1 / 19

Understanding Conditional Probability Through Color Sequences

This example explores conditional probability using color sequences to compute probabilities of events. We analyze the probability of "blue" appearing first given that it appears fourth, the joint events of colors, and prior probabilities. The framework relies on the relationship between joint probability and prior probability to derive conditional probabilities, illustrated through specific color combinations. Additionally, we compute probabilities for different scenarios involving "green" and "yellow" shows. The example highlights critical concepts in probability theory through practical applications.

urbana
Télécharger la présentation

Understanding Conditional Probability Through Color Sequences

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Conditional Probability – A Color Sequence Example

  2. Compute Pr{ blue shows 1st | blue shows 4th }

  3. Pr{ blue shows 1st | blue shows 4th} Identify Prior Event “Blue Shows 4th “

  4. Pr{ blue shows 1st | blue shows 4th}; Compute Prior Probability Pr{ Blue Shows 4th } = Pr{ exactly one of BBBB, BGGB, RYYB shows } = Pr{BBBB} + Pr{BGGB} + Pr{RYYB} = .10+.25+.15 = .50

  5. Pr{ blue shows 1st | blue shows 4th} Identify Joint Probability “Blue Shows 1st and 4th “

  6. Pr{ blue shows 1st | blue shows 4th} Compute Joint Probability Pr{ Blue Shows 1st and 4th } = Pr{ exactly one of BBBB, BGGB shows } = Pr{BBBB} + Pr{BGGB} = .10+.25 = .35

  7. Conditional Probability = Joint Probability / Prior Probability So, Pr{ Blue Shows 1st | Blue Shows 4th } = Pr{ Blue Shows 1st and 4th }/Pr{Blue Shows 4th } = .35/ .50= .70

  8. Compute Pr{ green shows 2nd or 3rd | yellow shows }

  9. Pr{ green shows 2nd or 3rd | yellow shows }; Identify Prior Event “Yellow Shows”

  10. Pr{ green shows 2nd or 3rd | yellow shows } Compute Prior Probability Pr{ Yellow Shows } = Pr{ exactly one of YYYY, BYRG, RYYB shows } = Pr{YYYY}+ Pr{BYRG}+ Pr{RYYB} = .30+.15+.15 = .60

  11. Identify Joint Event “Green Shows 2nd or 3rd and Yellow Shows” No sequences meet this requirement, so Pr{“Green Shows 2nd or 3rd and Yellow Shows”} = 0

  12. Conditional Probability = Joint Probability / Prior Probability So Pr{ Green Shows 2nd or 3rd | Yellow shows } = Pr{ Green Shows 2nd or 3rd and Yellow shows }/ Pr{Yellow shows } = 0 / .60= 0

  13. Compute Pr{ yellow shows | red shows }

  14. Pr{ yellow shows | red shows } Identify Prior Event “Red Shows”

  15. Compute Prior Probability Pr{ Red Shows } = Pr{ exactly one of RGGR, BYRG, RYYB shows } = Pr{RGGR} + Pr{BYRG} + Pr{RYYB } = .05+.15+.15 = .35

  16. Pr{ yellow shows | red shows } Identify Joint Probability “ Yellow and Red Show “

  17. Compute Joint Probability Pr{ Yellow and Red Show } = Pr{ exactly one of BYRG, RYYB shows } = Pr{BYRG}+ Pr{RYYB } = .15 + .15 = .30

  18. Conditional Probability = Joint Probability / Prior Probability So Pr{ Yellow Shows | Red Shows } = Pr{ Yellow and Red Shows }/Pr{Red Shows } = .30/.35 = 6/7 = .8571

More Related