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February 14th, 2016, 08:54 PM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  Add dad
One container holds the letters D A D and a second container holds the letters A D D. One letter is chosen randomly from the first container and added to the second container. Then a letter will be chosen from the second container. a) What is the probability that the second letter chosen is D if the first letter was A? If the first letter was D? b) What is the probability that the second letter chosen is A if the first letter was A? If the first letter was D? My attempt: a) If the first letter chosen is A, then the second container now contains A A D D. Therefore P(second letter chosen is D) = 2/4 = 1/2 If the first letter chosen is D, then the second container now contains A D D D. Therefore P(second letter chosen is D) = 3/4 b) If the first letter chosen is A, the second container now has A A D D. Therefore, P(the second letter chosen is A) = 2/4 = 1/2 If the first letter chosen is D, the second container now has A D D D. Therefore, the P(second letter is A) = 1/4 Am I correct? Is there a different/better/easier way to solve this? Thanks. 
February 15th, 2016, 01:12 AM  #2 
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 
You have given two different probabilities for each part. A single answer needs to be given for each part.

February 15th, 2016, 01:16 AM  #3 
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  
February 15th, 2016, 01:21 AM  #4 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
Both containers contain the letters A, D and D. Does their order have any relevance? This problem can be neatly solved with a table. 
February 15th, 2016, 01:22 AM  #5 
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 
You say that if the letter chosen is A or if the letter chosen is B. But, while choosing the letter, we don't know what has been chosen (because if we know, we don't need the concept of probability). I think that after finding the two possible outcomes, you should find their mean and write it as the final answer (in each part).

February 15th, 2016, 01:32 AM  #6 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
For the first container, you could say that: A_1, D_1 and D_2 exist. For the second container, you could say that: A_2, D_3 and D_4 exist.  If D_1 is selected from the first container and placed in the second container, in the second container: A_2, D_1, D_3 and D_4 would exist. If D_2 is selected from the first container and placed in the second container, in the second container: A_2, D_2, D_3, and D_4 would exist. If A_1 is selected from the first container and placed in the second container, in the second container: A_1, A_2, D_3 and D_4 would exist.  These are all possible scenarios. You can think of the A's and D's as unique objects.  Using the logic above you can deduce that: a.1) P(E_1)=1/2 a.2) P(E_2)=3/4 b.1) P(E_3)=1/2 b.2) P(E_4)=1/4 Last edited by perfect_world; February 15th, 2016 at 01:38 AM. 
February 15th, 2016, 01:38 AM  #7  
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230  Quote:
 
February 15th, 2016, 01:41 AM  #8 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
I'm just pedantic. @Shunya wanted a solid explanation as to why his results were correct. I gave him a detailed and robust response. I don't have to water anything down.
