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

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
Second one
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
.1756
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
so
P (first) + .1756 P( first) = 1
P(first) = .851
so
P(second) = .149 
1/36