probability

mst_1026

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You have two coins. one is fair and the other has a 3/4 chance of heads on each toss (the tosses are independent). You choose a coin randomly and flip it twice. If you get 1 head and 1 tail, what is the probability you chose the fair coin?


I know that the probability of having heads and tails given that it is a fair coin, is 1/4 and i also know the probability of having heads and tails given that is a biased coin is 3/16. but I dont know how to get P(F) probability a coin is Fair or P(B) the probability the coin is biased
 
I know that the probability of having heads and tails given that it is a fair coin, is 1/4

Not quite - remember that you could throw a head with the first toss and a tail with the second OR you could throw a tail with the first toss then a head with the second. (The same logic applies to the biased coin.)
 
@mst_1026 please reply with your thoughts on this problem. We ALL have to start somewhere. I myself have had to take many backward steps in my own studies. It's good to make sure you understand the pre-requisites before you try to build on them. I think you might have to do some other study before you get into Bayes' Theorem...

I think my post#2 explains why your probability of 1/4 was wrong. Do you have an improved answer for this? If you're still confused then please let me know - hopefully I (or others) can explain. To start off, can you show how you obtained the (incorrect) figure of 1/4? Have you studied probability tree diagrams?
 
@mst_1026 please reply with your thoughts on this problem. We ALL have to start somewhere. I myself have had to take many backward steps in my own studies. It's good to make sure you understand the pre-requisites before you try to build on them. I think you might have to do some other study before you get into Bayes' Theorem...

I think my post#2 explains why your probability of 1/4 was wrong. Do you have an improved answer for this? If you're still confused then please let me know - hopefully I (or others) can explain. To start off, can you show how you obtained the (incorrect) figure of 1/4? Have you studied probability tree diagrams?
the probability of getting 1 head and 1 tail giving that it is a fair coin Pr(HT∣F)= 1/2*1/2= 1/4
 
the probability of getting 1 head and 1 tail giving that it is a fair coin Pr(HT∣F)= 1/2*1/2= 1/4
This is wrong. Have you studied probability?

[MATH]\text {The probability of getting heads on the first toss and tails on the second} = \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}.[/MATH]
That was correct.

[MATH]\text {The probability of getting tails on the first toss and heads on the second} = \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}.[/MATH]
This also is correct.

But you can get one head and one tail by getting a head first and then a tail or by getting a tail first and then a head so

[MATH]\text {The probability of getting one head and one tail in two tosses} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}.[/MATH]
 
This is wrong. Have you studied probability?

[MATH]\text {The probability of getting heads on the first toss and tails on the second} = \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}.[/MATH]
That was correct.

[MATH]\text {The probability of getting tails on the first toss and heads on the second} = \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}.[/MATH]
This also is correct.

But you can get one head and one tail by getting a head first and then a tail or by getting a tail first and then a head so

[MATH]\text {The probability of getting one head and one tail in two tosses} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}.[/MATH]
oh okay i get it. am i also incorrect on my assumption that the probability of having heads and tails given that is a biased coin is 3/16, assuming that You choose a coin randomly and flip it twice.
 
This last response was funny. You say that you understood that you can get a head and a tail in two different orders but then you went and did not double your answer.
 
[MATH]\text {The probability of a head on the first flip and a tail on the second} = \dfrac{3}{4} * \dfrac{1}{4} = \dfrac{3}{16}.[/MATH]
[MATH] \text {The probability of a tail on the first flip and a head on the second} = \dfrac{1}{4} * \dfrac{3}{4} = \dfrac{3}{16}.[/MATH]
[MATH]\therefore \text {The probability of one head and one tail in two flips} \dfrac{3}{16} + \dfrac{3}{16} = \dfrac{3}{8}.[/MATH]
Let’s check.

By the same logic,

[MATH]\text {The probability of two heads on two flips} = \dfrac{3}{4} * \dfrac{3}{4} = \dfrac{9}{16}.[/MATH]
And

[MATH]\text {The probability of two tails on two flips} = \dfrac{1}{4} * \dfrac{1}{4} = \dfrac{1}{16}.[/MATH]
[MATH]\dfrac{3}{8} + \dfrac{9}{16} + \dfrac{1}{16} = 1.[/MATH]
Now do you know Baynes Theorem and what it means?

You still have not said how good you deem your grasp of probability theory.
 
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You have two coins. one is fair and the other has a 3/4 chance of heads on each toss (the tosses are independent). You choose a coin randomly and flip it twice. If you get 1 head and 1 tail, what is the probability you chose the fair coin?
See attached
 

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That's the answer I got. Using Bayes' theorem gives the same answer...

P(A|B) = P(B|A)*P(A) / P(B)

Let A be the event that the fair coin was used
Let B be the event of heads and tails

P(B|A) = 1/2
P(A) = 1/2
P(B) = 1/4 + 3/16 = 7/16

P(A|B) = (1/2) * (1/2) / (7/16) = (1/4) * (16/7) = 4/7

EDIT: Helpers are more likely to check your work if you post a picture, rather than attaching a pdf
 
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Yes that's the neat way of doing it.

Your way is good too, and probably more intuitive. But it's nice to be aware of both ways. Sometimes one method will be easier to use than the other.
 
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