Question

A specially made pair of dice has only one- and two-spots on the
faces. One of the dice has three faces with one-spot and three faces with two-spot.
The other die has two faces with one-spot and four faces with a two-spot. One of the
dice is selected at random and then rolled six times. If a two-spot shows up only once,
what is the probability that it is the die with four two-spots?

Now I have the answer and everything, however I do not understand the work.

chances of 1 two and 5 ones = 6 * 1/2^6 = 6/64 = 3/32 = 0.09375

If you roll the second one 6 times,
chances of 1 two and 5 ones = 6 * 2/3 * 1/3^5 = 12/(3*243) = 12/729 = 4/243 = 0.01646

So the probability it was the second die is
(4/243) / (4/243 + 3/32) = (0.01646 / 0.11021) = 0.14936 or about 15%

The part I don't understand it is the " 6 * 2/3 * 1/3^5" and "6 * 1/2^6", Everything else I understand.
Why is 6 multiplied by (1/2) to the power of 5? and for the second dice, why is 6 multiplied to 2/3 and 1/3 to the power of 5??

Thank you in advance