Conditional probability examples

 Conditional Probability 

 • Definition

         • The conditional probability of A given B, denoted P(A / B).

         • It is the probability of event A has when it is known that event B has already occurred. •  Rule for Condition 

         • al Probability

         • P(A / B)=P(A∩B) / P(B


Independent Events 

• Events A and B are independent if 

 • P(A∩B) = P(A)⋅P(B) 

 • If A and B are not independent then they are dependent

Dependent Events 

 But some events can be "dependent" ... which means they can be affected by previous events. 

 Example: Drawing 2 Cards from a Deck. After taking one card from the deck there are less cards available, so the probabilities change!


Examples:  • 

Q) Ninety percent of flights depart on time. Eighty percent of flights arrive on time. Seventy-five percent of flights depart on time and arrive on time.

 (a) A flight, has departed on time. What is the probability that it will arrive on time?

 (b) A flight has arrived on time. What is the probability that the flight departed on time?

 (c) Are the events, departing on time and arriving on time, independent? 

--> • Solution:

• A = {arriving on time} , D = {departing on time} 

 • We have: P {A} = 0.8 P {D} = 0.9 P {A ∩ D} = 0.75 

 (a) P {A | D} = P {A ∩ D} / P {D} = 0.75 / 0.9 = 0.8333. 

 (b) P {D | A} = P {A ∩ D} / P {A} = 0.75 / 0.8 = 0.9375.

 (c) Events are not independent because P {A | D} != P {A} , P {D | A} != P {D} , P {A ∩ D} != P {A}. P {D} 


Q) A fair die is rolled. •Find the probability that the number rolled is a five, given that it is odd.

 -->Solution:

      The sample space for this experiment is the set S={1,2,3,4,5,6} consisting of six equally likely outcomes. Let F denote the event “a five is rolled” F={5} 

 and let O denote the event “an odd number is rolled,” and O={1,3,5} 

P(F|O)=P(F∩O) / P(O) 

 Since F∩O= {5}∩{1,3,5} = {5}, P(F∩O)=1⁄6. 

 Since O= {1,3,5}, P(O) = 3⁄6.

 P(F|O)=P(F∩O)/P(O)= (1⁄6) / (3⁄6) =1/3 


Q) A fair die is rolled. •Find the probability that the number rolled is odd, given that it is a five. 

 •Solution: The sample space for this experiment is the set S={1,2,3,4,5,6} consisting of six equally likely outcomes. Let F denote the event “a five is rolled” F={5}  and let O denote the event “an odd number is rolled,” and O={1,3,5}

P(O|F)=P(O∩F) / P(F)

Obviously P(F)=1⁄6. In part  (a) we found that P(F∩O)=1⁄6. 

Thus P(O|F)=P(O∩F)P(F)= (1⁄6) / (1⁄6) =1 



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