This is a problem from ISI MStat 2019 PSA Problem 22 based on basic probability principles.
A shopkeeper has 12 bulbs of which 3 are defective. She sells the bulbs by selecting them at random one at a time.
What is the probability that the seventh bulb sold is the last defective one?
combination
Basic Probability
Multiplication principle
Answer: is 3/44
ISI MStat 2019 PSA Problem 22
A First Course in Probability by Sheldon Ross
Let’s algorithmify the whole procedure.
Select positions of two defectives and the last defective being already fixed. This can be done in \( {6 \choose 2} \)
Let’s count the number of ways the defectives can sit once their places are fixed. This can be done in 3! ways .
Now the position of the non-defectives are fixed.
Now let’s compute in how many ways they can sit there together in the remaining four places.
We have to arrange 9 non-defective objects in 4 places.
This can be done in 9 x 8 x 7 x 6 ways.
And without any restrictions we can select 7 bulbs in \( 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \) ways.
Hence the probability that the seventh bulb sold is the last defective one is \( \frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 } \)
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