I'm a bit stuck trying to figure out the combined probability from several discrete PDF's.

Lets say I have a bunch of different classes (Truck, Sports Car, Station Wagon, etc) and a bunch of different scenarios I've worked out from my dataset (Location, behaviour, size, heading, speed).

Probability that the vehicle at the position x,y (Golf course) is a:

  • Truck: 0.1
  • Sports Car: 0.7
  • Station Wagon: 0.2

Vehicle is white, probability that it is a:

  • Truck: 0.6
  • Sports Car: 0.1
  • Station Wagon: 0.3

Vehicle has 3 passengers, probability that it is a:

  • Truck: 0.2
  • Sports Car: 0.1
  • Station Wagon: 0.7

If I had info on an unknown vehicle (Has 3 passengers, painted white and is at the golf course) how would I create the combined probabilities that it is each class? How would I do this for n discrete distributions. Are these dependant or independent variables? What is the wording for what I'm trying to do so that I can look it up properly.

Sorry if this is a duplicate question, I don't know the correct jargon for this problem.


1 Answer 1


I ended up just taking the probabilities for each class and multiplied them together. After that I normalized back to 1. Probably not correct:

Prob that test-car (Has 3 passengers, painted white and is at the golf course) is class x:

  • Truck: (0.1 x 0.6 x 0.2)/Total = 0.012/(0.012+0.007+0.042)
  • Sports Car: (0.7 x 0.1 x 0.1)/Total = 0.007/(0.012+0.007+0.042)
  • Station Wagon: (0.2 x 0.3 x 0.7)/Total = 0.042/(0.012+0.007+0.042)

thus: - Truck: 0.20 - Sports Car: 0.11 - Station Wagon: 0.69

Hope someone comments with a better method than this

  • 2
    $\begingroup$ Multiplication is permitted only when the conditioning variables are independent. In general, you cannot obtain an answer with just the information given in your question. $\endgroup$
    – whuber
    Nov 19, 2019 at 18:30

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