Conditional probability: Difference between revisions

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=== Example ===
When [[Morse code]] is transmitted, there is a certain probability that the "dot" or "dash" that was received is erroneous. This is often taken as interference in the transmission of a message. Therefore, it is important to consider when sending a "dot", for example, the probability that a "dot" was received. This is represented by: <math>P(dot \text{dot sent \mid} dot| \text{ dot received}) = P(\text{dot \ received \mid} dot| \text{ dot sent}) \frac{P(\text{dot \ sent})}{P(\text{dot \ received})}.</math> In Morse code, the ratio of dots to dashes is 3:4 at the point of sending, so the probability of a "dot" and "dash" are <math>P(\text{dot \ sent}) = \frac {3}{7} \ and \ P(\text{dash \ sent}) = \frac {4}{7}</math>. If it is assumed that the probability that a dot is transmitted as a dash is 1/10, and that the probability that a dash is transmitted as a dot is likewise 1/10, then Bayes's rule can be used to calculate <math>P(\text{dot \ received})</math>.
 
<math>P(\text{dot \ received}) = P(\text{dot \ received \ } \cap \text{ dot \ sent }) + P(\text{dot \ received \} \cap \text{ dash \ sent})</math>
 
<math>P(\text{dot \ received}) = P(\text{dot \ received } \mid \text{ dot \ sent})P(\text{dot \ sent}) + P(\text{dot \ received } \mid \text{ dash \ sent})P(\text{dash \ sent})</math>
 
<math>P(\text{dot \ received}) = \frac{9}{10}\times\frac{3}{7} + \frac{1}{10}\times\frac{4}{7} = \frac{31}{70}</math>
 
Now, <math>P(\text{dot \ sent } \mid dot \text{ dot received})</math> can be calculated:
 
<math>P(\text{dot \ sent } \mid dot \text{ dot received}) = P(\text{dot \ received } \mid dot \text{ dot sent}) \frac{P(\text{dot \ sent})}{P(\text{dot \ received})} = \frac{9}{10}\times \frac{\frac{3}{7}}{\frac{31}{70}} = \frac{27}{31}</math><ref>{{Cite web|title=Conditional Probability and Independence|url=http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture4.pdf|access-date=2021-12-22}}</ref>
 
== Statistical independence ==