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Let M be the subset of members of a club that are married.
The number of members in this subset is n(M) = 32.
Let G be the subset of members of a club that are managers.
The number of members in this subset is n(G) = 20.
The intersection MG = (M & G) are those members that are married and are managers.
This subset has 8 persons : n(M & G) = 8.
(i) Find the probability that a person chosen at random is a married manager equals P = = = 0.1 = 10%. ANSWER
(ii) The set of those who is EITHER married OR manager is the UNION of the sets M and G, i.e. (M U G).
It has n(M U G) elements, n(M U G) = n(M) + n(G) - n(M & G) = 32 + 20 - 8 = 44. (1)
The rest of the members in the club are NEITHER married NOR managers,
and their number is 80 - 44 = 36.
Therefore, the answer to ii) is P = = = 0.45 = 45%. ANSWER
All questions are answered -- the problem is solved.
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The proof of the formula (1) is easy: the number of elements of the union of any two finite subsets of a universal set
is always equal to the sum of elements of the subsets minus number of elements in their intersection.
On this subject, see the lesson
- Counting elements in sub-sets of a given finite set
in this site.