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 June 13th, 2011, 03:02 AM #11 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Probability The original poster's question was "three letters are selected at random from the word BIOLOGY. Find the number of ways in doing this". If the original poster's intent is that BOO is the same as BOO and that BLY is the same as LYB, then the answer is 25. If the original poster's intent is that BOO is the same as BOO but that BLY is different from LYB, then the answer is 135. If the original poster's intent is that BOO is different from BOO and BLY is different from LYB and the answer is then 210.
June 13th, 2011, 05:07 AM   #12
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Re: Probability

Quote:
 Originally Posted by soroban And six women, three of whom are sisters:[color=beige] .[/color]$\{S_1,\,S_2,\,S_3,\,W_1,\,W_2,\,W_3\}$ There is 1 way to choose the 3 sisters.
hmm ... let's see if I have this straight. There is one way to choose Alice, Betty, and Carol, the 3 sisters ...
but there are two ways to choose OO.

June 14th, 2011, 02:57 AM   #13
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Re: Probability

Quote:
Originally Posted by mrtwhs
Quote:
 Originally Posted by soroban And six women, three of whom are sisters:[color=beige] .[/color]$\{S_1,\,S_2,\,S_3,\,W_1,\,W_2,\,W_3\}$ There is 1 way to choose the 3 sisters.
hmm ... let's see if I have this straight. There is one way to choose Alice, Betty, and Carol, the 3 sisters ...
but there are two ways to choose OO.
You're confusing these. No one said there were two ways to choose "OO." There are two ways to choose "O". In other words, there's two ways to choose one O out of two O's. There's only one way to choose two O's out of two O's. Same as the sisters: three ways to choose one sister out of three sisters, and one way to choose three sisters out of three sisters. The sisters are all different, and so are the O's.

Quote:
 If the original poster's intent is that BOO is the same as BOO and that BLY is the same as LYB, then the answer is 25.
This appears to be the intent. There isn't anything indicating to us that order matters, either directly or logically. But the answer is 35, sorry. Let me ask you something: do you think two coins coming up heads/tails or tails/heads is the same thing because you get one of each no matter what? (When order doesn't matter.) $\text{BO_1O_2}$ is the same as $\text{BO_2O_1}$, because that's just about order. But $\text{BO_1L}$ is not the same as $\text{BO_2L}$ cause that is involving two different letters. The fact that they look alike doesn't matter. Think of putting them into a hat, with two O's, and pulling them out. Pulling the first one out is a different option that pulling the second one out.

Quote:
Quote:
 Originally Posted by soroban Suppose a bowl contains 999 white marbles and 1 black marble. We draw one marble at random. What is the probability that it is black? According to your reasoning, the 999 white balls are identical . . so there are two outcomes: (1) the ball is white, (2) the ball is black.
and according to your reasoning the probability that it is white is 1/1000. The white ball you picked might not be the one I was looking at.
I think that was exactly his point - the white ball he picks might not be the one you're looking at. That's cause it's 999 different white balls. It's 999/1000 that it's white. This is the exact same thing as the O's. There's 2 O's and they are two distinct ones (tho not visually). Just like the 999 white balls are all distinct. You basically just said so yourself, so you're contradicting your own ideas.

June 14th, 2011, 03:16 AM   #14
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Re: Probability

Quote:
 You have four scrabble tiles N O O N. How many ways can you place them on your tray?
Since you speak of "placing" them, then order matters:
$\text{N_1N_2O_1O_2}$
$\text{N_1N_2O_2O_1}$
It's 4! = 24, and I don't feel like writing all those out, but you get the point that I'm saying the two's O's there aren't the same ones. And neither are the two N's.

Granted it will look alike if you flip those O's around. But I may not be looking at the same tile you are.

Let's say instead we each have an N. If we traded, they'd look the same, but they wouldn't be the same tiles, so that's a different way of placing them. They may look alike, and from that standpoint neither of us cares which tile it happens to be. If I desparately need an N to make a cool word, I really don't care which one I get. But if I want to know the probability of getting one, then they have to become distinctive -- if there's say 4 of them total, then I could pick any of the 4 and there would be 4 different ways to pick one.

June 14th, 2011, 04:43 AM   #15
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Re: Probability

Quote:
 Originally Posted by Erimess But $\text{BO_1L}$ is not the same as $\text{BO_2L}$ cause that is involving two different letters. The fact that they look alike doesn't matter. Think of putting them into a hat, with two O's, and pulling them out. Pulling the first one out is a different option that pulling the second one out.
I just put these scrabble tiles into a hat {B,I,O,L,O,G,Y}.

I just reached in and pulled out B,O,L.

Which one was it? The first O or the second O?

If you are calculating probability, then BOL has frequency 2 since there are two O's. However if you are just counting the ways that you can pull three tiles out of the hat then there is only the one outcome.

June 14th, 2011, 06:09 AM   #16
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Re: Probability

I think I see where the area of disagreement is arising.

