SOLUTION: 32 of the 80 members of a club are married and of these, 8 are managers. There are 20 managers in total in the club. Find the probability that a person chosen at random is; i) a

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