Expected Value
You can think of an expected value as a mean, or average, for a probability distribution.
Suppose random variable x can take different values x1,x2….,xn with probability p1,p2,p3…pn. Then Expected Value of experiment is given by
EV=x1*p1+x2*p2+x3*p3+…..xn*pn
Example 1.
Let X represent the outcome of a roll of a fair six-sided die. More specifically, X will be the number of values showing on the top face of the die after
toss. The possible values for X are 1, 2, 3, 4, 5, and 6, all equally likely (each having the probability of 1/6). The expectation of X is
Solution 1:-Our random variable x can take values 1,2,3,4,5 and 6 each having the
probability of 1/6 Expected Value of x =1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+6*1/6=3.5
Example 2.The weights (X) of patients at a clinic (in pounds), are: Sample Values= (108, 110, 123, 134, 135, 145, 167, 187, 199). Assume one of the patien
chosen at random. What is the Expected Value?
Solution: Find the mean. The mean is:
Expected Value=(108 + 110 + 123 + 134 + 135 + 145 + 167 + 187 + 199)/9 = 145.333.
It is not mandatory for the expected value to be present in sample space. As calculated above Expected Value of 145.33 is not present in our sample spa
,Joint Probability
Suppose we have event A which can occur with Probability P1 and event B which can occur with probability P2 .Then Joint probability is the situation when both event A and B occur
Joint Probability or P(A and B)=P(A)*P(B) when Event A and B are independent of each
other Example1). A coin is tossed 2 times. What is the probability of head occurring both
the times Ans1). Event(A)=Head occurring for the first time=1/2
Event(B)=Head occurring for the second time=1/2
P(A and B)=P(Head occurring both the times)=Probability(Event A)*Probability(Event B)=1/2*1/2=1/4
Joint Probability when Event B is dependent on Event A is given by P(A and B)=P(A)*P(B/A)
Where P(B/A) can be interpreted as probability of occurrence of event B given that event A had already occurred
Example 2). There are 3 Red Balls and 3 Blue balls in a box. What is the probability of withdrawing 2 Red balls from box in 2 attempts?
Ans 2).P(A)=Probability of Withdrawing Red Ball from Box in 1st attempt
P(A)=Number of Red Balls/Total No of Balls=3/6=0.5
P(B/A)= Probability of Withdrawing Red ball from Box in 2nd Attempt given that that we have already withdrawn red ball in the first attempt
P(B/A)=No of Red Balls Remaining/Total No of Balls Remaining=2/5=0.4
P(A and B)=P(A)*P(B/A)=0.5*0.4=0.20
Summary- In the First Example the probability of Event B does not change with the occurrence of Event A . Whereas in Example 2 the Probability of occurrence of Event B depends upon Even
, Conditional Probability
If events A and B are independent (where event A has no effect on the probability of event B), then the conditional probability of
event B given event A has occured is simply the probability of event B.
You can think of an expected value as a mean, or average, for a probability distribution.
Suppose random variable x can take different values x1,x2….,xn with probability p1,p2,p3…pn. Then Expected Value of experiment is given by
EV=x1*p1+x2*p2+x3*p3+…..xn*pn
Example 1.
Let X represent the outcome of a roll of a fair six-sided die. More specifically, X will be the number of values showing on the top face of the die after
toss. The possible values for X are 1, 2, 3, 4, 5, and 6, all equally likely (each having the probability of 1/6). The expectation of X is
Solution 1:-Our random variable x can take values 1,2,3,4,5 and 6 each having the
probability of 1/6 Expected Value of x =1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+6*1/6=3.5
Example 2.The weights (X) of patients at a clinic (in pounds), are: Sample Values= (108, 110, 123, 134, 135, 145, 167, 187, 199). Assume one of the patien
chosen at random. What is the Expected Value?
Solution: Find the mean. The mean is:
Expected Value=(108 + 110 + 123 + 134 + 135 + 145 + 167 + 187 + 199)/9 = 145.333.
It is not mandatory for the expected value to be present in sample space. As calculated above Expected Value of 145.33 is not present in our sample spa
,Joint Probability
Suppose we have event A which can occur with Probability P1 and event B which can occur with probability P2 .Then Joint probability is the situation when both event A and B occur
Joint Probability or P(A and B)=P(A)*P(B) when Event A and B are independent of each
other Example1). A coin is tossed 2 times. What is the probability of head occurring both
the times Ans1). Event(A)=Head occurring for the first time=1/2
Event(B)=Head occurring for the second time=1/2
P(A and B)=P(Head occurring both the times)=Probability(Event A)*Probability(Event B)=1/2*1/2=1/4
Joint Probability when Event B is dependent on Event A is given by P(A and B)=P(A)*P(B/A)
Where P(B/A) can be interpreted as probability of occurrence of event B given that event A had already occurred
Example 2). There are 3 Red Balls and 3 Blue balls in a box. What is the probability of withdrawing 2 Red balls from box in 2 attempts?
Ans 2).P(A)=Probability of Withdrawing Red Ball from Box in 1st attempt
P(A)=Number of Red Balls/Total No of Balls=3/6=0.5
P(B/A)= Probability of Withdrawing Red ball from Box in 2nd Attempt given that that we have already withdrawn red ball in the first attempt
P(B/A)=No of Red Balls Remaining/Total No of Balls Remaining=2/5=0.4
P(A and B)=P(A)*P(B/A)=0.5*0.4=0.20
Summary- In the First Example the probability of Event B does not change with the occurrence of Event A . Whereas in Example 2 the Probability of occurrence of Event B depends upon Even
, Conditional Probability
If events A and B are independent (where event A has no effect on the probability of event B), then the conditional probability of
event B given event A has occured is simply the probability of event B.
