The Puzzle

You have a jar containing:

  • 10 red candies
  • 20 blue candies
  • 30 green candies

You draw candies one at a time at random without replacement.

Question: What is the probability that, at the moment you remove the last red candy, there is still at least one blue and one green candy remaining in the jar?

Total Candies

Total number of candies: \( 10 + 20 + 30 = 60 \)

Approach

We aim to find the probability that when the last red candy is drawn, both blue and green candies are still present in the jar. This is equivalent to computing the probability that the last red candy is drawn before the last blue and last green candies.

Let:

  • \( T_r \): position (draw number) of the last red candy
  • \( T_b \): position of the last blue candy
  • \( T_g \): position of the last green candy

We are interested in:

\[ P(T_r < T_b \text{ and } T_r < T_g) \]

This event can occur in two mutually exclusive ways:

  1. \( T_r < T_b < T_g \)
  2. \( T_r < T_g < T_b \)

Therefore:

\[ P(T_r < T_b \text{ and } T_r < T_g) = P(T_r < T_b < T_g) + P(T_r < T_g < T_b) \]

Calculating the Probabilities

First Scenario: \( T_r < T_b < T_g \)

  • Probability that the last candy drawn is green: \( \frac{30}{60} \)
  • Given that, the probability that the last blue candy is drawn before the last green candy: \( \frac{20}{30} \)

So:

\[ P(T_r < T_b < T_g) = \frac{30}{60} \cdot \frac{20}{30} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \]

Second Scenario: \( T_r < T_g < T_b \)

  • Probability that the last candy drawn is blue: \( \frac{20}{60} \)
  • Given that, the probability that the last green candy is drawn before the last blue candy: \( \frac{30}{40} \)

So:

\[ P(T_r < T_g < T_b) = \frac{20}{60} \cdot \frac{30}{40} = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} \]

Final Probability

Adding both scenarios:

\[ \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \]

Answer

\[ \boxed{\frac{7}{12}} \]

Reference