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July 26th, 2018, 02:13 PM  #1 
Newbie Joined: Jul 2018 From: Florida Posts: 3 Thanks: 0  Conditional probability question  with beers!
Hi, I'm trying to solve the following question, but I'm not quite sure how to use Conditional probability rules (or Bayes Theorem): ***************************** It is known that 23 percent of young people like to drink German beer, 63 percent of young people like American beer and 28 percent like water. (for an individual person, it is possible to like all the mentioned drinks) What about elderly people? 65 percent of elderly people like to German beer 27 percent like American beer and 33 percent like water. A random chosen person likes water and American beer. What is the probability the he is young? ************************** Can anyone help pls? TNX 
July 26th, 2018, 03:20 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102 
I believe you'll need $P[young] = 1  P[old]$ to complete the problem 
July 27th, 2018, 08:25 AM  #3 
Newbie Joined: Jul 2018 From: Florida Posts: 3 Thanks: 0  hmm... ok, so if I do the following:
I actually need to calculate: P(young  American & water) = P(American & water  young) * P(young) / P(American & water) The problem is that I don't know how to calculate P(American & water  young) by using the given details... Also  P(young) is not given? Is it possible to solve the question without it? Thanks again! 
July 27th, 2018, 10:02 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102  Quote:
the second factor in the numerator is P(young) as I said you are going to need that to continue.  
July 27th, 2018, 03:25 PM  #5 
Newbie Joined: Jul 2018 From: Florida Posts: 3 Thanks: 0 
Ok, thanks, suppose P(young) is given and it's 0.4. How do I calculate P(American & water  young) and P(American & water)?

July 28th, 2018, 09:49 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102  Quote:
Looking at the given probabilities, and also assuming that $P[\text{young} \cup \text{old}] = 1$ and also obviously that $\text{young} \cap \text{old} = \emptyset$ we have $P[\text{American}] = P[\text{American}\text{young}]P[\text{young}] + P[\text{American}\text{old}](1P[\text{young}])$ $P[\text{American}] = (0.63)(0.4) + (0.27)(0.6) = 0.414$ You can do the same for water. The probability of both is then the product of the two individual probabilities as they are independent.  

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