ISI MStat 2019 PSA Problem 22 | Basic Probability

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 \( \frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 } \)

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 \( \frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 } \)