# S1 random varibable help!!!!!!!

The discrete random variable X has probability function

k(2 –x), x= 0, 1, 2,

P(X = x) = k(x–2), x= 3,

0, otherwise,

where k is a positive constant.

(a) Show that k = 0.25.
(b) Find E(X) and show that E(X ) = 2.5.
(c) Find Var(3X 2).

Two independent observations X1 and X2 are made of X.

(d) Show that P(X1 + X2 = 5) = 0.
(e) Find the complete probability function for X1 + X2.
(f) Find P(1.3 = X1 + X2 = 3.2).

Please can anyone tell me how to do the parts d,e,f? i have done the first 3 parts but these last three are confusing. Ive got the answers for this but in the MS they havent explained anything.
anyways this is wht written in the MS.
How and wht the hell they did? and wht is the logic behind it?

(d) P(X1 + X2) = P(X1 = 3 X2 = 2) + P(X1 = 2 X2 = 3) = 0 + 0 = 0
Let Y = X1 + X2 y 0 1 2 3 4 5 6
(e)
P(Y = y) 0.25 0.25 0.0625 0.25 0.125 0 0.0625

(f) P(1.3 X1 + X2 3.2) = P(X1 + X2 = 2) + P(X1 + X2 = 3)
= 0.0625 + 0.25 = 0.3125

k(2 &#8211;x), x= 0, 1, 2,

P(X = x) = k(x&#8211;2), x= 3,

This is quite confusing. do you work out "k" using k(2-x) or k(x-2), also can you confirm exaclt what values x can take, as im not sure whether it is "0,1,2 or 0,1,2,3.

Its actually :

............ { k(2 &#8211;x), x= 0, 1, 2,
..P(X=x) { k(x&#8211;2), x= 3
............ { 0, otherwise,

Forget the dots. The program wont allow me to leave spaces. so i just covered it with dots. I hope it makes sense now. X can take values of 0,1,2 usinf k(2-X) and can take 3 using k(X-2) .
What syllabus is this?
because I hope i dont have to know about "observations"!

I got the first three parts also, i think, but I'm hoping there is a letter missing in part b - E(X[here])

x............0...1...2...3....Total
P(X=x)....2k..k...0...k......4k
xP(X=x)...0...k...0..3k.....4k

a)

4k = 1 (because its the total)
therefore:
k=1/4

b) therefore I got E(X)=1

...just showed this so other people are clear...and to improve my confidence for tomorrow. Feel free to tell me if thats right...god i hope it is
for d, the probability of event x=2 and x=3 is 0 as one of the probability is 0. so when you times the probability of x=2(which is 0) and the probabilty of x=3, you'll get 0.

am I right?
for e, see attachment.
for f,

P(1.3<X1=X2<3.2)= P(1.3<Y<3.2)
=P(2<Y<3)
=0.625+0.25
=0.3125
I'm stuck at e, the marking scheme isn't helpful and I'm on vacation can't even ask the teacher..
Original post by Zaybraaa
I'm stuck at e, the marking scheme isn't helpful and I'm on vacation can't even ask the teacher..

Solution in post #6
Do you understand it?
Consider the possible values of $X_1 + X_2$
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
just seen the attached file, thank you sm <3

(Original post by BobbJo)Solution in post #6
Do you understand it?
Consider the possible values of $X_1 + X_2$
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
hey can ik why is x till 6 for e?
(Original post by BobbJo)Solution in post #6
Do you understand it?
Consider the possible values of $X_1 + X_2$
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
12 yrs late buddy
e.For example: when X1 X2=1 P{X1 X2}=P(x1=0)*P(X2=1)*2This is multiplied by two, because you can get X1=0 and X2=1 in any order, for example getting x1=0 and x2=1 or the other way round x1=1 and x2=0So, P{X1 X2}=P(x1=0)*P(X2=1)*2P{1}=P(x1=0)*P(X2=1)*2P(x1=0)=2*0.25=0.5P(X2=1)=0.25P{1}=0.25*0.5*2=0.25
damn i see no one knows the answer and its been 14 years
Original post by _zahra2608
damn i see no one knows the answer and its been 14 years

Plenty of people could help, but this isn't a do your homework for you site. See the maths forum posting guidelines - sticky thread at top of forum.

If you want some assistance, post your working so far, and say where precisely you're stuck.
(edited 3 years ago)
thank you !