If three dice are rolled and sum to 7, what is the probability at least one of the dice is a 1?
Here’s what I did:
$$\Omega = \{\{x,y,z\} : x,y,z \in N , 1 \le x,y,z \le 6\}$$
Define an event such that $A_n$ is a set that sums to 7
Sets that sum to seven
$A_1 = \{5,1,1\}$ , $N(A_1) = {3 \choose 1,2} = 3$
$A_2 = \{2,2,3\}$ , $N(A_2) = {3 \choose 1,2} = 3$
$A_3 = \{1,2,4\}$ , $N(A_3) = n! = 3! = 6$
$A_4 = \{1,3,3\}$ , $N(A_4) = {3 \choose 1,2} = 3$
Define an event such that $B_n$ is all sets that contain at least one die showing numeral 1.
$$P(B|A_n) = \frac{N(A \cap B)}{N(A_n)} = \frac{12}{15}$$
Is this correct?
This is correct. It would be better to say something to justify that you have all the combinations or to list them in an order that makes it clear you have been careful.