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ISI MStat 2019 PSA Problem 22 | Basic Probability

This is a beautiful problem from ISI MSAT 2018 PSA problem 13 based on basic counting principles . We provide sequential hints so that you can try .

This is a problem from ISI MStat 2019 PSA Problem 22 based on basic probability principles.

Basic Probability – ISI MStat Year 2019 PSA Problem 22

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

Key Concepts

combination

Basic Probability

Multiplication principle

ISI MStat 2019 PSA Problem 22

A First Course in Probability by Sheldon Ross

Try with Hints

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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