Question

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

Answer 1

For the two dice, we are given here as:
Dice 1:  Three faces with a​ one-spot and three faces with a​ two-spot

Dice 2: Two faces with a​ one-spot and four faces with a​ two-spot

Therefore, for the two dice, we have the probabilities here as:
P( one-spot | Dice 1) = 3/6 = 0.5
P( one-spot | Dice 2) = 2/6 = 1/3

Using law of total probability, we have the probability of getting 3 one-spot side out of the 6 throws as:
= 0.5*( (6c3)*0.5⁶ + (6c3)*(1/3)³*(2/3)³)  

= 0.2660

The probability that the dice 1 was selected given that there were three one spot dice in 6 throws is computed using Bayes theorem as:
P( dice 1 | 3 one-spot side out of the 6 throws)

= P( dice 1 and 3 one-spot side out of the 6 throws) / P( 3 one-spot side out of the 6 throws)

= 0.5*(6c3)*0.5⁶ / 0.2660

= 0.5874

Therefore 0.5874 is the required conditional probability here.