<s>[INST] 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?
 [/INST] You have to consider all the possible ways in which you can obtain $15$.
Those possibilities are:
$\{3, 6, 6 \}$, $\{6, 3, 6 \}$, $\{6, 6, 3 \}$
$\{5, 5, 5 \}$
$\{4, 5, 6 \}$, $\{4, 6, 5 \}$, $\{5, 4, 6 \}$, $\{5, 6, 4 \}$, $\{6, 4, 5 \}$, $\{, 5, 4 \}$
Total = $10$
There are only $2$ cases in which $4$ is the first element, so
$$P = \frac{2}{10} = \frac{1}{5}$$
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