Question

An input source for an electrical system is modeled by a discrete random variable X where X takes on a value at random from sample space S={1, 2, 3, and 4} volts with
corresponding CDF of this random variable, F X ( x ) = {0.2, 0.3, 0.7, and 1.0}.
(a) Find P X ( x ), the PMF of this random variable.
(b) Let event B be P [ X < 3]. Find P[B].
(c) Find the expected value of the function of the random variable, E [ X^2 ].
(d) Find P X ( x | B) .
(e) The power supply of the electrical system is turned on by a switch that sometimes fails. The
switch turns on the system with 0.3 chance. What is the chance that we attempt to turn on the
system 6 times and all fail but the switch ultimately turns on the system on the 7th attempt?
(f) Now, if we attempt switching on (and then quickly off) 10 times, what is the probability that
the system shows turned on in one or two of the attempts.

Answer 1

a)
The CDF is given by:
$F_X(x)=\begin{cases} 0 & \text{ if } x<1 \\ 0.2 & \text{ if } 1\leq x<2 \\ 0.3 & \text{ if } 2\leq x<3 \\ 0.7 & \text{ if } 3\leq x<4 \\ 1 & \text{ if } x\geq 4 \end{cases}$
Thus, the PMF is given by:
$f_X(x)=\begin{cases} 0.2 & \text{ if } x=1 \\ 0.1 & \text{ if } x=2 \\ 0.4 & \text{ if } x=3 \\ 0.3 & \text{ if } x= 4\\ 0 & \text{ o.w. } \end{cases}$




b)
B is the event that X<3
$P(B)=P(X<3)$
$=P(X=1)+P(X=2)$
$=0.2+0.1=0.3$



c)
We compute the following table

X	P(X=x)	X2*P(X=x)
1	0.2	0.2
2	0.1	0.4
3	0.4	3.6
4	0.3	4.8
Total	1	9

Hence,
$E(X^2)=\sum_{x=1}^{4}x^2*P(X=x)=9$




d)
We are to find PX(x|B)

It is known that,
$P_X(x|B)=\frac{P(X=x,B)}{P(B)}$
We construct the following table to get the probability distribution:

X	P(X=x)	P(X<3)	PX(x|B)
1	0.2	0.3	0.666667
2	0.1	0.3	0.333333

e)
p = P(turning on the system when switch is given) = 0.3
Let X is the random variable denoting the number of attempts required to switch on the system.
Clearly, X ~ Geom (p)
The pmf of X is given by:
$f_X(x)=(1-p)^{x-1}p, \text{ x= 0, 1, 2, ...}$

Hence,
$f_X(7)=(1-0.3)^{7-1}0.3=0.0352947$



f)
Let Y be a random variable denoting the number of times the system turns on in 10 attempts
Clearly, Y ~ Bin(10, p)
The pmf of Y is given by
$f_Y(y)=\binom{10}{y}p^y(1-p)^{10-y}, \text{ y = 0, 1, ...10}$

We are to find
$P(Y=1\bigcup 2)=P(Y=1)+P(Y=2)$
$=\binom{10}{1}0.3^1*0.7^9+\binom{10}{2}0.3^2*0.7^8$
$=0.121060821+0.2334744405$

$=0.3545352615$

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