Question 1039944
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Part a)


There is only one way to select a batch of 3 transistors where they're all defective. 


There are 15 C 3 = (15!)/(3!*(15-3)!) = 455 ways to select 3 resistors. This is where we don't care if they work or not. 


The notation 15 C 3 refers to the <a href="http://www.mathwords.com/c/combination_formula.htm">combination formula</a>


Dividing the two values gives 1/455 = 0.0021978021978. When we round to 3 decimal places, we get 0.002


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Part b)


There are 15-3 = 12 non-defective transistors. There are 12 C 3 = (12!)/(3!*(12-3)!) = 220 ways to pick 3 transistors where none of them are defective.


This is still out of the same total of 455 different combos (found earlier in part a)


220/455 = 0.48351648351649 which rounds to 0.484 (assuming your book wants you to round to 3 decimal places)


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Summary:
Answer to part a): <font color=red>0.002</font>
Answer to part b): <font color=red>0.484</font>


Answers are approximate. The values are rounded to 3 decimal places.  
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