Box I contains 5 red and 6 black marbles. Box II contains 4 red and 5 black marb

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Box I contains 5 red and 6 black marbles. Box II contains 4 red and 5 black marbles. One marble is drawn at random from Box I and dropped into Box II. A marble is then drawn at random from Box II and dropped into Box I. A marble is now picked at random from Box I. Find the probability that the last marble drawn from Box I is red.

Options:

A . 6231210

B . 7431320

C . 6211210

D . None of the above

Answer: Option C
:
C
A red marble can be drawn from Box I afterthe two rounds of transfers in the following four ways:
Case 1: R

\to

R

\to

R
Probability =

(\frac{5}{11} \times \frac{5}{110} \times \frac{5}{11})

Case 2: R

\to

B

\to

R
Probability =

(\frac{5}{11} \times \frac{5}{110} \times \frac{4}{11})

Case 3: B

\to

B

\to

R
Probability =

(\frac{6}{11} \times \frac{6}{110} \times \frac{5}{11})

Case 4: B

\to

R

\to

R
Probability =

(\frac{6}{11} \times \frac{6}{110} \times \frac{6}{11})

Thus resulting probability = sum of probabilities of individual cases
=

(\frac{5}{11} \times \frac{5}{110} \times \frac{5}{11})

+

(\frac{5}{11} \times \frac{5}{110} \times \frac{4}{11})

+

(\frac{6}{11} \times \frac{6}{110} \times \frac{5}{11})

+

(\frac{6}{11} \times \frac{6}{110} \times \frac{6}{11})

=

\frac{621}{1210}

Therefore, answer option (c) is the correct answer choice.

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