# Conditional Probability

- $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
- $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$

## Find the probability that somebody is healthy given that they have positive test result?

**Problem**

A disease is known to affect 1 in 10,000 people. A screening test for the disease shows a positive result for 99% of the people with the disease. The test also shows positive for 2% of people who do not have the disease. Find the probability that somebody is healthy given that they have positive test result?

A. 99.51% | C. 46.32% |

B. 12.32% | D. 78.36% |

**Problem**

The local weather forecaster says “no rain” and his record is 2/3 accuracy of prediction. But the Federal Meteorological Service predicts rain and their record is 3/4. With no other data available, what is the chance of rain?

A. 3/5 | C. 1/6 |

B. 1/4 | D. 5/12 |

## Probability

**Probability**

For outcomes that are equally likely to occur:

$P = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

If the probability of an event to happen is *p* and the probability for it to fail is *q*, then

$p + q = 1$

- Read more about Probability
- Add new comment
- 12102 reads