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13. probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths13. probability1 Mark
MCQ
Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P \left( E _{1} \mid E _{2}\right)=\frac{1}{2}$, $P \left( E _{2} \mid E _{1}\right)=\frac{3}{4}$ and $P \left( E _{1} \cap E _{2}\right)=\frac{1}{8}$. Then
  • A
    $P \left( E _{1} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
  • B
    $P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)= P \left( E _{1}^{\prime}\right) \cdot P \left( E _{2}\right)$
  • ✓
    $P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
  • D
    $P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
✓

Answer

Correct option: C.
$P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
c
(A) $P \left( E _{1}\right) \cdot P \left( E _{2}\right)=\frac{1}{6} \cdot \frac{1}{4}=\frac{1}{24} \neq P \left( E _{1} \cap E _{2}\right)$

(B) $P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)=1- P \left( E _{1} \cup E _{2}\right)$

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

$=1-\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{8}\right)=\frac{17}{24}$

$P \left( E _{1}^{\prime}\right) P \left( E _{2}\right)=\frac{5}{6} \times \frac{1}{4}=\frac{5}{24}$

(C) $P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right)- P \left( E _{1} \cap E _{2}\right)=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}$

(D) $P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{2}\right)- P \left( E _{1} \cap E _{2}\right)=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$

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