Posted by Anonymous on .
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?
First one, roll 6 times
p is a two = 1/2
p not a two = 1/2
p a two one out six rolls
= (6,1) (6!)/[(1)!(5!)] (1/2)^1(1/2)^5
= 6 (1/2)^6
p is a two = 2/3
p not a two = 1/3
p a two one out of 6 rolls
= 6 (2/3)^1 (1/3)^5
= 6 (2)/3^6
P second one / p first one = 2/3^6 /1/2^6
= 2 (2/3)^6
Now IN THIS EXPERIMENT where all other outcomes are thrown outthe probability that it is the first + the probability that it is the second = 1
P (first) + .1756 P( first) = 1
P(first) = .851
P(second) = .149