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