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Six identical-looking coins are in a box, of which five are unbiased, while the sixth comes up heads with probability 3/4 and tails with probability 1/4. Three coins are chosen from the box at random and removed. One of those three is chosen at random and tossed three times, coming up heads every time.

(a) What is the probability that the final coin selected was the biased coin, given that it came up heads each time?

(b) What is the probability that the biased coin is amongst the three coins removed from the box?

My answer is as follows;

To pick 3 fair coins at the start the probability is ${5 \choose 3}/{6 \choose3}$

to pick two fair and the biased is ${5\choose 2}/{6 \choose 3}$

probability of fair coin getting 3 heads is 1/8

probability of biased coin getting 3 heads is 27/64

to pick a fair coin from 3 fair coins is 1 to pick a fair coin from 2 fair and biased is 2/3 and to pick biased is 1/3

therefore to pick 3 fair and then get 3 heads is (10/20)(1/8) = 1/16

to pick 2 fair and a biased, then pick a fair and get 3 heads is (10/20)(2/3(1/8) = 1/24

to pick 2 fair and a biased then pick the biased and get 3 heads is similarly 9/128

This gives me an answer to part a of;

$\frac{9/128}{9/128 + 1/16 + 1/24}$

and part b is; 1/2

Is this way of working out the answer correct and also is there an easier way to do it?

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  • $\begingroup$ I think (b) is wrong if you take into account that the coin came up heads three times. $\endgroup$ – Pieter21 Nov 30 '15 at 12:44
  • $\begingroup$ Ah i read the question wrong, so it would be the same answer as a except the top of the fraction would be + 1/24 $\endgroup$ – KingJ Nov 30 '15 at 13:29
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(a)

First of all: there is no essential difference between "selecting $1$ out of $3$ that are on their turn selected out of $6$" and "selecting $1$ out of $6$".

If $B$ denotes the event that the biased coin was selected and $E$ the event that $3$ heads show up by $3$ tosses then: $$P(B\mid E)P(E)=P(B\cap E)=P(E\mid B)P(B)=\left(\frac34\right)^3\frac16=\frac{27}{384}$$

Also: $$P(E)=P(E\mid B)P(B)+P(E\mid B^c)P(B^c)=\left(\frac34\right)^3\frac16+\left(\frac12\right)^3\frac56=\frac{67}{384}$$ hence: $$P(B\mid E)=\frac{27}{67}$$

b)

Let $A$ denote the event that the biased coin was among the $3$ selected coins. Then: $$P(A\mid E)P(E)=P(A\cap E)=P(E\mid A)P(A)=P(E\mid A)\frac12$$

and it remains to find: $$P(E\mid A)=\left(\frac34\right)^3\frac13+\left(\frac12\right)^3\frac23=\frac{27}{192}+\frac{16}{192}=\frac{43}{192}$$

We end up with: $$P(A\mid E)=\frac{43}{67}$$

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Use https://en.wikipedia.org/wiki/Conditional_probability#Kolmogorov_definition

With A = coin is biased, B = three coin flips are all heads.

$P(A \cap B) = 1/6 * 27/64$

$P(B) = 5/6 * 1/8 + 1/6 * 27/64$

So $P(A|B) = 27/67$

(b) can be done in a similar way.

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