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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 13. probability4 Marks
MCQ
Let $E ^{ C }$ denote the complement of an event $E$. Let $E _{1}, E _{2}$ and $E _{3}$ be any pairwise independent events with $P \left( E _{1}\right) > 0$ and $P \left( E _{1} \cap E _{2} \cap E _{3}\right)=0$ Then $P \left( E _{2}^{ C } \cap E _{3}^{ C } / E _{1}\right)$ is equal to
  • ✓
    $\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}\right)$
  • B
    $\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right)$
  • C
    $\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)$
  • D
    $\mathrm{P}\left(\mathrm{E}_{3}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)$
✓

Answer

Correct option: A.
$\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}\right)$
a
Given $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$ are pairwise indepedent events $\operatorname{soP}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E}_{2}\right)$

and $\mathrm{P}\left(\mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E}_{3}\right)$

and $\mathrm{P}\left(\mathrm{E}_{3} \cap \mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{E}_{1}\right)$

$\& \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=0$

Now $\mathrm{P}\left(\frac{\overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}}{\mathrm{E}_{1}}\right)=\frac{\mathrm{P}\left[\mathrm{E}_{1} \cap\left(\overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}\right)\right]}{\mathrm{P}\left(\mathrm{E}_{1}\right)}$

$=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right)-\left[\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{3}\right)-\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)\right]}{\mathrm{P}\left(\mathrm{E}_{1}\right)}$

$=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right)-\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E}_{2}\right)-\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{3}\right)-0}{\mathrm{P}\left(\mathrm{E}_{1}\right)}$

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

$=\left[1-\mathrm{P}\left(\mathrm{E}_{3}\right)\right]-\mathrm{P}\left(\mathrm{E}_{2}\right)$

$=\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{c}}\right)-\mathrm{P}\left(\mathrm{E}_{2}\right)$

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