Martin throws two dice spontaneously.If the sum of the outcomes is 12,he offers lunch at a five star hotel with probability 2/3. If the sum is 7,he offers lunch with probability 1/2.In all other cases,he offers lunch with probability 1/3.Given that the lunch was offered,the probability that the sum of the outcomes equals 12 is a)1/18 b)1/20 c)1/24 d)1/36
this question is based on bayes theorem
We have to calculate that the lunch was offered, what is the probability that the sum of the outcomes equals 12.As lunch can be offered in three cases that sum was 12,that sum was 7 or else sum of number on the two dices.
Applying bayes theorem,
P(12/L)=probability that 12 was the sum when lunch was offered.
P(7/L)=probability that 7 was the sum when lunch was offered.
P(O/L)=probability that sum was other than 7,12 when lunch was offered.
Here, we have:
chance that sum will be 12=1/36,only(6,6) pair
chance that sum will be7=6/36, (6,1),(5,2), (4,3) ,(3,4),(2,5),((1,6)
Remaining cases=36-7=29 chance for these cases=29/36
P(12/L)= 2/3*1/36+ ½*6/36+1/3*29/36=1/20