Probability Calculation for Second Draw from Urn

This problem asks for the probability of drawing a red ball on the second draw from an urn, given specific initial conditions and that the first ball drawn is discarded without revealing its color.

Initial State of the Urn

The urn contains:

• 5 Red Balls
• 5 Black Balls
• Total Balls = 10

Understanding the Scenario

A single ball is drawn from the urn and set aside. We don't know if it was red or black. We then need to find the probability that the next ball drawn (the second ball) is red.

Method 1: Using the Law of Total Probability

Let R1 denote the event that the first ball drawn is red, and B1 denote the event that the first ball drawn is black. Let R2 denote the event that the second ball drawn is red.

The probability of the first ball being red is:

$$ P(R1) = \frac{\text{Number of Red Balls}}{\text{Total Balls}} = \frac{5}{10} = \frac{1}{2} $$

The probability of the first ball being black is:

$$ P(B1) = \frac{\text{Number of Black Balls}}{\text{Total Balls}} = \frac{5}{10} = \frac{1}{2} $$

Now, we consider the second draw conditional on the outcome of the first draw. After the first draw, there are 9 balls left.

• Case 1: First ball was Red (R1)

  If the first ball drawn was red, there are now 4 red balls and 5 black balls left.

  The probability of drawing a red ball second, given the first was red, is:

  $$ P(R2|R1) = \frac{\text{Remaining Red Balls}}{\text{Total Remaining Balls}} = \frac{4}{9} $$

• Case 2: First ball was Black (B1)

  If the first ball drawn was black, there are now 5 red balls and 4 black balls left.

  The probability of drawing a red ball second, given the first was black, is:

  $$ P(R2|B1) = \frac{\text{Remaining Red Balls}}{\text{Total Remaining Balls}} = \frac{5}{9} $$

Using the Law of Total Probability, the overall probability of the second ball being red is:

$$ P(R2) = P(R2|R1)P(R1) + P(R2|B1)P(B1) $$

Substitute the calculated probabilities:

$$ P(R2) = \left(\frac{4}{9}\right) \times \left(\frac{1}{2}\right) + \left(\frac{5}{9}\right) \times \left(\frac{1}{2}\right) $$

$$ P(R2) = \frac{4}{18} + \frac{5}{18} $$

$$ P(R2) = \frac{9}{18} $$

$$ P(R2) = \frac{1}{2} $$

Method 2: Using Symmetry

Since the color of the first ball removed is unknown, each of the original 10 balls has an equal chance of being the second ball drawn. Imagine all 10 balls are lined up in the order they are drawn. Any ball is equally likely to be in the second position.

Out of the 10 balls initially in the urn, 5 are red.

Therefore, the probability that the ball in the second position is red is simply the initial proportion of red balls:

$$ P(\text{Second ball is Red}) = \frac{\text{Initial Number of Red Balls}}{\text{Initial Total Number of Balls}} = \frac{5}{10} = \frac{1}{2} $$

Both methods confirm the result.