# probability questions

• Sep 14th 2006, 10:32 PM
air_razor
probability questions
Hi everyone I have this probability question that I don't understand what to do. Can someone give some help please. I would appreciate it if a detailed explaination if given so that I can follow along easily. Here are the questions.

What is the chance of winning the 1st prize of 649?

Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that [10 marks]

a) none is defective
b) exactly one is defective
c) at least one is defective

Thanks
• Sep 15th 2006, 12:00 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Hi everyone I have this probability question that I don't understand what to do. Can someone give some help please. I would appreciate it if a detailed explaination if given so that I can follow along easily. Here are the questions.

What is the chance of winning the 1st prize of 649?

Probability of success in a problem of this type is the ratio of the
number of outcomes considered a success divided by the total
number of possible outcomes, when each of the possible oucomes
is considered equally likley.

Here there is but one outcome considered a success that is winning.
There are a total of 649 possible equally likley outcomes, so the required
probability is:

1/649 ~= 0.001541 ~= 0.15%

RonL
• Sep 15th 2006, 12:09 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that [10 marks]

a) none is defective

That none are defective requires that:

1. The first chosen is not defective which has probability of 10/15, as there
are 10 non-defective bulbs out of a total of 15, and each is equally likely to be chosen.

2. That given that the first is not defective the second is not defective.
But if the first was not defective the second is drawn from 14 bulbs of
which 9 are non-defective, hence the probability is 9/14.

2. That given that the first two are not defective the third is not
defective. But if the first two were not defective the third is drawn from 13
bulbs of which 8 are non-defective, hence the probability is 8/13.

The final probability is the product of these three probabilities:

p=10/15 * 9/14 * 8/13.

RonL
• Sep 15th 2006, 12:15 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that [10 marks]

b) exactly one is defective

That exactly one is defective may be broken down into three
possible cases, that the first is defective and the others are
non-defective, that the second is defective and the others are
non-defective, and that the third is defective and the others are
non-defective.

The probability of each of these is calculated like the probability
in part a), and they are then added together.

First case: p1=5/15 * 10/14 * 9/13
Second case: p2=10/15 * 5/14 * 9/13
Third case: p3=10/15 * 9/14 * 5/13

p=p1+p2+p3=3* (5*10*9)/(15*14*13).

RonL
• Sep 15th 2006, 12:20 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that [10 marks]

c) at least one is defective

Thanks

That at least one is defective is the same as that they are not all
defective. The probability that they are not all defective is 1 minus
the probability that they are all defective, or

p(not all defective)=1-p(all defective).

Now we calculate the probability that they are all defective in the same
way that we calculated the probability that none were defective, except
where we used the number of non-defectives remaining to be chosen
we now use the number of defectives remaining to be chosen.

So:

p(all defective)=5/15 * 4/14 * 3/13

RonL
• Sep 15th 2006, 10:14 AM
air_razor
Thanks
Thanks Captain Black for your amazing responses. I have 2 more questions here you may not be able to get these because they are some what computer related but if you can or if you have some insight in them please let me know.

1)Adisadvantage of a broadcast subnet is the capacity wasted due to multiple hosts attempting to access the channel at the same time. As a simplistic example, suppose that time is divided into discrete slots, with each host of the n hosts attempting to use the channel with probability p during each slot. What fraction of the slots are wasted due to collisions? [From Tanenbaum]

2) Imagine that you have trained your St. Bernard dog, Bernie, to carry a box of three DAT tapes instead of a flask of brandy. Each DAT tape contains 80 Gbytes of information. (When your disk fills up, you consider that an emergency.) The dog can travel to your side, wherever you may be, at 15 km/hour. For what range of distance does Bernie have a higher data rate than 155 Mbps (OC-3) ATM line? [From Tanenbaum]

Can you take a shot!!??
Thanks
• Sep 15th 2006, 10:29 AM
CaptainBlack
Quote:

Originally Posted by air_razor
1)Adisadvantage of a broadcast subnet is the capacity wasted due to multiple hosts attempting to access the channel at the same time. As a simplistic example, suppose that time is divided into discrete slots, with each host of the n hosts attempting to use the channel with probability p during each slot. What fraction of the slots are wasted due to collisions? [From Tanenbaum]

You are asking what is the probability of two or more hits on a slot
form n shooters each with a probability p of hitting.

This is a binomial distribution problem.

The probability of exactly r hits from n shooters is:

B(r,n)=C(n,r)*p^r *(1-p)^(n-r).

Now we work the problem back wards (We do it this way as there
are fewer B's to compute this way, we use the fact that:

sum(B(r,n),r=0..n)=1

That the probability of some outcome is 1)

We compute B(0,n) and B(1,n), then our required probability is:

P(n,p)=1-[B(0,n)+B(1,n)]=1-[p^0*(1-p)^n)+np^1*(1-p)^(n-1)]

RonL

(if this were a real world problem I would also perform a simulation
to confirm the analysis).
• Sep 15th 2006, 10:44 AM
CaptainBlack
Quote:

Originally Posted by air_razor
2) Imagine that you have trained your St. Bernard dog, Bernie, to carry a box of three DAT tapes instead of a flask of brandy. Each DAT tape contains 80 Gbytes of information. (When your disk fills up, you consider that an emergency.) The dog can travel to your side, wherever you may be, at 15 km/hour. For what range of distance does Bernie have a higher data rate than 155 Mbps (OC-3) ATM line? [From Tanenbaum]

The transit time is:

t=d/15 hr

where d is the distance in km.

