AP EAMCET · Maths · Probability
If \(\mathrm{E}\) and \(\mathrm{F}\) are events such that \(P(\bar{F})=0.7\) and \(P(E \cap F)=0.2\), then \(\mathrm{P}(E \mid F)\) is
- A \(2 / 3\)
- B \(1 / 3\)
- C \(3 / 4\)
- D \(1 / 4\)
Answer & Solution
Correct Answer
(A) \(2 / 3\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(\overline{\mathrm{F}})=0.7\) and \(\mathrm{PE} \mathrm{nF})=0.2\) \(P\left(\frac{E}{F}\right)=\frac{P(E n F)}{P(F)}\) Here, \(\mathrm{P}(\mathrm{F})=1-\mathrm{P}(\overline{\mathrm{F}})\) \[ =1-0.7=0.3 \] \[ P\left(\frac{E}{F}\right)=\frac{0.2}{0.3}=\frac{2}{3} \]
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