Suppose that Monty Hall (on TV's Let's Make a Deal) asks you to choose between three doors: #1, #2, and #3. Behind a random door is a new Rolls Royce. Behind each of the other two doors is a goat. Let's assume that you would prefer a Rolls Royce to a goat. You choose a door. Now, Monty, who knows which door hides the Rolls Royce, shows you a goat behind one of the two doors that you did not choose. He then gives you the opportunity to change your choice. Assume that Monty always does this, regardless of your guess. Should you change your choice?
taken from here w/ solution and some misc. explainations
There are 10 black socks and 10 white socks(no left-right distinction) in the wardrobe. Your task is to draw the minimum number of socks at random to be sure you have a pair of a single color.
How many socks should you draw?
taken from here w/ solution
The local weatherman says No Rain, and his record is 2/3 accuracy of prediction. But the Federal Meteorological Service predicts rain, and their record is 3/4.With no other data available, what is the chance of rain?
taken from here w/ solution
5 comments:
I got the first two correct but I am unclear after looking at the solution for the third one.
Logically, you can't have both services be both right, nor both wrong as that would be a contradiction, since they predicted mutually exclusive outcomes.
However the solution link uses the combined probability of one right and the other wrong (e.g. 1/4 x 2/3 = 1/6) and the reverse case (3/4 x 1/3 = 1/4) as if these component probabilities are independent, but they do not add to 1. It then considers them together as the full range of possibilities and normalizes them to 1.
It's like saying that the component probabilities for the other two impossible cases (both right, both wrong) are invalid, but the components for the possible cases are valid. Take not notice that they don't add to 1, we can just re-ratio them and call that a probability. Isn't there something wrong here? Can anyone explain how this works better than the web solution?
too many thoughts... i'll separate them by a lot of dots ................
first you have to assume that they can both be right and can both be wrong
there are four total scenarios:
A and B are right (2/3 * 3/4 = 1/2)
A and B are wrong (1/3 * 1/4 = 1/12)
A is right, B is wrong (2/3 * 1/4 = 1/6)
A is wrong, B is right (1/3 * 3/4 = 1/4)
but it doesn't matter
Probability of rain is completely indeterminable from the above situations. It can rain 100% of the time and one of them can say it's not going to rain, but that doesn't change the chance of rain.
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what's going on in this question is that it may or may not be raining, but the forecasters disagree, which happens 5/12 of the time. that is, the probability that they'd disagree is 5/12. which is completely unrelated to whether or not it's raining.
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a better way to put it is to say that the forecasters have no bearing on the chance of rain
example: say you take a bowl and throw it at the ground. it has X chance of shattering. if you've thrown 10 bowls before, and i've been right 8 times, you could say i'm right 4/5 of the time. however, my record has no bearing on what's about to happen because, as they say, "probability has no memory." actually, what they say is "the dice have no memory," which is why casinos can stay in business. you can play roulette and have it land on 00 a million times in a row, but that has no bearing on the next throw.
it's the
i hope that makes sense
That made sense up until the "it's the" line. Then you lost me:P
did a LOLCAT ever say: i'm in ur attempts at posting, truncating your sentences?
i wonder what the rest of that line was ...
I like the second one about the socks. The answer was immediately obvious, so I thought for sure I had to be missing something. So I read it again, still seemed obvious. Read it again, still obvious. Finally I went to the solution cause I just couldn't believe that was all there was to it. That was all there was to it...
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