SOLUTION: The accompanying Venn diagram illustrates a sample space containing six sample points and three events, A, B, and C. The probabilities of the sample points are P(1) = 0.3, P(2) = 0

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Question 1160377: The accompanying Venn diagram illustrates a sample space containing six sample points and three events, A, B, and C. The probabilities of the sample points are P(1) = 0.3, P(2) = 0.2,
P(3) = 0.1, P(4) = 0.1, P(5) = 0.1 and P(6) = 0.2.

attach below is the link to view Venn Diagram

https://ibb.co/55ZBk35
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The probabilities of the sample points are
P%281%29+=+0.3
P%282%29+=+0.2
P%283%29+=+0.1
+P%284%29+=+0.1
P%285%29+=+0.1+
P%286%29+=+0.2
a)
find P%28A%29
elements of set A=1,3,5
add probabilities P%281%29+=+0.3, P%283%29+=+0.1, and+P%285%29+=+0.1
P%28A%29=0.3%2B0.1%2B0.1=0.5

b) P(B ∩ A)=0....reason: (B ∩ A={}, empty set )

c) P( A ∪ B ∪ C)=0.3%2B0.2%2B0.1%2B0.1%2B0.1%2B0.2=1.0
d)+P%28C%5Ec+%29=> C%5Ec+all except 6
P%28C%5Ec+%29=0.3%2B0.2%2B0.1%2B0.1%2B0.1=0.8
e) P( A ∩ C%5Ec ) -> A ∩+C%5Ec is 1,3,5
P( A ∩ C%5Ec )=0.3%2B0.1%2B0.1=0.5
e) P%28B+%2F+A%29-> B+%2F+A+is 2,4
P%28B%2FA%29=0.2%2B0.1=0.3
f) Are A and C mutually exclusive? Why?
Two sets are mutually exclusive (also called disjoint) if they do not have any elements in common; they need not together comprise the universal set.
so, in your case A and C are NOT mutually exclusive because they have one element in common


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!



I've replaced the points with their probabilities"



a) P(A) = sum of probabilities in circle A = 0.1+0.3+0.1 = 0.5

b) P(BᑎA) = sum of probabilities in the overlapping part of 
circle B and A.  This is totally empty so the probability = 0.

c) P(AᑌBᑌC) = sum of all probabilities in circles A, B and C =
0.1+0.3+0.1+0.2+0.2+0.1 = 1.0

d) P(C') = sum of all probabilities EXCEPT those in circle C = 
0.1+0.3+.1 = 0.5

e) P(AᑎC') = sum of probabilities in A which are not in circle C. 
0.1+0.3=0.4

f) P(B|A) = P(BᑎA)/P(A)

P(BᑎA) = 0, from b) above
P(A) = 0.5, from a) above

So P(B|A) = P(BnA)/P(A) = 0/0.5 = 0

f) Are A and C mutually exclusive?

No because the overlapping part of A and C is not empty.

Edwin