Consider Example 1.2.1. Let X denote the number of computer systems operable at
ID: 3024352 • Letter: C
Question
Consider Example 1.2.1. Let X denote the number of computer systems operable at the time of the launch. Assume that the probability that each system is operable is .9.
a) Use the tree of Fig.1.2 to find the density table.
b) There is a pattern to the probabilities in the density table. In particular,
f(x)= k(x)(.9)^x (.1)^(3-x)
where k(x) gives the number of paths through the tree yielding a particular value of X. Verify that k(x)=(n, x) for x=0,1,2,3
c) Find the table for F.
e) Use F to find the probability that at most one system is operable at the time of the launch.
Explanation / Answer
X represents the no. of computer systems operable at the time of the launch.
As per the tree given below X can take three values, i.e. X = 0 or 1 or 2 or 3 since in a trees there are three systems linked.
As per the data, Probability that the system is operable is 0.9 which also implies that the probability that any system is not operable at any point of time is 0.1.
a. The probability density function can be shown in the table below.
In this problem total no. of possible outcomes are 8. { yyy, yyn, yny, ynn, nyy, nyn, nny, nnn}
X = No. of system operable
P(X =x)=f(x)
REMARKS
P(X= 0) if X takes the value 0, that means none of the system is operable
0.001
No system is operable if the event ‘nnn’ occurs and the probability of occurrence is 0.1*0.1*0.1 = 0.001
P(X= 1) if X takes the value 1, that means only one of the system is operable
0.027
This event can happen if ‘ynn’,nyn, ‘nny’ happens. So the probability of occurrence is 0.9*0.1*0.1 + 0.1*0.9*0.1+0.1*0.1*0.9 = 0.027
P(X=2) ) if X takes the value 2, that means only two of the systems is operable
0.243
This event can happen if ‘yyn’,yny, ‘nyy’ happens. So the probability of occurrence is 0.9*0.9*0.1 + 0.9*0.1*0.9+0.1*0.9*0.9 = 0.243
P(X=3) if X takes the value 3, that means all of the system is operable
0.729
This event can happen if ‘yyy’ happens. So the probability of occurrence is 0.9*0.9*0.9 = 0.729
b. f(x) = k(x)*0.9^x*0.1^(3-x)
if x = 0, f(x) = k(x) * 0.9^0*0.1^(3-0) = 0.001 (From the table)
i.e. k(x) = 0.001/0.001 = 1
and (3,0) = 3!/0!*3! = 1
So the expression k(x) = (n,x) is true for x= 0
Similarly for x=1, f(x) = k(x)*0.9^1*0.1^(3-1) = k(x) *0.9*0.01 = 0.027(From the table)
i.e. k(x) = 0.027/0.009 = 3
and (3,1) = 3!/1!*2! = 3
So the expression k(x) = (n,x) is true for x= 1
Similarly for x=2, f(x) = k(x)*0.9^2*0.1^(3-2) = k(x) *0.81*0.1 = 0.243(From the table)
i.e. k(x) = 0.243/0.081 = 3
and (3,2) = 3!/2!*1! = 3
So the expression k(x) = (n,x) is true for x= 2
Similarly for x=3, f(x) = k(x)*0.9^3*0.1^(3-3) = k(x) *0.729*1 = 0.729(From the table)
i.e. k(x) = 0.729/0.729 = 3
and (3,2) = 3!/3!*0! = 1
So the expression k(x) = (n,x) is true for x= 3
c. the table for F
X
f(x)
F(x) = P(Xx) is also called as cumulative probability
0
0.001
0.001
1
0.027
0.028
2
0.243
0.271
3
0.729
0.729
d. The probability that at least one system is operable at lunch time is f(X1) = 1- F(X=0)
= 1- 0.001 = 0.999
e. The probability that at most one system is operable at the time of lunch = f(x1) = F(1)
= 0.028
e.
X = No. of system operable
P(X =x)=f(x)
REMARKS
P(X= 0) if X takes the value 0, that means none of the system is operable
0.001
No system is operable if the event ‘nnn’ occurs and the probability of occurrence is 0.1*0.1*0.1 = 0.001
P(X= 1) if X takes the value 1, that means only one of the system is operable
0.027
This event can happen if ‘ynn’,nyn, ‘nny’ happens. So the probability of occurrence is 0.9*0.1*0.1 + 0.1*0.9*0.1+0.1*0.1*0.9 = 0.027
P(X=2) ) if X takes the value 2, that means only two of the systems is operable
0.243
This event can happen if ‘yyn’,yny, ‘nyy’ happens. So the probability of occurrence is 0.9*0.9*0.1 + 0.9*0.1*0.9+0.1*0.9*0.9 = 0.243
P(X=3) if X takes the value 3, that means all of the system is operable
0.729
This event can happen if ‘yyy’ happens. So the probability of occurrence is 0.9*0.9*0.9 = 0.729
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