Understanding Conditional Probability with a Simple Marble Example

[Rajeev Bagra]

Why the Answer Is Two-Thirds: A Conditional Probability Marble Problem Explained with Tables

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Problem (Restated Simply)

A bag has one marble.
It is green or blue, each with a fifty-percent chance.

A green marble is added to the bag.
Now there are two marbles.

One marble is taken out at random.
The marble removed is green.

Question:
What is the chance that the marble left in the bag is also green?

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Step 1: List all possible outcomes

Only outcomes that can actually happen are listed.

Case	Initial marble	Marble removed	Marble remaining	Probability
1	Green	Green	Green	One half
2	Blue	Green	Blue	One quarter
3	Blue	Blue	Green	One quarter
Total				One

The probabilities add up to one, so the table is complete.

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Step 2: Use the given information

We are told that the marble removed is green.

So we discard the case where a blue marble was removed (Case 3).

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Step 3: Remaining possible cases

Case	Marble remaining	Original probability
1	Green	One half
2	Blue	One quarter

The total probability of these remaining cases is three quarters.

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Step 4: Adjust probabilities (conditioning)

Since we are now only looking at situations where a green marble was removed, we must rescale the probabilities so they add up to one.

Marble remaining	Final probability
Green	Two thirds
Blue	One third
Total	One

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Final Answer

The probability that the remaining marble is green is two-thirds.

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Key Insight (Plain English)

Seeing a green marble removed is more likely when both marbles were green than when there was only one green and one blue.
That is why the remaining marble is green two-thirds of the time.

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