..........

  • A
    $8$
  • ✓
    $6$
  • C
    $3$
  • D
    $9$
✓

Answer

Correct option: B.
$6$
b
Let $P \left( E _{1}\right)= P _{1} ; P \left( E _{2}\right)= P _{2} ; P \left( E _{3}\right)= P _{3}$

$P \left( E _{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)=\alpha= P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots$

$P \left(\overline{ E }_{1} \cap E _{2} \cap \overline{ E }_{3}\right)=\beta=\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)$

$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap E _{3}\right)=\gamma=\left(1- P _{1}\right)\left(1- P _{2}\right) P _{3} \ldots \ldots$

$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)= P =\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots$

Given that, $(\alpha-2 \beta) P =\alpha \beta$

$\Rightarrow\left( P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right)-2\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)\right) P = P _{1} P _{2}$

$\quad\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right)^{2}$

$\Rightarrow\left( P _{1}\left(1- P _{2}\right)-2\left(1- P _{1}\right) P _{2}\right)= P _{1} P _{2}$

$\Rightarrow\left( P _{1}- P _{1} P _{2}-2 P _{2}+2 P _{1} P _{2}\right)= P _{1} P _{2}$

$\Rightarrow P _{1}=2 P _{2} \quad \ldots \ldots(1)$

and similarly, $(\beta-3 \gamma) P =2 B \gamma$

$P _{2}=3 P _{3} \quad \ldots \ldots(2)$

So, $P _{1}=6 P _{3} \Rightarrow \frac{ P _{1}}{ P _{3}}=6$

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