Question

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={5, 6, 7, 8, 9}​, and event G={9, 10, 11, 12}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the number 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 complete your choice.A. F or G = { _____ }​(Use a comma to separate answers as​ needed.)B. F or G = { }Find P(F or G) by counting the number of outcomes in F or G. P(F or G) =________​(Type an integer or a decimal rounded to three decimal places as​ needed.)Determine P(F or G) using the general addition rule. Select the correct choice below and fill in any answer boxes within your choice.​(Type the terms of your expression in the same order as they appear in the original expression. Round to three decimal places as​ needed.)A. P(F or G)=_______+_______−________=_______B. P(F or G)=_______+_______=________

Answer 1

Answer:

A) F or G = {5, 6, 7, 8, 9, 10, 11, 12}; B) P(F or G) = 8/12 = 0.667

Step-by-step explanation:

The outcomes in event F are 5, 6, 7, 8 and 9.  The outcomes in event G are 9, 10, 11 and 12.  This means that the event F or G, containing all elements of both events, will be

{5, 6, 7, 8, 9, 10, 11, 12}.  Note that we do not count 9 twice; it only appears in the sample space once.  There are 8 elements out of 12 total in the sample space; this means the probability is 8/12 = 0.667.

Using the addition rule, we begin with P(F).  There are 5 elements in this event; this makes P(F) = 5/12.  

For P(G), there are 4 elements in G; this makes P(G) = 4/12.

To add them, we add P(F) and P(G) but then must take out the element counted twice; this gives us

5/12+4/12-1/12 = 9/12-1/12 = 8/12 = 0.667.

Answer 2

We have that

• F&G={5, 6, 7, 8, 9, 10, 11, 12}
• $P(FG)=0.667$
• $P(F or G)=5/12+4/12-(1/12)\\\\ P(F or G)=5/12+1/4$

From the question we are told

• S={ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14},
• F={5, 6, 7, 8, 9}
• G={9, 10, 11, 12}

a)

Generally

The P(F or G) by counting the number of outcomes in F or G is

F&G={5, 6, 7, 8, 9, 10, 11, 12}

b)

P(F or G) using the general addition rule

$P(FG)=\frac{N_f}{N_g}\\\\P(FG)=\frac{8}{12}$

$P(FG)=0.667$

c)

Generally Using addition Rue to attain P(F or G) we have

For P(F or G)

$P(F or G)=PF+PG-P(F+G)$

$P(F or G)=5/12+4/12-(1/12)\\\\ P(F or G)=5/12+1/4\\\\ P(F or G)=\frac{8}{12}$

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