stats question

JsBlk, I will attempt to lead you in the correct direction instead of simply providing the answer. If you have any questions, do not fret! Ask!

 

1) I will first deal with the problem where you are instructed to list the outcomes in \(\text{F or G}\) , and the answer you provided was \(\text{F or G}=\{11, 12\}\) . Yes, the elements listed fit the criterion, but there are so many others that you have omitted. It all has to do with the subtle discrepancy between "and" and "or."

 

     When a problem asks you to determine the outcomes in \(\text{F and G}\) , the question is only seeking for the overlap or commonality of both events. In other words, what events occur in both \(\text{F and G}\) ? A few examples may aid your understanding. 

 

• 4, which is in the sample space, occurs in neither \(\text{F}\) nor \(\text{G}\), so it does not fit the criterion \(\text{F and G}\).
• 8, which is in the sample space, occurs in \(F\) , but it does not occur in \(\text{G}\) , so it does not fit the criterion \(\text{F and G}\) . 
• 11, which is in the sample space, occurs in both \(\text{F}\) and \(\text{G}\) , so it does fit the criterion \(\text{F and G}\) .

Therefore, \(\text{F and G}=\{11, 12\}\). Both 11 and 12 occur in both events. No other event overlaps.

 

     When a problem asks you to determine the outcomes in \(\text{F or G}\) , the question is seeking for the union or the merging of both events. In other words, what events occur in either \(\text{F or G}\) ? I will use the previous examples to showcase the difference. 

 

• 4, which is in the sample space, occurs in neither \(\text{F}\) nor \(\text{G}\), so it does not fit the criterion \(\text{F or G}\).
• 8, which is in the sample space, occurs in \(F\) , but it does not occur in \(\text{G}\) , so it does fit the criterion \(\text{F or G}\) .
• 11, which is in the sample space, occurs in both \(\text{F}\) and \(\text{G}\) , so it does fit the criterion \(\text{F or G}\) .Therefore, \(\text{F or G}=\{7,8,9,10,11,12,13,14\}\) . Do you understand the difference?

 

2) Below is your thought process:

 

• In event F, there are 6 equally likely outcomes that can occur out of the 12 in the sample space or S. This means that there is a \(\frac{6}{12}\text{ or }\frac{1}{2}\) chance that an element in event F will occur.

 

• In event G, there are 4 equally likely outcomes that can occur out of the 12 in the sample space or S. This means that there is a \(\frac{4}{12}\text{ or }\frac{1}{3}\) chance that an element in event G will occur.

 

This is a great start, but you are forgetting one aspect of this problem. There is a commonality.  As aforementioned in the first problem, \(\text{F and G}=\{11, 12\}\) . You have essentially double counted because you counted {11,12} in both F and G as possible events. How do you think you should account for this error?