Posted by **Anonymous** on Thursday, July 31, 2014 at 4:53pm.

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?

- math -
**Damon**, Thursday, July 31, 2014 at 6:15pm
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

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