Question 1118494
.
In a survey of 200 employees of a company regarding their 401(k) investments, the following data were obtained.

<pre>
127		had investments in stock funds.
97		had investments in bond funds.
60		had investments in money market funds.
43		had investments in stock funds and bond funds.
36		had investments in stock funds and money market funds.
36		had investments in bond funds and money market funds.
22		had investments in stock funds, bond funds, and money market funds.

(a) What is the probability that an employee of the company chosen at random had investments in exactly two kinds of investment funds? 
  

(b) What is the probability that an employee of the company chosen at random had investments in exactly one kind of investment fund? 
  

(c) What is the probability that an employee of the company chosen at random had no investment in any of the three types of funds? 
</pre>

<B>Solution</B>


<pre>
We are given

S = 127
B =  97
M =  60

SB = 43
SM = 36
BM = 36

SBM = 22.


(c)  Then  

     n(S U B U M) = S + B + M - SB - SM - BM + SBM =

                  = 127 + 97 + 60 - 43 - 36 - 36 + 22 = 191

     is the number of those  who had investment at least in one of the three types of funds.

     Hence, the number of those who had NO investment in any of the three types of funds was 200 - 191 = 9.

     Thus the answer to the question (c) is  {{{9/200}}} = 0.045.


(a)  The number of those who had investments in exactly two kinds of investment funds is

     (SB - SBM) + (SM - SBM) + (BM - SBM) = 

   = (43 - 22)  + (36 - 22)  + (36 - 22) = 49.

      Hence, the answer to question (a) is  {{{49/200}}} = 0.245.



(b)  The number of those who had investments in exactly one kind of investment fund is 

      (S - SM - SB + SBM) + (B - SB - BM + SBM) + (M - SM - BM + SBM) = 

    = (127-43 - 36 + 22)  + (97 - 43 - 36 + 22) + (60 - 36 - 36 + 22) = 120. 

      Hence, the answer to question (b) is  {{{120/200}}} = 0.6.
</pre>

The solution is completed.