Counting Problem (ways of choosing 3 with conditions)

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SUMMARY

The problem involves selecting 3 light bulbs from a total of 13, consisting of 9 good and 4 defective bulbs, with the requirement of having exactly 1 defective bulb in the selection. The correct calculation involves using combinations, specifically C(4,1) for choosing 1 defective bulb and C(9,2) for choosing 2 good bulbs, resulting in a total of 144 valid combinations. The initial miscalculation of 288 arises from incorrectly applying the counting rule without considering the specific conditions of the selection.

PREREQUISITES
  • Understanding of combinations, specifically C(n, r)
  • Basic principles of counting and probability
  • Familiarity with combinatorial problems
  • Ability to differentiate between good and defective items in a selection
NEXT STEPS
  • Study the concept of combinations in depth, focusing on C(n, r) calculations
  • Learn about conditional probability in combinatorial contexts
  • Explore advanced counting techniques, such as the inclusion-exclusion principle
  • Practice similar problems involving selections with specific conditions
USEFUL FOR

Students studying combinatorics, educators teaching probability concepts, and anyone interested in solving counting problems with specific conditions.

mishima
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Homework Statement


Given 9 good light bulbs and 4 defective light bulbs, how many ways can you select 3 such that you get exactly 1 defective bulb?

Homework Equations


C(n,r)

The Attempt at a Solution


I understand total ways to select is C(13,3)=286, and that total ways to select only good bulbs is C(9,3)=84. Subtracting these would give 202 ways to get 1 or more defective bulb. Not sure how to further differentiate these cases...answer provided is 144.

If I just use rule of counting, I can pick from 4 bad, 9 good, then 8 bad in any order which is 288...double the provided answer.
 
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mishima said:

Homework Statement


Given 9 good light bulbs and 4 defective light bulbs, how many ways can you select 3 such that you get exactly 1 defective bulb?

Homework Equations


C(n,r)

The Attempt at a Solution


I understand total ways to select is C(13,3)=286, and that total ways to select only good bulbs is C(9,3)=84. Subtracting these would give 202 ways to get 1 or more defective bulb. Not sure how to further differentiate these cases...answer provided is 144.

If I just use rule of counting, I can pick from 4 bad, 9 good, then 8 bad in any order which is 288...double the provided answer.
In how many ways can you select 1 defective bulb?
In how many ways can you select 2 good bulbs?
 
C(4,1)*C(9,2). Thank you.
 

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