painfully easy I think the solution given in the tech interview site is wrong.
michele
Knowing "one" of the coins came up heads narrows it down to THREE choices:
Brad Corbin
I'm not convinced.
michele
You're making the mistake of thinking of HT and TH as the same outcome when it isn't.
Lou
Guys, I know that TH and HT are different (and, by the way, this is why flipping two coins is more probable to obtain one T and one F than TT or HH, 50% vs 25%). My point is that you have to guess the result of ONLY ONE COIN (the original question was "what is the chance that the other coin also came up heads?"), not of a combination of the two: so, since it can't be influenced from other launches, you have 50%.
michele
I think you are going to have to break out two quarters and a pad of paper to convince yourself. I'm serious, try it.
Brad Corbin
One final comment:
Brad Corbin
First I thought Michele is right because the two incidents (flip of coin 1 and flip of coin 2) are independent.
Steve Li
When you say the last/other one "is also a heads" there are two ways to interpret that:
saberworks
First, here's the riddle: "i flip a penny and a dime and hide the result from you. "one of the coins came up heads", i announce. what is the chance that the other coin also came up heads?"
Steve
The answer seems so obvious that I dont understand why people are discussing it in so much details. The two tosses are independent. Nowhere the problem says that the coins are biased or their results are independent. So the answer is 0.5 or 50% no matter how many coins you flip or how many times you flip a same coin. Meaning if I toss a coin 1000 times and each time I get a head, the probability that I get a head on 1001th toss is still 1/2 !!!
Mitesh Vasa
Nope. :)
Brad Corbin
Yup, the 50% theory is wrong. The chance of any one flip is 50% but that's not the scenario presented. Now I'm not silly enough to think the whole world speaks english which is why I posted code. Code is understandable by all.
Steve
Why do you people persist with this 50% nonsense? Let me rephrase it so that you better understand it. Here is the experiment. Try it yourself:
Aditya K. Prasad
It'e funny how the inevitable "try it yourself" argument always comes with much more detailed instructions than are implied by the problem in question. Where do you get step 2 from?
Ham Fisted
The second step comes from a process to actually set up the conditions of the test described. We are told that 'at least one of the coins is heds', which is logically equivalent to saying that 'not both of them are tails'. Cases where both coins are tails are therefore rejected from the sample set. We run the tests on the sample cases remaining.
David Clayworth
I like this problem. People get fooled because the wording is deceiving (not quite a "trick question", but close). Here's a clearer version of the problem:
Dave Howard
The question is *slightly* ambiguous. For instance, the interviewer may never say "one of" unless he means "only one of". But it's traditional in this sort of problem to ignore that sort of thing. IF SO, the answer is unambiguous. Possible rephrasings to make it COMPLETELY clear:
Jack
The question tests your understanding of conditional probabilities and reduced sample spaces.
Chuck Boyer
Glenn C. Rhoads
To make it clear, the way I read the problem is,
Jason Owen
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