Wrong, there is not a 50% chance of the other being female.
Look at it this way, on the first dog you have a 50/50 chance, from that you then have another 50/50 chance. This the chance of getting both males is the 2 probabilities times together, thus making it 25%.
Also, Temple, just because you know 1 it doesn't mean that the probability is reduced, there was still a 50% chance that dog could be female which must be taken into consideration.
EDIT 33%? Hmm, well, there are 3 a few possibilitiies of how to get that, if you count up all the possible pairings
Male Male Female Male Male Female Female Female Male Male Female Female
The first, FF, can be ruled out, since one of them was a male. Any of the remaining three will have "at least one male." And since only one out of three of the remaining cases gives you two Males, the probability that the other is a male as well is 1/3.
The two cases FM and MF are not the same thing. They are two different situations, and occupy distinct positions in the sample space. The chances that, out of two dogs, you will get one female and one male are 50 percent, because there are two different cases that yield that result. If you don't believe me, toss two coins a bunch of times, and see how many times you get one head and one tails.
mipegg: Wrong, there is not a 50% chance of the other being female.
Look at it this way, on the first dog you have a 50/50 chance, from that you then have another 50/50 chance. This the chance of getting both males is the 2 probabilities times together, thus making it 25%.
Also, Temple, just because you know 1 it doesn't mean that the probability is reduced, there was still a 50% chance that dog could be female which must be taken into consideration.
EDIT 33%? Hmm, well, there are 3 a few possibilitiies of how to get that, if you count up all the possible pairings
Male Male Female Male Male Female Female Female Male Male Female Female
Or, 33%.
This doesn't work. The chances that you will get two males in a random selection of two dogs are 25 percent, not 33 percent, as they would be here. There are only 4 possible outcomes. I'm not sure why you included male male and female female twice.
If you flipped two coins, the chances that you'll get 2 tails is only 25 percent, since the probability of both the first and second being tails is .5 * .5, hence, 25 percent.
So your answer to this problem is correct, just not for the reasons you stated.
Meh, I've never been all that great at math, my elementary school really sucked, so I went into 7th grade not knowing stuff that should have been review. I've been playing catchup ever since, and I've managed to start getting As consistently since freshman year (now a sophomore). But no doubt that if the Catholic school I went to through 6th grade had taught me better I'd be in Algebra 2 instead of Geometry this year.
I'm sad to hear your primary school screwed you over. That happened to me too (or maybe I just didn't pay attention - I can't remember): I was terrible at maths. Even the times tables confused the hell out of me. I have one suggestion, though: do they have a program called Kumon anywhere near where you live? It's a teaching program that goes from Year 1 level maths to Year 12 calculus stuff (and possibly beyond - I didn't go all the way), usingng rote learning to teach you speed and accuracy until the maths is imprinted on your eyelids. My only problem with the program was that it didn't really explain what you were doing or what it was for, (for example, I could solve quadratics 2 years before high school taught me what they meant) but if you already know this, or are willing to learn - or, as I did, just do what you're told - it'll completely change your maths ability. In primary school the times tables worried me; six months later I had caught up; by high school I was one of the "smart" kids. It had nothing to do with ability and everything to do with Kumon. I hated every minute of it, but now I wish I could go back to brush up on my calculus. Try it out if you have the money and the patience.
I remember having this huge discussion, and it was either 66% or 75%
I'll try to explain on what I can (vaguely... VERY vaguely) remember. You can have 3 situations: 2 males, 2 females, or 1 male and 1 female. Though this is actually 4 situations, with the interchangeable male and female in the last option.
Now, we have to add in the known dog, which is definitely male: the 2 female option is dropped out, form both sides. Now we have: 2 males (the option from the first 'filter'), 2 males (the second dog is male) , 1 male and 1 female (second dog is female), as our options. From this situation, it is concluded that there's a 66% chance of the second dog being male.
I know I skipped out on something (or did I?), but it was something like this.
Pardon me for the vagueness, I stopped caring about maths 9 years ago. I might have completely missed the point on this one (but I still can't get this idea out of my head)
EDIT: Right. Messed it up. Meant to have, in the second row of options: 2 males, 1 female and 1 male, 1 male and 1 female. So it's 33% indeed.
None of the above, it's 33 1/3% !=33%. The key point is that the question asked is is there at least one male, which, as has been pointed out above means that there are three options, not four, only one of which has both dogs as male.
Ignore everything else, the question is "Is the other dog male or female?" It doesn't matter if the other dog is male, female or a penguin in a dog suit.
