Originally posted by eldragonflythese solutions are getting tired, I think they are simplistic but best of all correct. Even running a simulation which I use as a last resort also demostrated the probability was 2/3. I am beginning to wonder if this is a mere windup.
Wrong, those are pemutations, not combinations. This is just a simple 2 coin toss, disguised as a poorly worded problem.
http://www.knowyourluck.com/coins2u.html
Your answers are ridiculous, and your tired solutions are simplistic.
Originally posted by eldragonfly1 Head and 1 Tail 50% 50%
Wrong, those are pemutations, not combinations. This is just a simple 2 coin toss, disguised as a poorly worded problem.
http://www.knowyourluck.com/coins2u.html
Your answers are ridiculous, and your tired solutions are simplistic.
Above is a quote from your link which is against your logic. Wouldn't you think that there are 3 possibilities: HH, TT, HT? so that's only 33% chance of achieving HT. You apply the same logic for reasoning your solution of the original problem.
Originally posted by mtthwUnless for the sake of argument, you pose an similar question, the youngest is a boy. The question is equivalent and satisfies both condition... and then your logic fails. Again a poorly worded problem, again the same profusion of thick incoherent nonsense.
Surely you can see that the following scenario:
"The youngest child is a boy, the oldest child is a girl"
Satisfies (2) but not (1).
Therefore, they are not equivalent questions. Therefore they can have a different answer.
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?
1) the youngest is either a boy or a girl - 50/50% odds.
2) 2 boys or one boy & one girl - again 50/50% odds.
Originally posted by eldragonflyThe problem isn't poorly worded. You poorly understand it.
Unless for the sake of argument, you pose an similar question, the youngest is a boy. The question is equivalent and satisfies both condition... and then your logic fails. Again a poorly worded problem, again the same profusion of thick incoherent nonsense.
[quote]What are the chances the family has a girl if...
1) the oldest is a boy?
2) they h ...[text shortened]... t is either a boy or a girl - 50/50% odds.
2) 2 boys or one boy & one girl - again 50/50% odds.
1)2 cases:
a)youngest is a boy
b)youngest is a girl
50/50
2)3 cases
a)youngest boy, oldest girl
b)youngest girl, oldest boy
c)both boys
in a and b the family has a girl, in c they don't, hence 2/3
Originally posted by deriver69Wrong. You are misinterpreting a simple word problem and then, almost as if on cue, are stumbling over yourselves to provide the "correct" solution, when the real question here is one of semantics and reasonable interpretation. After all this is just a simple coin toss problem. i kinda' feel sorry for you guys, your redundant and superficial reasoning skills and one-way explanations are quite a sight to behold.
these solutions are getting tired, I think they are simplistic but best of all correct. Even running a simulation which I use as a last resort also demostrated the probability was 2/3. I am beginning to wonder if this is a mere windup.
Originally posted by kbaumenYour second answer is a convenient mis-interpreation, in fact the problem is very poorly worded.
The problem isn't poorly worded. You poorly understand it.
1)2 cases:
a)youngest is a boy
b)youngest is a girl
50/50
2)3 cases
a)youngest boy, oldest girl
b)youngest girl, oldest boy
c)both boys
in a and b the family has a girl, in c they don't, hence 2/3
Pay attention.
At least one child is a boy implies ----->
one boy and another boy
or
one boy and one girl.
1st Child either a boy or girl
2nd Child either a boy or girl
2 outcomes for each birth, 2x2=4
there are 4 outcomes, the fact that two of the outcomes look similar (unless there are many years between the births) is irrelevant.
Leaving aside a girl both times (as we know this has not happened) :- boy then a boy, boy then a girl, girl then a boy
Out of these 3 outcomes two of them have a feminine slant
probability = 2/3
QED
But if you dont believe it try it. I believe it, know it and bloody tried it
Originally posted by eldragonflyone boy and one girl implies -->
Your second answer is a convenient mis-interpreation, in fact the problem is very poorly worded.
Pay attention.
At least one child is a boy implies ----->
one boy and another boy
or
one boy and one girl.
younger brother, older sister
or
younger sister, older brother.
P.S. Deriver69 explained it nicely.
Originally posted by deriver69Your explanation is totally bogus. ðŸ˜
1st Child either a boy or girl
2nd Child either a boy or girl
2 outcomes for each birth, 2x2=4
there are 4 outcomes, the fact that two of the outcomes look similar (unless there are many years between the births) is irrelevant.
Leaving aside a girl both times (as we know this has not happened) :- boy then a boy, boy then a girl, girl then a boy
...[text shortened]... ity = 2/3
QED
But if you dont believe it try it. I believe it, know it and bloody tried it
Might as well solve this stupid thing one more time, to see if I can make it clearer.
Q1. There are 3 cards are in a hat, one GG, one SG, and one SS. One of the cards is drawn at random (i.e. each card is equally likely to be drawn).
