Martin throws two dice simultaneously. 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 the 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 :

Question

apoorvk · Answer

hmm.

so the lunch is offered. well that can be three cases

sum is 12 --> only possible when both dice show 6 simultaneously
sum is 7 --> possible when the sum is 7 ---> can be in the following pairs -> (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)  --> 6 ways
all other outcomes ---> a total of 36 - (6 + 1) ways, that is 29 ways.

Now...

apoorvk · Answer

So, since the lunch was offered, the conditional probability of lunch being offered is -->
|dw:1334737041845:dw|


now the probability of the lunch being offered with the outcome being 12, is (the first line in the picture) 1/54


Now if you remember something called 'Bayes theorem", you can probably solve this? can you?? :)