The odds are still 50/50
Really... Why am I getting so annoyed. My old school days are creeping back to me, where I sit there quietly having done everything and then shouting at stupid kids in the class who can't understand simultaneous equations or entropy in thermodynamics.
Blind Punk Riot: male, female, hermaphrodite, one of those neutral babies that dont have anything down there, so something slightly less than 50% but more than 33%
Probably around 49.9997% or higher.
In one hand i have a carrot, what is the chance of landing a heads when flipping a coin? have the odds suddenly changed?
Okay, if the person who's bathing the dogs was asked "Is the first dog male?" then you'd be right, since the sex of the first dog doesn't affect the probability that the second one will be male. But:
You start off with four possibilities:
1.) Dog A is male, Dog B is male. 2.) Dog A is male, Dog B is female. 3.) Dog A is female, Dog B is male. 4.) Dog A is female, Dog B is female.
Each of these scenarios is equally likely. Then you are given the information that at least one of the dogs is male. This means that scenario 4 is impossible. It does not affect the probabilities of scenarios 1-3, which are still equally likely, i.e. there is a 33% chance of each.
So in scenario 1, the other dog is male. But in scenarios 2 and 3, the other dog is female. So there is a 33% chance that the other dog is male.
If the question asked had been "Is Dog A male?" then that would eliminate scenarios 3 and 4, making the probability that Dog B was male 50%.
EDIT: Does anyone have a pack of dogs and a bath? I'm just thinking us 33 percent-ers could make a pretty penny gambling with the 50 percent-ers over dog gender...
Guys, it would be 33.3(repeating)% if the question was asking for a set, but it's not. It's asking for the other puppy, not how they are together. As such, the first puppy might as well not even be there, as it has no bearing on the gender of the second. Therefore, it's 50%.
kailsar: Okay, if the person who's bathing the dogs was asked "Is the first dog male?" then you'd be right, since the sex of the first dog doesn't affect the probability that the second one will be male. But:
You start off with four possibilities:
1.) Dog A is male, Dog B is male. 2.) Dog A is male, Dog B is female. 3.) Dog A is female, Dog B is male. 4.) Dog A is female, Dog B is female.
Each of these scenarios is equally likely. Then you are given the information that at least one of the dogs is male. This means that scenario 4 is impossible. It does not affect the probabilities of scenarios 1-3, which are still equally likely, i.e. there is a 33% chance of each.
So in scenario 1, the other dog is male. But in scenarios 2 and 3, the other dog is female. So there is a 33% chance that the other dog is male.
If the question asked had been "Is Dog A male?" then that would eliminate scenarios 3 and 4, making the probability that Dog B was male 50%.
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
That is the call. What is the probability that the other one is a male?
The other information is unnecessary. Truely. I'm being honest.
Trust.
Edit; You only want A male dog, just the one. To be honest, it wouldn't matter to you if the other dog was even there or not.
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
That is the call. What is the probability that the other one is a male?
The other information is unnecessary. Truely. I'm being honest.
Trust.
Okay, the question is 'What is the probability that the other one is a male?'. But which dog is the 'other' one? The one that hasn't already been identified as male. As I said, if you eliminated a dog at random, then the probability of the other being a male is 50%. Not that I'm advocating eliminating dogs at random. But you're not eliminating a dog at random, you're eliminating a dog, knowing that it's a male. So the only way that the other one is a male is if they were both male to start with. We know that they weren't both female to start with, so the chances that they were both male is one-third.
There are 4 options: one is tossed out because it has no males in it (female female), which is proven faulty by the shopkeeper This leaves you with 3 options: male female, female male, male male (DING DING DING DING)
There are 4 options: one is tossed out because it has no males in it, which is proven faulty by the shopkeeper This leaves you with 3 options: male female, female male, male male (DING DING DING DING)
1/3 chance that the other one is male
Yeah... I can't put it any simpler than that
Sorry what was the question again? "What is the percentage that the other dog is male?"
Not; "what is the percentage that they are both male?" and even if that was how it was said. You have been told that one of them is male
So it can only be male/female or male/male. Why are people repeating male/female as female/male since you have been told that one is Male already? That is the same thing.
It is 50%
seriously, when I found this forum, I felt it was sound minded people, admittedly a bit touchy about freedom of expression. As in youre not allowed a view if it conflicts with anyone elses.
This is really annoying now, I might have to go out back and shoot myself. And old yella'.
