Two coin tossed simultaneously If two coins are tossed simultaneously 4 times, What will be the probability of getting exactly two heads? 
Suppose that one knows that there is at least one toss that yields two heads, compute the probability that there is exactly one toss that yields two heads?
The sample space is too large to do counting. Any shorter way to solve this?
 A: Two (presumably fair) coins being thrown four times each is equivalent to one coin being thrown eight times - order is irrelevant as each event is independent. We can model this with a binomial distribution: the number of heads, $X$, is given by $X \sim B(8, 0.5)$.
$P(X=2) = \binom{8}{2}0.5^2(1-0.5)^6 = 28\times\frac{1}{256} = 0.109375$ (or $\frac{7}{64}$).
If you're not familiar with the binomial distribution, the $\binom{8}{2}$ is the number of ways that two heads out of 8 could be tossed (i.e. 1st and 2nd; 1st and 3rd; 1st and 4th; ...; 7th and 8th). Then $0.5$ is the probability of a head - this event must occur twice, so we square it, and $1-0.5$ is the probability of a tail - this occurs six times, so we raise it to that power.
For the next question, let's say $Y$ is the number of times we roll two heads with our two dice. I believe you're asking for $P(Y=1\;|\;Y\ge1)$. We can again model $Y$ with a binomial distribution, noting that the probability that both coins are heads is a quarter: $Y\sim B(4, 0.25)$.
Now, $P(Y=1) = \binom{4}{1}0.25^1(1-0.25)^3 = \frac{27}{64}$.
$P(Y\ge1) = 1 - P(Y = 0) = 1 - (1-0.25)^4 = 1-\frac{81}{256} = \frac{175}{256}$.
So the probability that the former is true, given that the latter is, is: $\frac{27}{64} \div \frac{175}{256} = \frac{27\times256}{64\times175} = \frac{108}{175}\approx0.6171$.
