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 October 13th, 2012, 04:45 AM #1 Newbie   Joined: Oct 2012 Posts: 4 Thanks: 0 Probability Problem hello, can you solve this problem: there're 15 bulbs. 3 of them are broken. the bulbs are checked one by one. what's the chance of getting the third broken bulb on the fifth check
 October 13th, 2012, 11:55 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 628 Thanks: 92 Math Focus: Electrical Engineering Applications Re: Probability Problem Here are the possible picks and their probabilities: $BBGGB=\frac{3}{15} \cdot \frac{2}{14} \cdot \frac{12}{13} \cdot \frac{11}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ $BGBGB=\frac{3}{15} \cdot \frac{12}{14} \cdot \frac{2}{13} \cdot \frac{11}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ $BGGBB=\frac{3}{15} \cdot \frac{12}{14} \cdot \frac{11}{13} \cdot \frac{2}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ $GBBGB=\frac{12}{15} \cdot \frac{3}{14} \cdot \frac{2}{13} \cdot \frac{11}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ $GBGBB=\frac{12}{15} \cdot \frac{3}{14} \cdot \frac{11}{13} \cdot \frac{2}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ $GGBBB=\frac{12}{15} \cdot \frac{11}{14} \cdot \frac{3}{13} \cdot \frac{2}{12} \cdot \frac{1}{11}=\frac{3 \cdot 2}{15 \cdot 14 \cdot 13}=\frac{6}{2730}$ The total is $\frac{36}{2730}=0.013187$ Adding a little intelligence: You have six possibilities. In each numerator there will be $1 \cdot 2 \cdot 3$ because of the 3 bad bulbs that will not be cancelled in the denominator and $11 \cdot 12$ that will be cancelled because of the 2 good bulbs. In each denominator there will be $15 \cdot 14 \cdot 13$ that will not be cancelled in the numerator and $11 \cdot 12$ that will be cancelled.
 October 15th, 2012, 02:31 AM #3 Newbie   Joined: Oct 2012 Posts: 4 Thanks: 0 Re: Probability Problem Thanks! now can i understand it..

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