In laboratory 1, there are 20 computers, of which 14 are Dell, 6 are HP. In laboratory 2, there are 20 Dell computers and 10 HP computers. A new laboratory is being formed, for which 9 computers from lab 1 are taken, and 6 computers from lab 2 into the new laboratory. From this new laboratory, we randomly select a computer. What is the probability that the chosen computer is a Dell computer?
How I would solve this:
Let $A$ denote the event that a Dell computer is chosen. Let $M_i$ denote the event that the $i$-th laboratory is chosen.
$P(M_1) = 9/15$ because we pick $9$ computers from laboratory $1$.
$P(A\mid M_1) = 14/20$.
$P(M_2) = 6/15$ because we pick $6$ computers from laboratory $2$.
$P(A\mid M_2) = 20/30 = 2/3$.
$P(A) = P(M_1) \times P(A\mid M_1) + P(M_2) \times P(A\mid M_2) = 9/15 \times 14/20 + 6/15 \times 20/30 = 103/150$
Now, knowing how tricky probability tasks are, I am not quite confident this is correct. Can someone please go through my answer, and if wrong correct it? Thanks in advance.