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Equally Likely Events

If P A ∩ B =7...

Question

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

\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 $\frac{14}{17}$
$P (A | B) = \frac{P (A \cap B)}{P (B)}$
$= \frac{\frac{7}{10}}{\frac{17}{20}} = \frac{14}{17}$

Suggest Corrections

Similar questions

P (A \cap B) = 7 / 10

P (B) = 17 / 20

P

P (A | B)

P (A) = \frac{4}{5}, P (B^{'}) = \frac{2}{5}

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

P (A \cap B^{'})

P (A) = 0.7, P (B) = 0.4, P (A \cap B) = 0.3

P (A \cap B^{'})

P (A \cup B) = \frac{4}{5}, P (¯ ¯¯ ¯ A \cup ¯ ¯¯ ¯ B) = \frac{7}{10}

P (B) = \frac{2}{5}

P (A)

P (A \cup B) = P (A) + P (B) - P (A \cap B)

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (A \cap C) + P (A \cap B \cap C)

\frac{2}{5}

\frac{4}{7}

\frac{2}{3}

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