Names out of a hat

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JohnStudent

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If I have a class of 26 students and put all of their names in a hat, and I pull 2 names out of the hat at the same time, what is the probability for one student's name to be pulled?
 
JohnStudent said:
If I have a class of 26 students and put all of their names in a hat, and I pull 2 names out of the hat at the same time, what is the probability for one student's name to be pulled?
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What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you! ;)
 
Yes, I have thought about this. I was think 2/26 = 1/13, but then if you pull both names at the same time, are you still actually pulling one at a time? And each student' name is in there only once. So then I was thinking would it be 2/26 * 1/25?
 
JohnStudent said:
If I have a class of 26 students and put all of their names in a hat, and I pull 2 names out of the hat at the same time, what is the probability for one student's name to be pulled?
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First of all, whether you chose both names at once or one after the other, the results will be the same. What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

HINT: Questions like this are sometimes best answered by rephrasing the question. That is, the chance of a particular student being chosen is one minus the chance that the student is not chosen. Of course you could go through all the possibilities and add the up and that might be easier for a small number of students chosen..
 
Ishuda said:
First of all, whether you chose both names at once or one after the other, the results will be the same. What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

HINT: Questions like this are sometimes best answered by rephrasing the question. That is, the chance of a particular student being chosen is one minus the chance that the student is not chosen. Of course you could go through all the possibilities and add the up and that might be easier for a small number of students chosen..
Click to expand...

Thanks for posting that. I did not realize there was a delay in the posting. So the chance that a particular student isn't chosen would be 24/26 or 12/13 = 92% chance. That means the chance of your name being drawn out of the hat would be about 8% chance?
 
Denis said:
No. About 4%.

2 possible ways:

1: drawn 1st
chance = 1/24 = .0416666.....

2: drawn 2nd
1st draw: 23/24 (23 out of 24 that another is drawn)
2nd draw: 1/23
23/24 * 1/23 = 1/24 = .0416666.....
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Denis,
I think you misunderstood (or maybe I do). I think the about 8% was for two draws which is what you show: 4.17% for drawn first and 4.16% drawn second for a total of 8.34% which is about 8%.
 
Denis said:
No. About 4%.

2 possible ways:

1: drawn 1st
chance = 1/24 = .0416666.....

2: drawn 2nd
1st draw: 23/24 (23 out of 24 that another is drawn)
2nd draw: 1/23
23/24 * 1/23 = 1/24 = .0416666.....
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Ok, but then do you have to add these together? Because your name could be pulled first, or it could be pulled second.
 
And if it were out of 26, wouldn't the denominator be 26? Just making sure we switched the total to 24 in this particular example.
 
Ishuda said:
Denis,
I think you misunderstood (or maybe I do). I think the about 8% was for two draws which is what you show: 4.17% for drawn first and 4.16% drawn second for a total of 8.34% which is about 8%.
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Also, why do we have to multiply by the probability that it won't happen on the second time? Why couldn't we do 1/26 + 1/25 ?
 
Personally, I think the best way to tackle this question (or any math question really) is to break it down and figure out what it's really asking. There's two ways of looking at this problem, and I'll tackle both. The first way is that the problem is asking "What's the odds that the either the first or the second name you draw will be the given student?" Let's look at that further by examining a similar, but easier situation Suppose you rolled a fair six sided dice, with a 1/6 chance of any number coming up. Now, what's the odds that your roll will be a 2 or a 5? Well, there's a 1/6 chance that it will be a 2, and a further 1/6 chance that it will be a 5. So, how do you think you'd figure out the odds of rolling a 2 or a 5?

The other way of looking at the problem is to say that the odds of something happening are one minus the odds of the thing not happening. So, now the question we're solving is, "What's the odds that both the first and the second name won't be the given student?" So, here we'll also examine an easier situation. Suppose you roll two fair dice, one blue and one red. And we want to know the odds of the blue dice coming up 2 and the red dice coming up 5. Well, as before, the blue dice has a 1/6 chance of coming up 2, and the red dice has a 1/6 chance of coming up 5. But remember that both of the conditions must be true. If the blue dice comes up 3, we don't care what the red dice came up. In other words, we only need to consider the number on the red dice 1/6 of the time. So, how do you think you'd figure out the odds of rolling a blue 2 and a red 5?

Now, can you apply the above logic to your problem?
 
ksdhart said:
Personally, I think the best way to tackle this question (or any math question really) is to break it down and figure out what it's really asking. There's two ways of looking at this problem, and I'll tackle both. The first way is that the problem is asking "What's the odds that the either the first or the second name you draw will be the given student?" Let's look at that further by examining a similar, but easier situation Suppose you rolled a fair six sided dice, with a 1/6 chance of any number coming up. Now, what's the odds that your roll will be a 2 or a 5? Well, there's a 1/6 chance that it will be a 2, and a further 1/6 chance that it will be a 5. So, how do you think you'd figure out the odds of rolling a 2 or a 5?

