Conditional Probability with coins (Edited with Progress) You have two coins that look identical, but one of them is fair and the other is weighted. The weighted coin has a 3/4 probability to flip heads and a 1/4 probability to flip tails. You forget which coin is which, so you test each coin by flipping it once. If one coin flips heads while the other flips tails, you will guess that the coin that flipped heads is the weighted coin. If both coins flip the same result, then you will guess at random which coin is weighted. Using this method, what is the probability that you will guess correctly?
I already know the answer is 5/8, but my textbook didn't offer a great solution, so I would like to see how other people solve it.
Progress: Let P(X) = P(Fair Head intersect Unfair Head) + P(Fair Tail intersect Unfair Tail) = 1/2. The complement of P(X) is also 1/2. The probability of guessing given X occurs is 1/2.  The probability of guessing given X does not occur is 3/4. Therefore the probability of guessing is (1/2)(1/2) + (1/2)(3/4) = 5/8.
 A: Let the probability of flipping a heads for the fair and unfair coins be $p_1=\frac{1}{2}$ and $p_2=\frac{3}{4}$ respectively.
Let 'SR' be taken to mean 'same results', 'DR' for 'different results' and 'CG' for 'correct guess'.
Flipping the coins,
$$P(\textrm{SR}) = p_1p_2 + (1-p_1)(1-p_2)=\frac{1}{2}$$
$$P(\textrm{DR}) =\frac{1}{2} $$
Now,
$$P(\textrm{CG})=P(\textrm{CG and DR})+P(\textrm{CG and SR})$$
$$=P(\textrm{CG}\,|\,\textrm{DR})P(\textrm{DR})+P(\textrm{CG}\,|\,\textrm{SR})P(\textrm{SR})$$
$$=P(\textrm{unfair is Heads}\,|\,\textrm{DR}).\frac{1}{2}+\frac{1}{2}.\frac{1}{2}$$
$$={\frac{3}{4}.\frac{1}{2}\over\frac{1}{2}}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}$$
$$=\frac{5}{8}$$
A: I think you could use a good bit more detail ...
$FH$: fair coin comes up heads.  $P(FH) =\frac{1}{2}$
$FT$: fair coin comes up tails. $P(FT) =\frac{1}{2}$
$UH$: unfair coin comes up heads. $P(UH) =\frac{3}{4}$
$UT$: unfair coon comes up tails. $P(UT)=\frac{1}{4}$
$X$: same outcome. $X=(FH \cap UH) \cup (FT \cap UT)$.
$$P(X)= P(FH \cap UH) \cup (FT \cap UT) = \text{ (mutually exclusive events) } $$
$$=P(FH \cap UH) + P(FT \cap UT) = \frac{1}{2}*\frac{3}{4}+\frac{1}{2}*\frac{1}{4}=\frac{1}{2}$$
Hence, $P(X^C)=1-P(X)=\frac{1}{2}$
$G$: guess correctly
$P(G|X)=\frac{1}{2}$ since if both come up same then random pick.
$$P(G|X^C) = \frac{P(FT \cap UH)}{P(X)}=\frac{\frac{1}{2}*\frac{3}{4}}{\frac{1}{2}}=\frac{3}{4}$$
So:
$$P(G) = P(G \cap X) +P(G \cap X^C)= P(X)*P(G|X)+P(X^C)*P(G|X^C)=$$
$$ \frac{1}{2}*\frac{1}{2}+\frac{1}{2}*\frac{3}{4}=\frac{5}{8}$$
A: $P(\text{the same})=\frac12$ and $P(\text{guess right }|\text{ the same})=\frac12$
$P(\text{different})=\frac12$ and
$\overbrace{\left.\begin{align}
\text{weighted heads: }\tfrac34\cdot\tfrac12&=\tfrac38\text{ (guess right)}\\  
\text{weighted tails: }\tfrac14\cdot\tfrac12&\textstyle=\tfrac18\text{ (guess wrong)}
\end{align}\right\}
P(\text{guess right }|\text{ different})=\frac34}^{\text{assuming different}}$
$\begin{align}
P(\text{guess right})
&=P(\text{the same})\cdot P(\text{guess right }|\text{ the same})\\
&+P(\text{different})\cdot P(\text{guess right }|\text{ different})\\[6pt]
&=\frac12\cdot\frac12+\frac12\cdot\frac34\\[6pt]
&=\frac58
\end{align}$
