If $P (A \cap B) = 7 / 10$ and $P (B) = 17 / 20$ , where $P$ stands for probability then $P (A | B)$ is equal tp

\frac{7}{8}

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\frac{17}{20}

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\frac{14}{17}

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\frac{1}{8}

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Solution

The correct option is C $\frac{14}{17}$
Using conditional property formula $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$

$= \frac{\frac{7}{10}}{\frac{17}{20}}$ , substitute the given values

$= \frac{14}{17}$ ,

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Similar questions

P (A \cap B) = \frac{7}{10}

P (B) = \frac{17}{20},

P

P (A | B)

If A and B are two events such that $P (A) = \frac{2}{5}$ and $P (A \cap B) = \frac{3}{20}$ , then $P (A | (A^{'} \cup B^{'}))$ is equal to

P (A) = \frac{1}{2}

P (B) = \frac{3}{8}

P (A \cap B) = \frac{1}{5}

\begin{matrix} 2 & 14 & 17 & 11 & 15 7 & 20 & 8 & 17 & 10 8 & 20 & 1 & 8 & 3 \end{matrix}

P (A) = 1 / 5, P (B) = 1 / 6;

P (A \cap C) = 1 / 20. P (B \cup C) = 3 / 8

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