SOLUTION: I'm stuck at a probability Question and I don't know how to post it here so I have attached a picture! please help with the question in red block .
http://imgur.com/a/h5RXu
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Question 1086535: I'm stuck at a probability Question and I don't know how to post it here so I have attached a picture! please help with the question in red block .
http://imgur.com/a/h5RXu
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
 = 1-P(\overline{A}))
 = 1-\frac{2}{3})
 = \frac{1}{3})
---------
Now create a Venn diagram. I'll call this figure 1.
The overlap is important here (as you'll see in figure 2). The fact that )
is nonzero indicates that overlap is needed between sets A and B.
Note: U is the universal set.
---------
Now turn to figure 2.
The change here is that I've added two shaded regions. The green region is the portion inside circle A but it is not in circle B. The area of this green region will be exactly equal to the probability your book wants for part (iii). In other words,
area of green region = )
Note how the orange region is the remaining part inside circle A, and its also in circle B as well. The orange region is the area corresponding to the probability . So this means,
area of orange region =
The green and orange regions do not overlap. The regions combine to compose set A. They completely partition set A.
This is an example of The Law of Total Probability
Because of this fact, we can say
 + P(A \cap B) = P(A))
 + \frac{1}{4} = \frac{1}{3})
 + \frac{1}{4} - \frac{1}{4} = \frac{1}{3} - \frac{1}{4})
 = \frac{1}{3} - \frac{1}{4})
 = \frac{4}{12} - \frac{3}{12})
 = \frac{4-3}{12})
 = \frac{1}{12})
Answer: 1/12 (which is approximately 0.0833 in decimal form)
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