Fire Daemon: The chance that both dogs are male is 25%. Flip two coins, whats the chance of both being heads or both being tales? 25% with a 50% that one will be heads one will be tails.
The same thing applies here. I think. I might have read the question wrong.
Your basic theory is wrong. Each coin toss is individual. If a coin is tossed and lands heads up, if it is tossed again the chance of it falling on heads again is 50%, then if tossed again it is still 50%. This is because each time it is tossed it is equally likely to be heads or tails, every flip is independent of the next.
Back to the question, the chance of the other beagle being female is 50%. If one is male then the other possible combinations are:
Male Female Male Male
As such there is a 50% chance that it will be male female.
So it can only be male/female or male/male. Why are people repeating male/female as female/male since you have been told that one is Male already? That is the same thing.
It is 50%
seriously, when I found this forum, I felt it was sound minded people, admittedly a bit touchy about freedom of expression. As in youre not allowed a view if it conflicts with anyone elses.
This is really annoying now, I might have to go out back and shoot myself. And old yella'.
Yes, but it can't be male/female, since if it were, you would have the male in your hand, and the other one would be female.
And I'm sorry if you feel that we're not allowing you your opinion, but I'm just giving you mine. And since the 50% option is still rising in the poll, where the 33% option is staying where it is, it appears that most people are agreeing with you.
Okay, the question is 'What is the probability that the other one is a male?'. But which dog is the 'other' one? The one that hasn't already been identified as male. As I said, if you eliminated a dog at random, then the probability of the other being a male is 50%. Not that I'm advocating eliminating dogs at random. But you're not eliminating a dog at random, you're eliminating a dog, knowing that it's a male. So the only way that the other one is a male is if they were both male to start with. We know that they weren't both female to start with, so the chances that they were both male is one-third.
Ok, for total clarity;
It is not asking for them to be both male. Even if it was, that would be 50% too.
You know one is male.
So. The starting odds are; Dog 1 - Male, Dog 2 - Female Dog 1 - Female, Dog 2 - Male Dog 1 - Male, Dog 2 - Male, Dog 1 - Female, Dog 2 - Female.
You are told "Dog 1" is Male.
So you cancel out; Dog 1 - Female, Dog 2 - Male And; Dog 1 - Female, Dog 2 - Female.
Now that thats sorted, I can go back to my non-paid job of being infuriated at things people don't understand.
seriously, when I found this forum, I felt it was sound minded people, admittedly a bit touchy about freedom of expression. As in youre not allowed a view if it conflicts with anyone elses.
This is really annoying now, I might have to go out back and shoot myself. And old yella'.
Oh, the hypocrisy in this post.
Also, you are presuming that you KNOW which one of the 2 dogs is male. You do NOT. Dog A can be either male or female, Dog B can be either male or female. All that we DO know, is that they can't both be male. The 50% option is derived from the fact that everybody starts from Dog A being male, then calculating odds of Dog B being male or not. Dog A can be female, however, the very dog you picked to start with.
This is also how I messed up in my first post in this thread, making 2 '2 male' options.
It is not asking for them to be both male. Even if it was, that would be 50% too.
You know one is male.
So. The starting odds are; Dog 1 - Male, Dog 2 - Female Dog 1 - Female, Dog 2 - Male Dog 1 - Male, Dog 2 - Male, Dog 1 - Female, Dog 2 - Female.
You are told "Dog 1" is Male.
So you cancel out; Dog 1 - Female, Dog 2 - Male And; Dog 1 - Female, Dog 2 - Female.
Now that thats sorted, I can go back to my non-paid job of being infuriated at things people don't understand.
You're not told that 'Dog 1' is male. You're told that a dog is male. So you don't cancel out 'Dog 1 - Female, Dog 2 - Male', because one of those dogs is male.
It's not that simple. You don't know that dog one is male and the other is unknown. You know that at least 1 or two is male. That really makes a difference. Suppose the gender of each dog is determined by a coin toss. We have 4 outcomes.
actually i do know it's that simple, and you only have 2 different outcomes not 4 :)
Dog 1 male - from the problem, at least one is male, so this one will always be male, if you move it to dog 2 it doesn't change
Dog 2 - unknown at this time male female
so by following the possibilities we get these outcomes
dog 1 dog 2 male male male female
Every possibility is equal
Female/female is out. That leaves us with male/male and the possibility that only one of them is male. The second option is twice as probable because the male dog can be either one. If you don't believe it, just try it with some coins and keep the record.
yes but with only 2 options it's 50/50 :)
lrn2math (just kidding, many people have problems with stochastic)
English is not my first language so there might be some grammarmistakes or something...
yeah i'm actually pretty good at math problems and this is more of a logic word problem than anything else
i do understand statistics pretty well
way too many ppl are looking too deeply into this. it's not that hard to think of the answer, yes you could possibly get 33% HOWEVER there's a logical flaw in it, not to mention a biological one as well, dogs don't have "sets" of babies in their litters, they don't have equal amounts of male and female babies
We start with no knowledge of the dogs' gender. I'll name the dogs: Sparky and Spot.
