Question 1189344
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1. A shipment of 10 TV sets contains 3 defective units. If three units are taken for inspection, 
then what is the probability that:
a. all the defective TV sets are included;
b. no defective TV set {{{highlight(cross(shall_be))}}} <U>are</U> included;
c. only one of the defective TV sets {{{highlight(cross(shall_be))}}} <U>are</U> included.
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<pre>
(a)  P = {{{(3/10)*(2/9)*(1/8)}}} = {{{6/720}}} = {{{1/120}}}.



(b)  P = {{{(7/10)*(6/9)*(5/8)}}} = {{{210/720}}} = {{{7/24}}}.



(c)  P = {{{(C[3]^1*C[7]^2)/C[10]^3}}} = {{{(3*21)/120}}} = {{{21/40}}}.
</pre>

Solved.