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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?

- 3/44
- 9/44
- 13/44
- 7/44

combination

Basic Probability

Multiplication principle

But try the problem first...

Answer: is 3/44

Source

Suggested Reading

ISI MStat 2019 PSA Problem 22

A First Course in Probability by Sheldon Ross

First hint

Let’s algorithmify the whole procedure.

Select positions of two defectives and the last defective being already fixed. This can be done in \( {6 \choose 2} \)

Second Hint

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.

Final Step

Hence the probability that the seventh bulb sold is the last defective one is \( \frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 } \)

Content

[hide]

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?

- 3/44
- 9/44
- 13/44
- 7/44

combination

Basic Probability

Multiplication principle

But try the problem first...

Answer: is 3/44

Source

Suggested Reading

ISI MStat 2019 PSA Problem 22

A First Course in Probability by Sheldon Ross

First hint

Let’s algorithmify the whole procedure.

Select positions of two defectives and the last defective being already fixed. This can be done in \( {6 \choose 2} \)

Second Hint

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.

Final Step

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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