(a) What is the probability that the GG was picked?
(b) What is the probability that the SG was picked?
(c) What is the probability that the SS was picked?
A1. Since there are 3 cards, and each card is equally likely to be drawn, we have:
P(GG) + P(SG) + P(SS) = 1
P(GG) = P(SG) = P(SS)
Therefore P(GG) = P(SG) = P(SS) = 1/3.
Q2. The chosen card is now placed on a table in a random manner such that one side is exposed and one is covered (i.e. either side of the card is equally likely to be exposed). Without looking, what is the probability that the exposed side is:
(a) silver?
(b) gold?
A2. There are 3 cards with 2 sides each, so there are 6 possibilities:
P(GG card chosen, G1 face up) = P(GG card chosen)*P(G1 face up) = (1/3)*(1/2) = 1/6
P(GG card chosen, G2 face up) = P(GG card chosen)*P(G2 face up) = (1/3)*(1/2) = 1/6
P(SS card chosen, S1 face up) = P(SS card chosen)*P(S1 face up) = (1/3)*(1/2) = 1/6
P(SS card chosen, S2 face up) = P(SS card chosen)*P(S2 face up) = (1/3)*(1/2) = 1/6
P(SG card chosen, G face up) = P(SG card chosen)*P(G face up) = (1/3)*(1/2) = 1/6
P(SG card chosen, S face up) = P(SG card chosen)*P(S face up) = (1/3)*(1/2) = 1/6
(a) Of the above possibilities, 3 meet the criteria of having a silver side exposed. Therefore, the probability that the exposed side is silver is 3/6 = 1/2.
(b) Of the above possibilities, 3 meet the criteria of having a gold side exposed. Therefore, the probability that the exposed side is gold is 3/6 = 1/2.
Q3. The exposed side is revealed to be silver. Given this information:
(a) what is the probability that the chosen card was SS?
(b) what is the probability that the chosen card was SG?
(c) what is the probability that the chosen card was GG?
A3. For brevity, let's rename the probabilities given in Q2 above as follows:
P(GG card chosen, G1 face up) = P(GG,G1) = 1/6
P(GG card chosen, G2 face up) = P(GG,G2) = 1/6
P(SS card chosen, S1 face up) = P(SS,S1) = 1/6
P(SS card chosen, S2 face up) = P(SS,S2) = 1/6
P(SG card chosen, G face up) = P(SG,G) = 1/6
P(SG card chosen, S face up) = P(SG,S) = 1/6
(a) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, only P(SS,S1) and P(SS,S2) meet the criteria of being the SS card. Therefore, the probability that the card chosen was SS given that a silver side has already been exposed is given as:
P(SS chosen given silver side exposed) = (P(SS,S1) + P(SS,S2)) / (P(SS,S1) + P(SS,S2) + P(SG,S))
= (1/6 + 1/6) / (1/6 + 1/6 + 1/6)
= 2/3.
(b) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, only P(SG,S) meets the criteria of being the SG card. Therefore, the probability that the card chosen was SG given that a silver side has already been exposed is given as:
P(SG chosen given silver side exposed) = P(SG,S) / (P(SS,S1) + P(SS,S2) + P(SG,S))
= (1/6) / (1/6 + 1/6 + 1/6)
= 1/3.
(c) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, none of them meet the criteria of being the GG card. Therefore, the probability that the card chosen was GG given that a silver side has already been exposed is given as:
P(GG chosen given silver side exposed) = 0 / (P(SS,S1) + P(SS,S2) + P(SG,S))
= 0 / (1/6 + 1/6 + 1/6)
= 0
Originally posted by kbaumenWrong. You tried to "fancy" up a simple coin toss problem, but you have failed. And your faked explanations are doomed to self-induced mediocrity.
one boy and one girl implies -->
younger brother, older sister
or
younger sister, older brother.
P.S. Deriver69 explained it nicely.
Originally posted by eldragonflyThat is pure rhetoric. Can you explain away the fact that when I actually simulated this (despite knowing I didnt need to) the results were consistent with 67% not 50%?
Your explanation is totally bogus. ðŸ˜
Incidently I have a friend who was teased by his older sister, another friend who teased his younger sister, they are clearly two different things.
(PS have another friend who loved her younger brother but she is another story)
Originally posted by PBE6Now you are supplying randomness, when before you couldn't be bothered. In a word super-bogus. 😉
Might as well solve this stupid thing one more time, to see if I can make it clearer.
Q1. There are 3 cards are in a hat, one GG, one SG, and one SS. One of the cards is drawn at random (i.e. each card is equally likely to be drawn).
(a) What is the probability that the GG was picked?
(b) What is the probability that the SG was picked?
(c) What is t given silver side exposed) = 0 / (P(SS,S1) + P(SS,S2) + P(SG,S))
= 0 / (1/6 + 1/6 + 1/6)
= 0