Describe disjoint events
Describe a conditional probability
Define independent events
Law of Total Probability
Baye’s Theorem
Two events A and B are considered disjoint if \(P(A\cap B)=0\). In general terms, only one event can occur, not both.
Let there be 2 events A and B. Given that B has occurred, what is the probability that A occurs? The conditional probability requires there to be at least 2 events and one event must have occurred. Additionally, the events cannot be disjoint. Conditional probabilities are denoted as \(P(A|B)\), the probability of A given B has occurred.
\[ P(A|B)=\frac{P(A\cap B)}{P(B)} \]
| Uses Eye Glasses | ||
| Needs Glasses | Yes | No |
| Yes | 44 | 14 |
| No | 2 | 40 |
Find the probability of needing glasses
Find the probability of not using glasses
Find the probability of not using glasses and needing glasses
Find the probability of not using glasses, given they need glasses
Events A and B are considered independent if \(P(A\cap B)=P(A)P(B)\). In other words, The occurrence of one event will not have an effect on the occurrence of the other event.
| Uses Eye Glasses | ||
|---|---|---|
| Needs Glasses | Yes | No |
| Yes | 44 | 14 |
| No | 2 | 40 |
The law of total probability allows you to compute the probability of an event A given that the sets {\(B_1\), … , \(B_n\)} partitions event A. The law of total probability is given as
\[ P(A)= \sum^n_{i=1}P(A\cap B_i) \]
\[ P(A)=\sum^n_{i=1}P(A|B_i)P(B_i) \]
The probability of an individual having a disease, given that they test positive for the disease, is 0.82. The probability of an individual having a disease, given they tested negative, is 0.14. The probability of testing positive is 0.6. What is the prevalence of a disease (probability of having a disease)?
Baye’s theorem computes the probability of an event \(B_i\) given event A
\[ P(B_i|A) = \frac{P(A\cap B_i)}{P(A)}=\frac{P(A|B_i)P(B_i)}{\sum^n_{i=1}P(A|B_i)P(B_i)} \]
The probability of an having a disease, given that they test positive for the disease, is 0.82. The probability of and individual having a disease, given they tested negative, is 0.14. The probability of testing positive is 0.6. What is the probability of resulting in a false negative?