The original poster's question was

Quote:
 Three letters are selected at random from the word BIOLOGY. Find the number of ways in doing this. Hence,find the probability that the selection a)Does not contain letter O b)Contains both of the letter O
The OP started with a combinatorial question and then changed to a probability question. I certainly agree that for calculating probabilities, you must count the frequencies of identical letters in order to make outcomes in the sample space equally likely. But if you are strictly enumerating (combinatorially) the possibilities, then what I have said is correct.

As far as NOON goes, the answer is $\frac{4!}{2!2!}= 6$. This is a standard high school problem. For the technique, I refer you to Applied Combinatorics by Fred Roberts, Applied Combinatorics by Alan Tucker, or Enumerative Combinatorics by Richard Stanley.

June 14th, 2011, 06:16 PM   #17
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Re: Probability

Quote:
Originally Posted by mrtwhs
I think I see where the area of disagreement is arising.

The original poster's question was

Quote:
 Three letters are selected at random from the word BIOLOGY. Find the number of ways in doing this. Hence,find the probability that the selection a)Does not contain letter O b)Contains both of the letter O
The OP started with a combinatorial question and then changed to a probability question. I certainly agree that for calculating probabilities, you must count the frequencies of identical letters in order to make outcomes in the sample space equally likely. But if you are strictly enumerating (combinatorially) the possibilities, then what I have said is correct.
Maybe you just don't understand how problems in texts are usually addressed then. When it asks to find the number of ways to do something, it's not just looking for the possibilities of what things are by the way they look, which is what you are doing. If you simply look at them, then a BIO with O #1 and a BIO with O #2 will look the same, yes. But it didn't ask that. It's asking for the number of ways to "select" them, not the number of different ways we can make them look. It's a typical question, and the answer is 35. Selecting one O is different than selecting the other O. And we don't need the sentence asking about the probability to know that - that's what it means.

Quote:
 As far as NOON goes, the answer is $\frac{4!}{2!2!}= 6$.
You're trying to pick two out of four with that! Furthermore, if you're "placing" them somewhere, then order matters and it's not a combination at all. If you're asking where I'm "placing" them, is NOON the same as ONON? No, cause I placed the first two letters in a different place. You didn't say how many ways to "pick" them - you said "place" them, which implies a position.

Quote:
 This is a standard high school problem. For the technique, I refer you to Applied Combinatorics by Fred Roberts, Applied Combinatorics by Alan Tucker, or Enumerative Combinatorics by Richard Stanley.
I'm not an expert at this, but I'm thinking it's you who needs to do the reading. Seems to me you're still talking about what they LOOK like and how many different ways they can LOOK. If that's what you mean, then that's what you need to say. But that isn't what the problem asked for.

Let me ask you one more question: If there were 3 white marbles in a bag, how many ways are there for you to pick one white marble?

June 14th, 2011, 06:41 PM   #18
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Re: Probability

Quote:
Originally Posted by mrtwhs
Quote:
 Originally Posted by Erimess But $\text{BO_1L}$ is not the same as $\text{BO_2L}$ cause that is involving two different letters. The fact that they look alike doesn't matter. Think of putting them into a hat, with two O's, and pulling them out. Pulling the first one out is a different option that pulling the second one out.
I just put these scrabble tiles into a hat {B,I,O,L,O,G,Y}.

I just reached in and pulled out B,O,L.

Which one was it? The first O or the second O?
A silly question since you know that I don't know, and that you don't know, unless you mark them in some way. But there IS another way to pick B,O,L - with the other O as well, regardless of which one you got this time. Would you agree that you could have picked the OTHER O? And therefore, is that not another way you could pick BOL? I'm not asking if there's another way to make it look, but another way to "pick" those letters. Yes, there is. You could have picked the other O.

Quote:
 If you are calculating probability, then BOL has frequency 2 since there are two O's. However if you are just counting the ways that you can pull three tiles out of the hat then there is only the one outcome.
A probability is:
$\text{\frac{number of ways to get the desired outcome}{number of ways to get any outcome}$

Why does "number of ways" suddenly mean something completely different if we aren't doing a probability? In this problem, when you go to figure the probabilities presented, that 35 has to go in the denominator. In most problems I've seen, if they ask the "number of ways" something can be done prior to asking for a probability, they're kind of taking it one step at a time and "setting you up" for getting the probability. The "number of ways" to pick 3 letters has to go into the denominator. Why should it mean something different when they first asked the "number of ways" to do this? Where and how did that meaning change? (Which is a good thing to ask whoever put that question/solution together.)

 June 14th, 2011, 07:05 PM #19 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Probability Applied Combinatorics by Fred Roberts, Prentice-Hall (ISBN:0-13-039313-4) P. 50 #8 How many different "words" can be formed using all the letters of the word renegotiate? Answer: $\dfrac{11!}{3!2!}= 3326400$

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