The other way of looking at the problem is to say that the odds of something happening are one minus the odds of the thing not happening. So, now the question we're solving is, "What's the odds that both the first and the second name won't be the given student?" So, here we'll also examine an easier situation. Suppose you roll two fair dice, one blue and one red. And we want to know the odds of the blue dice coming up 2 and the red dice coming up 5. Well, as before, the blue dice has a 1/6 chance of coming up 2, and the red dice has a 1/6 chance of coming up 5. But remember that both of the conditions must be true. If the blue dice comes up 3, we don't care what the red dice came up. In other words, we only need to consider the number on the red dice 1/6 of the time. So, how do you think you'd figure out the odds of rolling a blue 2 and a red 5?

Now, can you apply the above logic to your problem?
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For my own personal enlightenment, would please explain in some detail what the heck the above has to do with the original question? Either one understands counting or one does not. The question is about counting, is it not?
 
Denis said:
Yes, I misunderstood, screwed up, goofed, ....
Not only did I multiply once where I should have added, I also used 24 instead of 26 :(
Correcting my post:
2 possible ways:

1: drawn 1st
chance = 1/26

2: drawn 2nd
1st draw: 25/26 (25 out of 26 that another is drawn)
2nd draw: 1/25

So: 1/26 + 25/26 * 1/25 = ~.076923...
(so don't quite agree with you, Sir Ishuda)

Coming to join you in the corner, Subhotosh!
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7.69% is still about 8%. I guess I should have checked your numbers:oops:
 
pka said:
For my own personal enlightenment, would please explain in some detail what the heck the above has to do with the original question? Either one understands counting or one does not. The question is about counting, is it not?
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Hi,

Thank you for your response. I am following on the first example of rolling a 2 or a 5 on 1 die. The odds of a 2 are 1/6 + the odds of a 5 which is also 1/6 so we would get 2/6 = 1/3 or 33% chance of it happening. The odds of it not happening are 1 - 4/6 or 1 - 2/3 which gives us our same answer.

Ok so on the second example, I follow it somewhat. The odds of the first blue die coming up a 2 are 1/6. Doesn't that mean we need to look at the red die 5/6 of the times? Not 1/6? So would the second roll be the odds of it not happening times the odds of it happening? (1 - 5/6)(1/6) which would be 1/36? Then would we add the odds of the blue die coming up a 2 and the red coming up a 5? This would be 7/36 which is about 19%.
 
JohnStudent said:
Also, why do we have to multiply by the probability that it won't happen on the second time? Why couldn't we do 1/26 + 1/25 ?
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Sorry back to the roll 2 dice, 1 blue, 1 red. The probability that the blue one comes up a 2 would be 1/6. Then the probability of the red coming up a 5 would be the probability that it doesn't happen the first time (5/6) times the probability that it happens this time (1/6)? So would we add 1/6 + 5/36 and get 11/36? Or was I right the first time?

Thanks!
 
JohnStudent said:
Sorry back to the roll 2 dice, 1 blue, 1 red. The probability that the blue one comes up a 2 would be 1/6. Then the probability of the red coming up a 5 would be the probability that it doesn't happen the first time (5/6) times the probability that it happens this time (1/6)? So would we add 1/6 + 5/36 and get 11/36? Or was I right the first time?

Thanks!
Click to expand...

My two examples were meant to tackle precisely this question. You appeared to be confused about why you sometimes multiply probabilities and sometimes add them. There's a rule about this and it's actually quite simple. If two or more events are occurring, and you need them all to happen, you multiply the probabilities. Returning to the dice example, you want both the red dice to come up 5 and the blue dice to come up 2. The red dice has a 1/6 chance of coming up 5. If that happens, then we look at the blue dice to see if it's a 2, which also happens 1/6 of the time. So, overall, the two dice will give us our desired rolls 1/6 of 1/6 of the time. And do you remember what the word "of" means in the context of fractions?

Similarly, if you have two (or more) events but you only need one to happen, then you add the probabilities. With the dice example, your answer of 1/6 + 1/6 = 1/3 is correct. In the case of the students, the probability actually changes for the second draw, because the first student's name is not replaced, but that doesn't change the underlying principle. The given student's name can come up on either the first draw or the second draw, so we add the probabilities.

Hopefully that clears a few things up for you.
 
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