Question 1159141
.
There are 50 iPads available for check. Of them, 10 are defective. 
A sample of 3 iPads is randomly selected without replacement. 
What is the probability that at least one of the iPads is defective? 
round to the nearest thousand
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



I edited the original text to make it grammatically correct.
Also,  I removed all the words that are irrelevant,  to transform 
from the  " bla-bla-bla "  style to normal mathematical style.



- - - - - - - - <U>S o l u t i o n &nbsp;&nbsp;1</U> - - - - - - - -


<pre>
An event "At least one is defective" is the complement to the event "No one is defective".

An event "No one is defective" means that our 3 randomly selected iPads are from the set 
of 50-10 = 40 good iPads.


So, we write

    P(at least one is defective = 1 - P(no one is defective) = 1 - {{{(40/50)*(39/49)*(38/48)}}} = 0.496 (rounded).


At this point, the solution 1 is complete.  The answer is  P = 0.496  (rounded).
</pre>



- - - - - - - - <U>S o l u t i o n &nbsp;&nbsp;2</U> - - - - - - - -


<pre>
An event "At least one is defective" is the complement to the event "No one is defective".

An event "No one is defective" means that our 3 randomly selected iPads are from the set 
of 50-10 = 40 good iPads.


So, we write

    P(at least one is defective = 1 - P(no one is defective) = 1 - {{{C(40,3)/C(50,3)}}} = 1 - {{{9880/19600}}} = 0.496 (rounded).


Here we relate  C(40,3) = 9880 triples (combinations), consisting of good iPads only, 
to C(50,3) = 19600, the number of all possible triples (combinations).


At this point, the solution 2 is complete.  The answer is  the same:  P = 0.496  (rounded).
</pre>

Solved completely in two different ways for your better understanding.