A bag contains 99 red marbles and 99 blue marbles. Taking two marbles out of the bag, you:

• put a red marble in the bag if the two marbles you drew are the same color (both red or both

blue), and

• put a blue marble in the bag if the two marbles you drew are different colors.

Repeat this step (reducing the number of marbles in the bag by one each time) until only one

marble is left in the bag. What is the color of that marble?

Number of Marbles in the Bag = [ 99 red marbles + 99 blue marbles]

Now, When you take two marbles out of the Bag, following are the conditions given

⇒put a red marble in the bag if the two marbles you drew are the same color (both red or both blue)

⇒ put a blue marble in the bag if the two marbles you drew are different colors.

Now, Following are the possibilities

1. When you draw two red marble each time from the bag, after each draw number of marbles left in the bag

(99 R+99 B) ⇒ (98 R+99 B) ⇒ (97 R+99 B) ⇒ (96 R+99 B) ⇒ ... (1R+99B) ⇒Then there are two Possibilities

(2R+97 B) - -Both are Blue.

or

(99B) - -If one is red and one is Blue.

2. When you draw two Blue marble each time from the bag, after each draw number of marbles left in the bag

(99 R+99 B) ⇒ (100 R+98 B) ⇒ (101R+97 B) ⇒ (102 R+96 B) ⇒ ... (197R+1B) ⇒Then there are two Possibilities

(196R+1B) - -Both are Red.

or

(196R + 1B) - -If one is red and one is Blue.

3. When you draw One Blue and One Red, marble each time from the bag, after each draw number of marbles left in the bag

(99 R+99 B) ⇒ (98 R+99 B) ⇒ (97 R+99 B) ⇒ (96 R+99 B) ⇒ ... (1R+99 B) ⇒ (0R+99 B) ⇒ (1R+97B) ...

In this case also, there are two possibilities.

R=Red

B=Blue

Can't Be predicted what will be the color of last Marble in all the three cases.