bonanova
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Everything posted by bonanova
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If this helps the discussion ... and I'm not sure it will ... For itachi-san.. embolden: verb: give encouragement to In other words, you make a person bold by emboldening them. To refer to words that were put into bold type, I don't know of a word for that. For LIS.. embold/ed - not a word, as you point out. On the other hand, no one said that it was. Great. That's settled. Back to the debate.
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Nice one,
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Hi ash013, Try thinking of it this way. For families with two children, opposite genders occur half the time. But two boys happen only 1/4 of the time, and two girls happen 1/4 of the time. So the odds of two boys can never be the same as the odds of one boy and one girl. Do you see how that is different from just asking the gender of one particular child?
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Yup. Bonus points awarded for correcting the oops before any of us other geniuses did.
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I guess we all were had? The "one key" that will help solve this puzzle turned out in fact not to be the Shift key [heh] but the ASCII. <_< I bow to the OP's superior wit.
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grey cells has it.
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Nope. Altho there does seem to be a pattern. I'd try something different.
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Nope. Good thought tho.
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So, echo ... guess you looked the formula up after all.
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Here's a hint. With the assumption of equal probability of boy/girl at each birth, families with two children will have opposite gender children half the time and same gender children half the time. [boys 1/4 and girls 1/4]. That should help you with the math.
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Using that approach, you come up with the result that of the population of families with two children, at least one of which is a boy, [only] half of them have one boy and one girl. Are you comfortable with that result?
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1 3 1 2 2 8 3 3 1 4 3 0 5 3 1 6 3 0 7 3 1 8 3 . . . . What is the next term? Is the sequence infinite? If not, how many terms are there?
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It may be easier to see this way.... 6 9 7 73 67 71 9 19 6 13 8 ?[/code][spoiler=looks like]15[/spoiler]
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A valid way to test an hypothesis about probability is to calculate a sample space and count the favorables. You'd need to be able to write a simple program and have access to a random number generator. Sometimes just translating the conditions of the OP into code, without even running the code, clarifies things. I'm very interested to hear some of the claims being made in this thread tested in this way. Post your results and your code. Corollary to proof by computer simulation: Take your theories of probability to Las Vegas. The credence of your theory stands in direct correlation to the length of time that your net worth increases.
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Stacking and/or folding get you more pieces. They're not forbidden in the OP. Imran, [post 25] I'm not sure how you get your results for 4 and 5 cuts. Can you make a sketch?
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Think about your case 1 and case 2. You are counting the same case [she has two boys] twice. If "he" has a younger brother [case 2], then his younger brother [who has equal right to be called "he"] has an older brother [= case 1]. Think about how the OP differs from this question: [this is the question that you answered; try to see how it differs from the OP] The woman has a son, named John. John has a sibling. What are the odds that John's sibling is a girl? Take your four cases, substitute John for he, and you get the correct answer. ----------------- Step 1. Enumerate the set of all equally likely outcomes Step 2. Count the favorable outcomes. Sept 3. Take the ratio. Everyone "gets" this procedure, but Step 1 must be done with a sharp eye for all and an unbiased eye for equally likely. shaunarenee combined two equally likely cases BG and GB into one [twice as likely] case. You took one of the equally likely cases BB and made it into two.
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Ask yourself: of the families that have two children, what are the odds that there is a boy and a girl? I would say there are four equally likely cases: BB BG GB GG and two of them are favorable, so the odds are 50% You would say, When working probability Girl-Boy and Boy-Girl is the same thing. so there are 3 cases - two boys, two girls, and one each. Three cases, and one of them is favorable, so the odds are 1/3. Are you still comfortable with your reasoning?
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It's been a while, but I think
