FRST231 Discrete Random Variable For Single Policy

1. A shipment of 7 computers contains 2 defective units. A forestry company makes a random purchase of 3 of the units. If X is the number of defective units purchased by the company, find the probability distribution of X. Express the results graphically as a probability bar graph.

2. A potential customer for a $90,000 fire insurance policy possesses a home in an area which, according to experience, may sustain a total loss in a given year with probability of 0.001and a 50% loss with probability 0.01. Ignoring all other partial losses, what premium should the insurance company charge for a yearly policy to make 10% above the break-even point?

Answer:

Here are 7C3 ways of selecting 3 units from 7 

there are 5C3 ways of selecting 3 good units 
so the probability of

no defects (X = 0) is    

There are 5C2 ways of selecting 2 good units , and 2C1 ways of selecting 1 defective unit 
so the probability of

One defect (X = 1)    

There are 5C1 ways of selecting 1 good units , and 2C12ways of selecting 2 defective unit 
so the probability of

 Two defect (X = 2)    

There are only 2 defective , so probability of X=3 is zero   

Let Y be the discrete random variable that represents the amount of money that  insurance company will have to pay on a single policy for a home in the area. The set of possible values of Y is {10, 42500, 85000). According to the statement, we have the following probability distribution for Y: 

Thus, the expected amount of money paid on a single policy is  

The company should charge $510 for a yearly policy on a home in this area in order to break even (i.e. make no profit).  

Given

2 Douglas-fir

3 hemlock

 1 balsam seedlings

 Let X denote the number of hemlocks, and Y the number of Douglas-firs.

a)

Joint probabilities 

Marginal probabilities

P(X)=3/6

P(Y)=2/6  

To be independent the following must be true P(X)=P(XlY)

Event X  is pulling a Hemlock

Event Y is pulling a DF

P(X)=3/6

P(Y)=2/6  

Hence not independent 

X	Y	XY	Pr
1	2	2	1/3
2	1	2	1/3
3	0	0	1/3

Expected number of successful wells 0.1+0.4*1+2*0.3+3*0.1+4*0.1 = 1.8

X	P(X)
0	0.1
1	0.4
2	0.3
3	0.1
4	0.1

first 15 shots. 

You want to find out the probability of 4 red, 1 black and 1 white. This can be written as P (RRRRBW) but RRRRBW is the same as WRRRRB and many others.

And these possibilities must be considered, because it's not necessary for the FIRST FOUR to be red, the NEXT ONE black and the last white.

They can be in any order, as long as they're there. 

If the 2 firs were the first two chosen, the chance would be

But there are 10 ways of arranging the 2 firs (viz. positions 1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5 and 4-5), so the answer we want is 10 times the above   

(a) By the symmetry of the bell curve, the proportion under 24.95cm is the same as the proportion over 35.05cm, which in turn is    

So you need to turn to the normal dist. table and look up the p value giving z < about 10/6, then subtract that from 1 to get  

(b) Note that we need to exclude those under 24.95cm and those greater than 40.05cm. Aim is to obtain each of these, then simply subtract both from 1 to get final probability. 

Replace only 3% of the saws that fail,

How long a guarantee should you offer

Normal distribution with an average of 12 years and a standard deviation of 2 years.

Now

Hence it will take 8.74 years .