Probabilty problem

**K A D C Dilshan** · May 18th 2009, 10:18 PM

A heavy-equipment salesperson can contact either one or two customers per day with probability 1/3 and 2/3 ,respectively .Each contact will result in either no sale or a 50,000$ sale ,with the probabilities 0.9 and 0.1 ,respectively .Give the probability distribution for daily sales .Find the mean and standard deviation of the daily sales .

I am completely confused with this Qn because I was unable to apply E(X) and V(X) here .
Can somebody explain me in detail ?

What is the meaning of E(X+Y) provided that X and Y are two random variables ?
What is the meaning of E(X-Y) ?

Thank You

**matheagle** · May 18th 2009, 11:06 PM

Let X=$ in sales per day.
It seems X can be 0, 50k or 100k based on 0,1 or 2 sales.
You need to partition this, i.e.

$P(X=0)=P(X=0, 2 customers) +P(X=0, 1 customer)$

$= P(X=0|2 customers) P(2 customers) + P(X=0|1 customer) P(1 customer)$

$=(.9)(.9)(2/3) + (.9)(1/3)$

Since $P(X=0)+P(X=50k)+P(X=100k)=1$

I'll only compute $P(X=100k)=P(X=100k, 2 customers)$

$=P(X=100k| 2 customers)P(2 customers)=(.1)(.1)(2/3)$

Then you compute E(X) by its definition and $V(X)=E(X^2)-(EX)^2$ .

And we need to assume independence between the customers too, when the salesperson see two of them.

**Moo** · May 18th 2009, 11:11 PM

Hello,

Firstly, you need to find which values a daily sales can take.

In the case where there is only one customer :
0 or 50000
In the case where there are two customers :
0 (both contacts result in no sale), 50000 (exactly one of the contacts), 100000 (both contacts)

So finally, the possible values are :
0,50000,100000

Now I have to ask you... do you know about conditional expectation ?

Otherwise, you'll have to calculate the different probabilities associated to each value.
For example for value 0 :
$\mathbb{P}(X=0)=\mathbb{P}(X=0|N=1)\mathbb{P}(N=1) +\mathbb{P}(X=0|N=2)\mathbb{P}(N=2)=$ $0.9/3+0.9^2\times 2/3$

What is the meaning of E(X+Y) provided that X and Y are two random variables ?
What is the meaning of E(X-Y) ?

Imagine you throw two dices. X represents the number on the first dice, Y on the second. What's the average of the sum of numbers ? That's E(X+Y).
What's the average for the difference of numbers ? That's E(X-Y).

**K A D C Dilshan** · May 18th 2009, 11:25 PM

can you please moo show me the steps to achieve the answers ? ..
are those 0.9 and 0.1 are conditional probabilities ?
ifso is this right ?
0 -0.9
50000 -1/30
100000-1/15

what is conditional expectation ?

**matheagle** · May 18th 2009, 11:31 PM

I have the conditionals above
P(AB)=P(A|B)P(B) after I partitioned this via the number of customers.

**Moo** · May 18th 2009, 11:46 PM

Originally Posted by K A D C Dilshan

can you please moo show me the steps to achieve the answers ? ..
are those 0.9 and 0.1 are conditional probabilities ?
ifso is this right ?
0 -0.9
50000 -1/30
100000-1/15

what is conditional expectation ?

0.9 just means that there is no sale.

In other words, if you have N=1, what's the probability that the number of sales is 0 ? It's 0.9
If you have N=2, what's the probability that the number of sales is 0 ? It's 0.9x0.9, because you need two contacts to give no sale.

Conditional probability...that's not for now if you've never heard of it

**K A D C Dilshan** · May 18th 2009, 11:59 PM

please moo show the steps to solve the problem ?
and how to prove E(X+Y)=E(X)+E(Y)

**matheagle** · May 19th 2009, 12:06 AM

$E(X+Y) =\int\int (x+y)f(x,y)dxdy$

$= \int\int xf(x,y)dxdy +\int\int yf(x,y)dxdy=E(X)+E(Y)$

which also works in the discrete and mixed settings.

**K A D C Dilshan** · May 19th 2009, 12:26 AM

I am really sorry everybody ..I couldn't see your all answers as you all have used latex code here ..sorry for making same reply again again ..thank you eagle and moo giving me the idea of conditional expected values ..

I have another problem
how to perform the above proof for a discreet distribution ..

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