Question 1206336
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You need to schedule three clubs for picture day in the spring. 
Of the 124 students in clubs, 28 of them are in NHS, 54 of them are in student governement 
and 20 are in both. While trying to avoid double scheduling, if the photographer chooses 
a student in clubs at random, what is the probability that the student 
is in either NHS or student government but not both?
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<pre>
The formula is 

    P = ({{{28/124}}} + {{{54/124}}} - {{{20/124}}}) - {{{20/124}}}.


The aggregate of the first three terms is the probability that 
the randomly chosen student is in the union of NHS and government
(belong to one or another set). The fourth term in the formula is to 
exclude the intersection (to provide "but not in both").


Simplify and get a final number

    P = {{{(28+54-20-20)/124}}} = {{{42/124}}} = {{{21/62}}} = 0.33871  (rounded).
</pre>

Solved.