So the data transfer rate is:

DataRate=240/t = 240*15/d Gbytes/hr,

so in units of Mega bits per second this is:

DataRate1=DataRate*1000*8/60/60 =(240*15*1000*8)/(60*60*d) Mbps
..............=8000/d Mbps

And I'm sure you can finish it from there.

RonL
• Sep 16th 2006, 07:25 AM
air_razor
Thanks Captain Black but..
Thanks Captain Black but for the transit time formula what happened to the kilometers? Is it t=d/15 km per hour?... It's not 15 hours right? As you have it here t=d/15 hr. Because the question has it as 15 km/hr speed.

Also I'm still confused...can you finish the question all the way please?

Thanks
• Sep 16th 2006, 07:29 AM
air_razor
Thanks Captain Black but the question I think is asking if you have two or more hits on a slot that means that other slots aren't being used. What is the fraction of slot not being used.
• Sep 16th 2006, 07:42 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Thanks Captain Black but the question I think is asking if you have two or more hits on a slot that means that other slots aren't being used. What is the fraction of slot not being used.

"What fraction of the slots are wasted due to collisions?"

a slot is wasted when there is a collision in it is my reading of this.

A slot is not wasted due to a collision if there is no host attempting to
use it.

Also it is not wasted if it is in use.

RonL
• Sep 16th 2006, 07:45 AM
CaptainBlack
Quote:

Originally Posted by air_razor
Thanks Captain Black but for the transit time formula what happened to the kilometers? Is it t=d/15 km per hour?... It's not 15 hours right? As you have it here t=d/15 hr. Because the question has it as 15 km/hr speed.

Also I'm still confused...can you finish the question all the way please?

Thanks

d is in units of km. the speed (call it v km/hr, but it will be 15 km/hr in the
problem)is in units of km/hr, so the units of d/v are hours.

RonL
• Sep 16th 2006, 07:54 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
The transit time is:

t=d/15 hr

where d is the distance in km.

So the data transfer rate is:

DataRate=240/t = 240*15/d Gbytes/hr,

so in units of Mega bits per second this is:

DataRate1=DataRate*1000*8/60/60 =(240*15*1000*8)/(60*60*d) Mbps
..............=8000/d Mbps

And I'm sure you can finish it from there.

RonL

Quote:

Originally Posted by air_razor

Also I'm still confused...can you finish the question all the way please?

Thanks

Bernie data rate is:

DataRate1=8000/d Mbps,

we want DataRate1>155 Mbps, writing this out in long form we have:

8000/d>155,

so:

d<8000/155 km,

simplifying this gives:

d<51.61.. km

So Bernies data rate is greater that a 155 Mbps (OC-3) ATM line for all
distances less that about 51.61 km.

RonL
• Sep 17th 2006, 01:47 AM
air_razor
slots question

"What fraction of the slots are wasted due to collisions?"

a slot is wasted when there is a collision in it is my reading of this.

A slot is not wasted due to a collision if there is no host attempting to
use it.

:)
Thanks Captain Black I read this question as when collisions occur on a slot the other slot could be and should be used but are not and these are wasted. For example if there were anyways 3 slots available and there were 3 hosts trying to access the channel. When one host uses a slot and the other 2 hosts accidentally collide with the first host there are 2 other slots not being used that could be. I think I need a fraction to reflect that occurance, when slots that aren't being used should be used because of collisions. Something like the summation of all slots not being used divided by slots being used with collisions occuring. I'm not sure if this is possible to show but I hope so.

Thanks
• Sep 17th 2006, 03:17 AM
CaptainBlack
Quote:

Originally Posted by air_razor

"What fraction of the slots are wasted due to collisions?"

a slot is wasted when there is a collision in it is my reading of this.

A slot is not wasted due to a collision if there is no host attempting to
use it.

:)
Thanks Captain Black I read this question as when collisions occur on a slot the other slot could be and should be used but are not and these are wasted. For example if there were anyways 3 slots available and there were 3 hosts trying to access the channel. When one host uses a slot and the other 2 hosts accidentally collide with the first host there are 2 other slots not being used that could be. I think I need a fraction to reflect that occurance, when slots that aren't being used should be used because of collisions. Something like the summation of all slots not being used divided by slots being used with collisions occuring. I'm not sure if this is possible to show but I hope so.

Thanks

Think of this as a design excercise, empty slots, and slots with a host
accessing it are normal operation, a colloision is the type of event
which you wish to minimise the occurence of, and so is what you are
interested in the probability of.

RonL