A probability experiment is conducted in which the sample space of the experiment is S= {3,4,5,6,7,8,9,10,11,12,13,14}, event F= {7,8,9,10,11,12}, and event G = {11,12,13,14}. Assume that eeach outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the nummber of outcomes in F or G. Determine P(F or G) using the general addition rule.

List the outcomes in F or G. Select the correct choice below and, if necessary, fill in the answer box to compute your choice.

F or G = {11, 12}

Find P(F or G) by counting the number of outcomes in F or G.

P(F or G) = [__] round to the 3rd decimail place (I think the answer is .833)

(Because 6\12 outcomes = 1\2 divide the numerator and the denominator = .5

4\12 outcomes = 1\3 divide the numerator and the denominator = .33333

add .5+.33333 = .833 I think I have it all wrong though so I could use some help)

Now I need ro determine the answer using the addition rule P(F or G) = __ + __ - ___= ? Round to 3rd decimal place.

I dont know the corret numbers to input in the equation in order to do the math

JsBlk Jun 2, 2018

#1**+3 **

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?

TheXSquaredFactor Jun 2, 2018

#2**+1 **

should i not count 11 and 12 in the out come?

If i do that it would mean F= 4\12=.33333 and G= 2\12= 16667 and that = 5

am i correct

JsBlk
Jun 3, 2018

#3**+2 **

No, you should consider the overlap in your answer. Just do not count them twice.

By the way, I did not tell you the addition rule. It may help you. \(P(\text{F or G})=P(\text{F})+P(\text{G})-P(\text{ F and G})\).

- P(F)=6/12
- P(G)=4/12
- P(F and G)=2/12 <-- This is the overlap

Plug this into the formula, and you have the answer.

TheXSquaredFactor
Jun 3, 2018

#4**+2 **

Here is a way to visualize it. I am going to use a Venn Diagram.

- In F, 6/12 are equally likely outcomes
- In G, 4/12 are equally likely outcomes

If you were to add these together, like you did as your first attempt, then you would get 0.833 again. Notice, though, that you have 11 and 12 in both, and you had counted them in both times for both F and G. You can only count them once!

TheXSquaredFactor
Jun 3, 2018