Probability Confusion: Find Fran & Ron's Coin Toss Result

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The discussion revolves around a probability problem involving three coins: two fair coins and one two-headed coin. Fran and Ron each select a coin at random, and the task is to find the probability that both get heads when they toss their coins. The correct probability calculation is based on the likelihood of selecting the unfair coin, which is 2/3, compared to the 1/3 chance of selecting two fair coins. The confusion arises from misunderstanding the distribution of coin types, where the presence of two fair coins increases the chances of selecting at least one unfair coin. Clarifying these probabilities helps resolve the confusion regarding the selection process.
barneygumble742
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hi...this question is from a sample midterm that we went over in class today. i still don't understand certain things. I'm hoping someone can explain it to me.

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consider three identical-looking coins. Two of the coins are ordinary fair coins (H and T are equally likely to occur on any toss), but the third coin is 2-headed (both sides of the coin show H). Fran and Ron each choose a coin at random, and the remaining coin is discarded. Suppose that Fran and Ron toss their coins simultaneously.

Find the probability that Fran gets H and Ron gets H.
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the answer is (1/3)*(1/2)*(1/2)+(2/3)*(1)*(1/2)
the explanation is...you have 1/3 chance of getting a fair coin (heads and tails)
once you get a fair coin, the possibility of getting heads on the first coin is 1/2
and the possibility of getting heads on the second coin is also 1/2
plus
you have 2/3 chance of getting an UNfair coin (heads and heads)
once you get an UNfair coin, the possibility of getting heads on one coin is 1
and the possibility of getting heads on the second coin is also 1/2

could someone please explain why the possibility of choosing the UNfair coin is 2/3 and not 1/3?
you have 2 fair coins so i think that you have double the chances of getting a fair coin than an UNfair coin.

to me it would make perfect sense if the 1/3 and the 2/3 are switched.

thanks,
mark
 
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They are comparing the probability of getting AT LEAST ONE unfair coin out of 2, to getting two fair coins. The probabilities of those events are 2/3 and 1/3 respectively.
If the three coins are labelled U, F1, F2 for unfair, fair1, fair2
then the possible combinations you can get are
U, F1
U, F2
and
F1, F2
So the probability you get at least one unfair coin is 2/3 and the probability of 2 fair coins is 1/3.
 
F R neither
f1 f2 u
f1 u f2
f2 f1 u
f2 u f1
u f1 f2
u f2 f1
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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