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

$p + q = 1$

 

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