So before we call, we have four possibilities for the dogs' gender. All are equally probably. 1 in 4: Sparky Male, Spot Male 1 in 4: Sparky Male, Spot Female 1 in 4: Sparky Female, Spot Male 1 in 4: Sparky Female, Spot Female
These probabilities add up to one, representing all possible combinations.
Now the shopkeeper makes the phone call. We eliminate the Female/Female pairing, so that means we now have only three equally probably situations. In all three cases that remain, the statement "at least one dog is male" is true. We also have no reason to assume that any of the possibilities is more likely than any of the others. We have to adjust our probabilities to add up to 1, since the sum of the probabilities of all possible outcomes equals 1. That 1 in 4 initially given to Sparky and Spot both being female is divided into the three remaining equally likely possibilities.
So before we make the adjustment, we're at: 1 in X: Sparky Male, Spot Male 1 in X: Sparky Male, Spot Female 1 in X: Sparky Female, Spot Male 0: Sparky Female, Spot Female
Dividing 1/4 by 3, we get 1/12. 1/4 + 1/12 = 4/12, or 1/3.
The assumptions in the problem are that male and female are equally likely (not true), the chances of some oddball offspring are zero, and that the fellow bathing the dogs is an idiot who only answers the question literally and doesn't think to say, "Oh, yeah, one is, but the other is a girl," or "Both are."
Blind Punk Riot: I'm sorry, but if any of you go to school still, or college, or university.
Please ask your teacher/tutor this question.
Write the question down word for word and show them it. They will say 50% and laugh at you, depending on how mean they are.
They would laugh at me, but only for bringing them such a simple question.
The sample space has already been said : M/M, M/F, F/M, F/F
Now the events must be sorted correctly; m1: the event that one dog is male. m2: The event that the other dog is male.
You are told that m1 has occured, therefore the probabilty you want is
P(m2 | m1)= P(m2&m1)/P(m1)=0.25/0.75=1/3
The only way m2&m1 occur is the M/M outcome, which has the probability of 0.25. The ways that m1 occur alone are M/M, M/F, F/M, meaning it has a probabiliy of 0.75.
Dog 1 male - from the problem, at least one is male, so this one will always be male, if you move it to dog 2 it doesn't change
That's exactly where the problem comes from. You're assuming that you can call dog A male, since in the situation 'dog A is female, dog B is male', you can just swap the dogs around. Well you can, but you've altered the probability that dog B is male. Because one third of the time dog A was male, and dog B was female. Another third of the time, dog A was female, and dog B was male, but you swapped them, so dog B is now female. The last third, they were both male. So the probability that the other dog was male is 33%.
Blind Punk Riot: I'm sorry, but if any of you go to school still, or college, or university.
Please ask your teacher/tutor this question.
Write the question down word for word and show them it. They will say 50% and laugh at you, depending on how mean they are.
And they would be wrong.
Seriously, Riot. How can you just say that everybody should be entitled to his/her own opinions, then blatantly say that everybody NOT saying 50% is wrong?
Also, WE HAVE EXPLAINED IT. READ IT. TRY TO UNDERSTAND IT.
This will be my last post in here, because some people simply can't think for themselves and go with the mob, and this is simply infuriating me.
Just ADMIT that you're wrong for a change, there's NOTHING wrong with that.
Am I being a hypocrite now? No. This is math, not ethics. Math is an exact science: this means that there is one answer, and ONLY one, if you're asking a simple question, at least. 1+1 is not either 2 or 3, it clearly is 2.
I understand the matter of this question, I understand the answer, and professors have calculated this and say this is a fact.
SO.
2 dogs. 4 possibilities: 1) A is male, B is female 2) A is female, B is male 3) A is male, B is male 4) A is female, B is female.
