Suppose I flip an unfair coin, weighted such that it has a 75% chance to land on heads, twice. Given that it is the same result both times (both heads or both tails), what is the probability that the outcome was heads both times?
This is a fairly simple question, but I can't remember how to solve it (and others like it). I know that I can't simply find the probability of both flips being heads (which is .5625), so how do I go about solving this?
There are four possible outcomes to two coin flips: HH, HT, TH, TT. Given that the same outcome happened twice, we are left with HH and TT. However, before removing HT and TH we will calculate the probabilities to get said rolls.
$$P(HH)=\alpha^2$$
$$P(HT)=\alpha(1-\alpha)$$
$$P(TH)=(1-\alpha)\alpha$$
$$P(TT)=(1-\alpha)^2$$
where in your case $\alpha=.75$
Now, to get the probabilities when we know that HT and TH weren't rolled, simply do
$$P(HH|HH\text{ or TT})=\frac{P(HH)}{P(HH)+P(TT)}=\frac{\alpha^2}{\alpha^2+(1-\alpha)^2}=\frac{9}{10}$$