probability of single event given probability of multiple events If the probability of an unfair coin landing heads 2 times out of 3 tosses is (0.9^2)(0.1)(3) find the probability of heads, i.e., the probability of success.
--I'm having trouble approaching this question.
 A: $\binom{3}{2}\cdot 0.9^2\cdot 0.1$ is the probability of two successes (and one failure) in a Binomial Distribution with parameters $n=$what? and $p=$what?.
A: If we denote the probability of heads by $p$, then the probability of two heads out of three tosses is
$$
P = \binom{3}{2}p^2(1-p) = 3p^2(1-p)
$$
We wish to set this equal to $3(9/10)^2(1/10)$.  One way is to do this by inspection, which yields an obvious value of $p = \, ?$
As Henry points out in the comments, however, there is another solution, which isn't surprising since the equation
$$
3p^2(1-p) = 3\left(\frac{9}{10}\right)^2\left(\frac{1}{10}\right)
          = \frac{243}{1000}
$$
is a cubic and may have as many as three solutions.  However, to correspond to a physical solution, we must also impose the constraint $0 \leq p \leq 1$.  If we add $3p^2(1-p)$ to both sides and expand, we get
$$
3p^3-3p^2+\frac{243}{1000} = 0
$$
Cubics are, in general, a messy matter to solve, but in this case, we are fortunate to already know a value by inspection.  Call this value $p_1$, and then to find other solutions, we assert $p \not= p_1$, so we can divide both sides by $p-p_1$:
$$
\frac{3p^3-3p^2+\frac{243}{1000}}{p-p_1} = 0
$$
Carrying out this division yields
$$
3p^2-\frac{3}{10}p-\frac{27}{100} = 0
$$
(Technically, I've given away $p_1$ by giving the result of the division, but I think that won't be much of a spoiler.)  This can be solved by means of the quadratic formula, and yields the other two roots:
$$
p_{2,3} = \frac{-10 \pm \sqrt{100-4(100)(9)}}{200}
        = \frac{-1 \pm \sqrt{37}}{20}
$$
Of these $p_3 < 0$ and therefore can't be used as a physical solution, but $p_2 \doteq 0.35414$ and therefore satisfies the conditions.
In fact, since $3p^2(1-p)$ has a simple maximum at $p = 2/3$ and is otherwise increasing in $(0, 2/3)$ and decreasing in $(2/3, 1)$, it follows that any coin (except for one with a $p = 2/3$ probability of falling heads) has a "twin" that produces two heads in three flips with the same probability.  If one coin comes up heads with probability $p_1$, the other comes up heads with probability
$$
p_2 = \frac{1-q+\sqrt{1+2q-3q^2}}{2}
$$
Note that this other coin is a "twin" only in the sense of having the same probability of generating two heads in three tosses, and otherwise has a much different distribution.