Shopkeeper says that 1 dog is male. What most people do at this point, is assign the 'male' attribute to Dog A automatically, taking only options 1 and 3 for their 1/2 solution. (That includes you) It is however, never said that Dog A is that male. The option that Dog A is female, and Dog B was the male the shopkeeper talked about, still exists and should NOT be discarded.
So the only real option that is tossed out the window, is option 4.
In the remaining 3 options, namely: 1) A is male, B is female 2) A is female, B is male 3) A is male, B is male
Only one suits the male-male bill. ONE. Out of. THREE.
They would laugh at me, but only for bringing them such a simple question.
The sample space has already been said : M/M, M/F, F/M, F/F
Now the events must be sorted correctly; m1: the event that one dog is male. m2: The event that the other dog is male.
You are told that m1 has occured, therefore the probabilty you want is
P(m2 | m1)= P(m2&m1)/P(m1)=0.25/0.75
And you would be correct. If you weren't told that one of them is male to begin with.
ONE OF THEM IS MALE.
Therefore you only have the probablity of the other one.
One is male, therefore you discount the odds of that one. Thats what you do.
Go to them and tell them what you just said aswell they will seriously lolololololol
But you don't know which one is male, therefore you have to take into account the probability of the other one. The two probabilities are not independant.
They would laugh at me, but only for bringing them such a simple question.
The sample space has already been said : M/M, M/F, F/M, F/F
Now the events must be sorted correctly; m1: the event that one dog is male. m2: The event that the other dog is male.
You are told that m1 has occured, therefore the probabilty you want is
P(m2 | m1)= P(m2&m1)/P(m1)=0.25/0.75
And you would be correct. If you weren't told that one of them is male to begin with.
ONE OF THEM IS MALE.
Therefore you only have the probablity of the other one.
One is male, therefore you discount the odds of that one. Thats what you do.
Go to them and tell them what you just said aswell they will seriously lolololololol
But you don't know which one is male, therefore you have to take into account the probability of the other one. The two probabilities are not independant.
You're telling me, that because the first dog, you can hold that one in your hand. Thats male.
Ok. You know that is male.
How many dogs are left there. theres one there.
It can either be male, or female. the dog youre holding can only be male.
why would you put the male dog down, and pick up the female dog. proclaiming that that makes a different odd?
Seriously, Riot. How can you just say that everybody should be entitled to his/her own opinions, then blatantly say that everybody NOT saying 50% is wrong?
Also, WE HAVE EXPLAINED IT. READ IT. TRY TO UNDERSTAND IT.
This will be my last post in here, because some people simply can't think for themselves and go with the mob, and this is simply infuriating me.
Just ADMIT that you're wrong for a change, there's NOTHING wrong with that.
Am I being a hypocrite now? No. This is math, not ethics. Math is an exact science: this means that there is one answer, and ONLY one, if you're asking a simple question, at least. 1+1 is not either 2 or 3, it clearly is 2.
I understand the matter of this question, I understand the answer, and professors have calculated this and say this is a fact.
SO.
2 dogs. 4 possibilities: 1) A is male, B is female 2) A is female, B is male 3) A is male, B is male 4) A is female, B is female.
Shopkeeper says that 1 dog is male. What most people do at this point, is assign the 'male' attribute to Dog A automatically, taking only options 1 and 3 for their 1/2 solution. (That includes you) It is however, never said that Dog A is that male. The option that Dog A is female, and Dog B was the male the shopkeeper talked about, still exists and should NOT be discarded.
So the only real option that is tossed out the window, is option 4.
In the remaining 3 options, namely: 1) A is male, B is female 2) A is female, B is male 3) A is male, B is male
Only one suits the male-male bill. ONE. Out of. THREE.
Ok, feeling the need to quote myself now, so people would read this. It landed as the last post on page 4, so odds of that being read are pretty low.
Also, don't pm me telling me I'm wrong, Riot. I'm not trying to annoy you, YOU're the one annoying me and winding ME up. Read the post I just quoted. It's in bold. That is where you went wrong in your reasoning.
Wrong, there is not a 50% chance of the other being female.
Look at it this way, on the first dog you have a 50/50 chance, from that you then have another 50/50 chance. This the chance of getting both males is the 2 probabilities times together, thus making it 25%.
Also, Temple, just because you know 1 it doesn't mean that the probability is reduced, there was still a 50% chance that dog could be female which must be taken into consideration.
EDIT 33%? Hmm, well, there are 3 a few possibilitiies of how to get that, if you count up all the possible pairings
Male Male
Female Male
Male Female
Female Female
Male Male
Female Female
Or, 33%.