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Two identical boxes contain biased coins. Each coin in box $1$ shows head with probability $1/4$, and tails with probability $3/4$, while each coin in box $2$ shows heads with probability $1/3$, and tails with probability $2/3$. Twice we choose one box at random and take a coin from that box. Each of the two coins is then tossed once. If the two coins show the same side, what is the probability that the two coins were taken from the same box?

Currently, I have the following.

Let

$A = \{\text{Box 1 is chosen}\}$, $B = \{\text{Box 2 is chosen}\}$, $C = \{\text{A coin shows heads}\}$, $D = \{\text{A coins shows tails}\}$, $E = \{\text{Both coins show the same side}\}$, $F = \{\text{Both coins are taken from the same box}\}$

Then we must determine the following. $$P(F|E) = \frac{P(E|F)P(F)}{P(E)}$$

Did I solve the following correctly?

$P(F) = 1/2$

$P(C)=P(C|A)P(A)+P(C|B)P(B) = 1/8 + 1/6 = 7/24$

$P(D)=P(D|A)P(A)+P(D|B)P(B) = 3/8 + 1/3 = 17/24$

$P(E)=P(C)P(C) + P(D)P(D) = \frac{7}{24}\frac{7}{24}+\frac{17}{24}\frac{17}{24}=169/288$

I am having trouble solving the following.

$P(E|F)=P(C|A)P(C|A)+P(D|A)P(D|A)+P(C|B)P(C|B)+P(D|B)P(D|B)=1/16+9/16+1/9+4/9=85/72$

What am I doing wrong?

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2 Answers

up vote 2 down vote accepted

This is how I would have done the question, to find $P(F|E)$ rather than $P(E|F)$:

The probability both coins come from the same box and both show the same side $P(E,F)$ is $$\left(\frac14\right)^2\left(\frac12\right)^2 +\left(\frac34\right)^2\left(\frac12\right)^2 + \left(\frac13\right)^2 \left(\frac12\right)^2 +\left(\frac23\right)^2\left(\frac12\right)^2 = \frac{85}{288}.$$

The probability both coins come from the different boxes and both show the same side $P(E,F^C)$ is $$2\left(\left(\frac14\right)\left(\frac12\right)\left(\frac13\right)\left(\frac12\right)+ \left(\frac34\right)\left(\frac12\right)\left(\frac23\right)\left(\frac12\right)\right) = \frac{84}{288}.$$

So you are correct when you say $P(E)=\frac{169}{288}$ as it is the sum of those two terms.

Given both coins show the same side, the probability both coins come from the same box $P(F|E)$ is $$\frac{\frac{85}{288}}{\frac{85}{288}+\frac{84}{288}}=\frac{85}{85+84}=\frac{85}{169}.$$

If you really wanted to calculate $P(E|F)$ then, using $P(F) = \frac12$, it would be ${\frac{85}{288}}/{\frac12} = \frac{85}{144}$ which as leonboy points out is half of your result.

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Where it says "Given both coins show heads", you mean "Given both coins show the same side"? –  joriki Oct 15 '12 at 7:44
    
@joriki: Yes, thank you - I started off in one direction and then changed the numbers –  Henry Oct 15 '12 at 7:52
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The first term in your computation of $P(E|F)$ corresponds to the events "both heads, both from box 1". The probability of that event is

$ P( A_1 C_1 A_2 C_2 | F) = P(C_1 C_2 | A_1 A_2 F) P(A_1 A_2 | F) = P(C|A) P(C|A) \frac{1}{2}$

The same goes for the other terms

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