document.write( "Question 1039944: A box contains 15 transistors, 3 of which are defective. If 3 are selected at​ random, find the probability that...
\n" ); document.write( "a. All are defective.
\n" ); document.write( "b. None are defective.
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Algebra.Com's Answer #654700 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "Part a)\r
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\n" ); document.write( "\n" ); document.write( "There is only one way to select a batch of 3 transistors where they're all defective. \r
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\n" ); document.write( "\n" ); document.write( "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. \r
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\n" ); document.write( "\n" ); document.write( "The notation 15 C 3 refers to the combination formula\r
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\n" ); document.write( "\n" ); document.write( "Dividing the two values gives 1/455 = 0.0021978021978. When we round to 3 decimal places, we get 0.002\r
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\n" ); document.write( "Part b)\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "This is still out of the same total of 455 different combos (found earlier in part a)\r
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\n" ); document.write( "\n" ); document.write( "220/455 = 0.48351648351649 which rounds to 0.484 (assuming your book wants you to round to 3 decimal places)\r
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\n" ); document.write( "Summary:
\n" ); document.write( "Answer to part a): 0.002
\n" ); document.write( "Answer to part b): 0.484\r
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\n" ); document.write( "\n" ); document.write( "Answers are approximate. The values are rounded to 3 decimal places.
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