Hello, I would like to get help on this probability problem that has a and b. At a sit-down restaurant, 40% of the tables have children at them. Among the tables with children, 30% of the orders include a glass of milk. Among the tables without children, only 10% of the orders include a glass of milk. Suppose we randomly choose an occupied table from this restaurant. a) What is the probability that order from that table includes a glass of milk? b) Given that the order includes a glass of milk, what is the probability there are children at the table?

Take the number of tables 100. 

Now these tables are divided into 2 parts of 40 and 60 

40 tables are occupied and 60 are not occupied 

Now it is given that 30% of the occupied tables orders a glass of milk 

This means 30% of 40 orders a glass of milk  = 12 

and 10% of the non-occupied tables orders a glass of milk this means = 6 

Randomly choose an occupied table from this restaurant. 

a) What is the probability that order from that table includes a glass of milk?

Probability = 12/40

b) Given that the order includes a glass of milk, what is the probability there are children at the table

P(E1) = P(occupied table) = 40/100 

and P(E2) = P(non-occupied table) = 60/100 

P(A/E1) = P(milk order from occupied table) = 12/40 

P(A/E2) = P(milk order from non-occupied table) = 6/ 60 

Now, the probability of order of milk from occupied tables P(E1/A) 

P(E1/A) = P(E1)P(A/E1)/( P(E1)P(A/E1) + P(E2)P(A/E2)) = 40/100 x 12/40 / (40/100 x 12/40 + 60/100 x 6/60) 

= 12/(12+6) = 2/3