homework - probability problem A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?
My Attempt:
The sample space for the above problem, $S = 6^3$.
Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5, 6\}, \{4, 6, 5\}]$
Thus, the total number of possible sets are 2. Thus, the probability for the required event (say $A$),
$$P(A) = \frac{2}{216}$$
$$\therefore P(A) = \frac{1}{108}$$
In my book, the answer given is $\frac{1}{5}$. How's is my answer?

Since the sum of the 3 numbers thrown is 15 it follows that the sample space is given by 10 elements: 15=6+6+3 ($3!/2!=3$ ways), 15=6+5+4 (3!=6 ways), 15=5+5+5 ($3!/3!=1$ way).
There are two cases when $4$ is the first value 4,5,6 and 4,6,5. 
Therefore the probability is $p=2/10=1/5$.
