SOLUTION: Three sets A,B and C and the universal set, U, are such that n(A)=21, n(B)=30, n(C)=30, n(A∩B)=11, n(B∩C)=17, n(A∩C)=13, n(A U B U C)'=13, n(U)=60 (U

Algebra ->  Probability-and-statistics -> SOLUTION: Three sets A,B and C and the universal set, U, are such that n(A)=21, n(B)=30, n(C)=30, n(A∩B)=11, n(B∩C)=17, n(A∩C)=13, n(A U B U C)'=13, n(U)=60 (U      Log On


   

Question 73262: Three sets A,B and C and the universal set, U, are such that
n(A)=21, n(B)=30, n(C)=30, n(A∩B)=11, n(B∩C)=17, n(A∩C)=13, n(A U B U C)'=13, n(U)=60 (U=union, '=complement)
if P(X) is the probability that a randomly chosen element from the universal set is in set X determine:
a) P(A∩B∩C)
b) P(A U B)
c) P(A U C)'
d) P(A∩B∩C|A U B U C)
e) P(A|(B U C)')
f) P(A'|B')
THANKYOU
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
There are 8 disjoint regions:
A∩B∩C, A∩B∩C', A∩B'∩C, A∩B'∩C', A'∩B∩C, A'∩B∩C', A'∩B'∩C, A'∩B'∩C'. 

The "sieve" formula for 3 sets:

n(AUBUC) = n(A)+n(B)+n(C)-n(A∩B)-n(A∩C)-n(B∩C)+n(A∩B∩C)

[If you haven't had this sieve formula, please post again, or email
me AnlytcPhil@aol.com and I can show you how to do it without that
formula.  But it's easier if you have it.]

We also know

n(A∩B∩C) = n(U) - n[(A∩B∩C)'] = 60 - 13 = 47

Substituting what we have in the sieve formula:

47 = 21 + 30 + 30 - 11 - 13 - 17 + n(A∩B∩C)

47 = 40 + n(A∩B∩C)

n(A∩B∩C) = 7  

From that we can get the number of elements in each of the other
seven disjoint regions:

1. n(A∩B∩C) = 7
2. n(A∩B∩C') = n(A∩B)-n(A∩B∩C) = 11-7 = 4 
3. n(A∩B'∩C) = n(A∩C)-n(A∩B∩C) = 13-7 = 6
4. n(A'∩B∩C) = n(B∩C)-n(A∩B∩C) = 17-7 = 10
5. n(A∩B'∩C') = n(A)-n(A∩B∩C')-n(A∩B'∩C)-n(A∩B∩C) = 21-4-6-7 = 4
6. n(A'∩B∩C') = n(B)-n(A∩B∩C')-n(A'∩B∩C)-n(A∩B∩C) = 30-4-10-7 = 9
7. n(A'∩B'∩C) = n(C)-n(A∩B'∩C)-n(A'∩B∩C)-n(A∩B∩C) = 30-6-10-7 = 7

a) P(A∩B∩C)

n(A∩B∩C)=47, so P(A∩B∩C) = 47/n(U) = 47/60 

----------------------

b) P(AUB)

n(AUB)=n(AUBUC)-n(A'B'∩C) = 47-7 = 40, so

P(AUB) = n(AUB)/n(U) = 40/60 = 2/3

-------------------------

c) P[(A U C)'] 

n(AUC) = n(AuBuC)-n(A'∩B∩C') = 47 - 9 = 38, so

P(AUC) = 38/40 = 19/20, and

P[(AUC)'] = 1 - P(AUC) = 1 - 19/20 = 1/20
 
------------------------

d) P(A∩B∩C|A U B U C) =

 P[(A∩B∩C)n(AUBUC)]      P(A∩B∩C)     n(A∩B∩C)/n(U)     7/60 
-------------------- = ----------- = --------------- = ------- = 7/47
     P(AUBUC)            P(AuBuC)     n(AuBuC)/n(U)     47/60


------------------------

                  P(An(BUC)')                            P(A∩B'∩C')
e) P[A|(BUC)'] = ------------- = (by DeMorgan's law) = ------------- = 
                   P[(BUC)']                              P(B'∩C')

   n(A∩B'∩C')/n(U)        4/60            4
  ----------------- = ------------- = ----------
     P(B'∩C')          n(B'∩C')/60     n(B'∩C')

Since n(B'∩C') = n(A∩B'∩C')+n(A'∩B'∩C') = 4+13 = 17, the above = 4/17

------------------------

               P(A'∩B')                           P[(AUB)']
f) P(A'|B') = ---------- = (by DeMorgan's law) = ----------- =  
                 P(B')                              P(B')

 1 - P(AUB)                        1 - 2/3           1/3
------------ = [By prob (b)] = --------------- = ----------- = 
  1 - P(B)                      1 - n(B)/n(U)     1 - 30/60 

   1/3       1/3 
--------- = ----- = (1/3)(2/1) = 2/3 
 1 - 1/2     1/2 

Edwin