Probability Worksheet

[EoinFar]

Hey guys, first time poster.

Am currently working through a probability worksheet for revision purposes. I've got an exam on prob. tomorrow and was looking to get some clarification.

My questions ask simple probability, but I just want to get a bit of a clearer picture of if I'm right or not. Here's one of the questions.

A programming team contains seven men and three woman, each of whom buys one ticket for the team raffle. What is the probability that the second prize goes to a woman, given that the first prize does?

To break that down, obviously I'd have to find the probability of the first event happening which would be 3/10. Now the likelihood of a woman winning a second time would just be the first event squared as the second time would have the same probability, right? The question doesn't state the first prize winner can't be in to win again so I'm going with the assumption she can win again.

Another one I'm unsure of is

A computer lab has four PC's, three mainframes and two workstations. Every time the lab is used, one is randomly picked. Last week the lab was used four times. What is the probability:

(a) I used a PC once and a mainframe three times?

So my answer to this was multiplying the likelihoods of each outcome in that 4/9 x 3/9 cubed which got me 4/243. Was this the right method?

(b) I used a workstation at least once?

My method to this was doing 7/9 to the power of 4, then taking the result from one which got me 4160/6561.

I've been slowly but surely working my way through probability but the interpretation is where I'm getting a little hung up. Thanks for any help offered!

 

[Dr.Peterson]

A programming team contains seven men and three woman, each of whom buys one ticket for the team raffle. What is the probability that the second prize goes to a woman, given that the first prize does?

To break that down, obviously I'd have to find the probability of the first event happening which would be 3/10. Now the likelihood of a woman winning a second time would just be the first event squared as the second time would have the same probability, right? The question doesn't state the first prize winner can't be in to win again so I'm going with the assumption she can win again.

I don't think I've heard of a raffle in which your ticket is put back in if you win. If it were, then the prizes would be independent, and you would be right; but don't assume that.

Instead, assume that after a woman won the first prize, there are 7 men and 2 women still in. What is the probability a woman will win now?

A computer lab has four PC's, three mainframes and two workstations. Every time the lab is used, one is randomly picked. Last week the lab was used four times. What is the probability:

(a) I used a PC once and a mainframe three times?

So my answer to this was multiplying the likelihoods of each outcome in that 4/9 x 3/9 cubed which got me 4/243. Was this the right method?

(b) I used a workstation at least once?

My method to this was doing 7/9 to the power of 4, then taking the result from one which got me 4160/6561.

Here it is clear that the selections are "with replacement" (that is, one can be picked more than once, so they are independent).

But you've missed the fact that it isn't specified which time you use the PC. The order could be any of PMMM, MPMM, MMPM, and MMMP. So what is the probability really?

You are correct on (b): the probability of using a workstation at least once is 1 minus the probability of not using one at all.

 

[EoinFar]

I don't think I've heard of a raffle in which your ticket is put back in if you win. If it were, then the prizes would be independent, and you would be right; but don't assume that.

Instead, assume that after a woman won the first prize, there are 7 men and 2 women still in. What is the probability a woman will win now?


Here it is clear that the selections are "with replacement" (that is, one can be picked more than once, so they are independent).

But you've missed the fact that it isn't specified which time you use the PC. The order could be any of PMMM, MPMM, MMPM, and MMMP. So what is the probability really?

You are correct on (b): the probability of using a workstation at least once is 1 minus the probability of not using one at all.

Thanks for your help!

So you're telling me that in the first one I've considered the possibility of a PC, then the mainframe 3 times in THAT order but not necessarily all orders? I've found all possible outcomes to match my sample space of 4 uses to be 4 over 36 or 1/9 as my answer of using a PC once and mainframe 3 times. Is that the correct method?

And you're right. I've tried to rework it to with no repetition.

Is the 3/10 x 2/9 method viable?

If I'm missing any fundamental rules, I'd appreciate the assistance but I understand that's not the point of the forum. I'm using Khan Academy and a few other resources to learn.

 

[Dr.Peterson]

So you're telling me that in the first one I've considered the possibility of a PC, then the mainframe 3 times in THAT order but not necessarily all orders? I've found all possible outcomes to match my sample space of 4 uses to be 4 over 36 or 1/9 as my answer of using a PC once and mainframe 3 times. Is that the correct method?

I'm not quite sure what you are saying your answer is. It isn't just 1/9.

Is the 3/10 x 2/9 method viable?

Can you explain that, too? That doesn't give the right answer (for either question).

 

[EoinFar]

I'm not quite sure what you are saying your answer is. It isn't just 1/9.

Can you explain that, too? That doesn't give the right answer (for either question).

My apologies,

I've considered, is what you're saying in the first question that I've considered one permutation but not all 4 possible permutations and therefore it's

4/9 x 3/9 cubed x 4?

And my apologies, I'm not really grasping how to answer the second question using what I've tried to find on it. Khan Academy and other sources use coins and dice, but applying those examples has been difficult for me.

 

[Dr.Peterson]

I've considered, is what you're saying in the first question that I've considered one permutation but not all 4 possible permutations and therefore it's

4/9 x 3/9 cubed x 4?

Yes, that's correct: Just multiply what you did before by 4.

You're referring to the second problem, part (a), right? Have you answered the first problem, about the raffle?

I'm not really grasping how to answer the second question using what I've tried to find on it. Khan Academy and other sources use coins and dice, but applying those examples has been difficult for me.

You had part (b) correct. Are you asking about the raffle problem here? Please say more, so I can be sure what we are talking about!

 

[Steven G]

Hey guys, first time poster.

Am currently working through a probability worksheet for revision purposes. I've got an exam on prob. tomorrow and was looking to get some clarification.

My questions ask simple probability, but I just want to get a bit of a clearer picture of if I'm right or not. Here's one of the questions.

A programming team contains seven men and three woman, each of whom buys one ticket for the team raffle. What is the probability that the second prize goes to a woman, given that the first prize does?

To break that down, obviously I'd have to find the probability of the first event happening which would be 3/10. Now the likelihood of a woman winning a second time would just be the first event squared as the second time would have the same probability, right? The question doesn't state the first prize winner can't be in to win again so I'm going with the assumption she can win again.

Another one I'm unsure of is

A computer lab has four PC's, three mainframes and two workstations. Every time the lab is used, one is randomly picked. Last week the lab was used four times. What is the probability:

(a) I used a PC once and a mainframe three times?

So my answer to this was multiplying the likelihoods of each outcome in that 4/9 x 3/9 cubed which got me 4/243. Was this the right method?

(b) I used a workstation at least once?

My method to this was doing 7/9 to the power of 4, then taking the result from one which got me 4160/6561.

I've been slowly but surely working my way through probability but the interpretation is where I'm getting a little hung up. Thanks for any help offered!

re 1st problem: I would mark you correct. The reason is that you assumed something reasonable, something that you truly thought could happen--That the 2nd place winner could also be the 1st place winner. This is a probability problem not a question of your knowledge of raffle rules. This is what makes some problems have more than one answer. The author of the problem knew what she/he meant but readers might interpret it differently. Never worry about making these mistakes. If it is on an exam, then I think that it is extremely fair to ask the instructor to make it clear if one could win both prizes. If you can't ask this question then note on your exam paper that you are assuming that one can win both prices (or that one can win only one prize) OR if time permits then solve it both ways. Don't stress out because you don't know the rules of a raffle contest!!