Q:

Probability: Confused between probability of event vs sequence

I am taking online Probability MOOC by MIT. While solving question 2.b assignment 4
The newest invention of the 6.041/6.431 staff is a three-sided die with faces numbered 1, 2, and 3. The PMF for the result of any one roll of this die is,
$p_X(x) = \begin{cases}
    1/2   & \quad \text{if } x = 1\\
1/4  & \quad \text{if } x = 2\\
1/4  & \quad \text{if } x = 3\\
0  & \quad \text{otherwise}\\
  \end{cases}$
Consider a sequence of six independent rolls of this die, and let Xi be the random variable corresponding to the ith roll.
(a) What is the probability that exactly three of the rolls have result equal to 3?
(b) What is the probability that the first roll is 1, given that exactly two of the six rolls have
result of 1?
My answer for a is ${6 \choose 3} * (1-1/4)^3 * (1/4)^3$, but similar logic for b fails as my answer is $1/2 * {5 \choose 1} * (1-1/2)^4 * (1/2)$ but correct answer is $\frac{5}{{6 \choose 2}}$.
Can someone please help me understand why am I wrong?

A:

For a) this is just a binomial distribution. Probability of a $3$ is $p=1/4$. Then the probability of getting exactly three times a $3$ in six rolls is
$$P[X=3]=\binom{6}3 p^3(1-p)^{6-3}=20\cdot\frac1{4^3}\cdot\frac{3^3}{4^3}\approx 0.132$$
For b) this is just a matter of combinatorial theory. You have already a sequence with two $1$'s and you want to know how many of them start by a $1$. There are $\binom{6}2=15$ possible distinct combinations with exactly two $1$'s, of them there are $5$ that start with a $1$ (fix the first number and count how many combinations of exactly one $1$, in a sequence of five numbers, exists), thus the probability is just $5/15=1/3$.

