Probability Puzzle of Colored Balls and Urns

An urn contains b black balls and r red balls. One of the balls is drawn at random, but when it is put back in the urn c additional balls of the same color are put in with it. Now suppose that we draw another ball. Show that the probability that the first ball drawn was black given that the second ball drawn was red is b/(b + r + c).

Probability puzzles are always intriguing and challenging. In this blog post, we will explore a problem related to colored balls and urns, and how to solve it using Bayes' Theorem. The problem is about drawing a ball from an urn, adding some additional balls of the same color to the urn, and then drawing another ball from it. We need to find the probability of the first ball being black, given that the second ball drawn was red.

Let's define some variables to represent the events:

B: First ball drawn was black

R: Second ball drawn was red

C: Additional balls of the same color added to the urn

b: Number of black balls in the urn before adding additional balls

r: Number of red balls in the urn before adding additional balls

The probability of drawing a black ball in the first draw is b/(b + r). After adding c additional balls of the same color, the total number of black balls in the urn becomes b + c, and the total number of balls becomes (b + c) + (r + c) = b + r + 2c. Similarly, the total number of red balls in the urn becomes r + c. Therefore, the probability of drawing a red ball in the second draw, given that the first ball drawn was black, is:

P(R|B) = (r + c) / (b + r + 2c)

Now, we need to find the probability of the first ball being black, given that the second ball drawn was red. This is the conditional probability of B given R, denoted as P(B|R).

Using Bayes' theorem, we can write:

P(B|R) = P(R|B) * P(B) / P(R)

We have already calculated P(R|B) and P(B). Now, we need to calculate P(R), the probability of drawing a red ball in the second draw, irrespective of the color of the first ball drawn.

P(R) = P(R|B) * P(B) + P(R|not B) * P(not B)

where not B means the first ball drawn was not black. The probability of drawing a red ball in the second draw, given that the first ball drawn was not black, is:

P(R|not B) = (r + c) / (b + r + 2c)

The probability of the first ball not being black is:

P(not B) = r / (b + r)

Substituting the values, we get:

P(R) = (r + c) / (b + r + 2c) * b / (b + r) + (r + c) / (b + r + 2c) * r / (b + r)

= (r + c) / (b + r + 2c)

Now, we can calculate P(B|R) as:

P(B|R) = P(R|B) * P(B) / P(R)

= (r + c) / (b + r + 2c) * b / (b + r) / (r + c) * (b + r + 2c) / (r + c)

= b / (b + r + c)

Therefore, the probability of the first ball being black, given that the second ball drawn was red, is b/(b + r + c).