A trial consists of tossing a fair coin twice and noting H = number of heads observed. What is the probability that if 5 trials are performed, we will note H=0 two times, H=1 one time, and H=2 two times?
So I'm deciding what kind of distribution this questions falls under. I cross about bernoulli trials and think it is some type of combination type question, but I'm still stuck on going further. I know that in 1 trial, H has the following possibilities H=0, H=1, or H=2. But moving forth to 5 trials I'm stuck. Any ideas?
The random variable $H$ is clearly follows a binomial distributed with parameters $n = 2$, $p = 0.5$. But the random variable $\boldsymbol X = (N_0, N_1, N_2)$ where $N_i$ is the number of times $H = i$ in $5$ trials, follows a multinomial distribution with parameters $\eta = 5$, $\pi_i = \Pr[H = i]$. That said, it is not necessary to know the form of the distribution of $\boldsymbol X$ in order to calculate $\Pr[\boldsymbol X = (2,1,2)]$. The idea is to first compute the individual probabilities $\pi_i = \Pr[H = i]$, assume the outcome of distinct trials are independent, and determine the number of ways you can observe $N_0 = 2$, $N_1 = 1$ and $N_2 = 2$. THat is to say, if I have a 3-sided die numbered $0$, $1$, and $2$, then how many distinct sequences of length $5$ can I roll such that the number of zeroes is $2$, the number of ones is $1$, and the number of twos I roll is $2$?