Question

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.

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is 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.