There are two identical boxes A and B filled with equal number of balls which are identical
in all aspects except for their color.
There are n + 1 number of white balls and n + 1 number of black balls in the box A. And
also, there are n + 2 number of white balls and n number of black balls in the box B.
There is a biased coin. The probability of obtaining head when tossed is 1/3 . When the coin is
tossed, if the head is obtained then a ball is taken out from the box A and if the side with
the value is obtained, then a ball is taken out from the box B. Here n ∈ Z
+, n > 1.Two balls are taken out from the box randomly one after the other without any
replacements. If the probability of the balls taken out be a black color ball and a
white color ball is 5/9, find the value of n.

Let's denote the events and probabilities:

- Let \( P(A) \) be the probability of selecting a ball from box A.
- Let \( P(B) \) be the probability of selecting a ball from box B.
- \( P(A) = \frac{1}{3} \) since it's given that the probability of obtaining a head is \( \frac{1}{3} \), which corresponds to choosing from box A.
- Therefore, \( P(B) = 1 - P(A) = \frac{2}{3} \), which corresponds to choosing from box B.

Now, let's consider the probabilities of drawing the first and second balls from each box:

### For Box A:
- The total number of balls in box A is \( 2n + 2 \) (since there are \( n + 1 \) white balls and \( n + 1 \) black balls).
- The probability of drawing a black ball from box A on the first draw is \( \frac{n+1}{2n+2} \).
- The probability of drawing a white ball from box A on the first draw is \( \frac{n+1}{2n+2} \).

If a black ball is drawn first, then the probability of drawing a white ball next is \( \frac{n+1}{2n+1} \) (since one black ball has been removed, leaving \( n+1 \) white balls and \( n \) black balls).

Thus, the probability of drawing a black ball and then a white ball from box A is:
\[
P(\text{black and white from A}) = \frac{n+1}{2n+2} \times \frac{n+1}{2n+1}
\]

### For Box B:
- The total number of balls in box B is \( 2n + 2 \) (since there are \( n+2 \) white balls and \( n \) black balls).
- The probability of drawing a black ball from box B on the first draw is \( \frac{n}{2n+2} \).
- The probability of drawing a white ball from box B on the first draw is \( \frac{n+2}{2n+2} \).

If a black ball is drawn first, then the probability of drawing a white ball next is \( \frac{n+2}{2n+1} \) (since one black ball has been removed, leaving \( n+2 \) white balls and \( n-1 \) black balls).

Thus, the probability of drawing a black ball and then a white ball from box B is:
\[
P(\text{black and white from B}) = \frac{n}{2n+2} \times \frac{n+2}{2n+1}
\]

### Combined Probability:
The total probability of drawing a black ball and then a white ball (considering both boxes) is:
\[
P(\text{black and white}) = P(A) \times P(\text{black and white from A}) + P(B) \times P(\text{black and white from B})
\]
Substituting the values:
\[
P(\text{black and white}) = \frac{1}{3} \times \frac{(n+1)(n+1)}{(2n+2)(2n+1)} + \frac{2}{3} \times \frac{n(n+2)}{(2n+2)(2n+1)}
\]
This is given to be equal to \( \frac{5}{9} \), so:
\[
\frac{1}{3} \times \frac{(n+1)^2}{(2n+2)(2n+1)} + \frac{2}{3} \times \frac{n(n+2)}{(2n+2)(2n+1)} = \frac{5}{9}
\]
Multiplying through by 9 to clear the fractions:
\[
3 \times \frac{(n+1)^2}{(2n+2)(2n+1)} + 6 \times \frac{n(n+2)}{(2n+2)(2n+1)} = 5
\]
Now, simplify and combine the terms:
\[
\frac{3(n+1)^2 + 6n(n+2)}{(2n+2)(2n+1)} = 5
\]

Simplifying further:
\[
3(n^2 + 2n + 1) + 6n^2 + 12n = 10n^2 + 15n + 3
\]
This simplifies to:
\[
9n^2 + 18n + 3 = 10n^2 + 15n + 5
\]
Bringing everything to one side:
\[
n^2 - 3n - 2 = 0
\]
This can be factored as:
\[
(n - 2)(n + 1) = 0
\]

So, \( n = 2 \) or \( n = -1 \).

Since \( n > 1 \), the valid solution is \( n = 2 \).

So, the value of \( n \) is \( \boxed{2} \).

Would you like more details or have any questions? Here are some related questions:

1. How does the probability change if the coin is fair instead of biased?
2. What happens if there were replacements after drawing the balls?
3. How does the solution change if \( n = 1 \)?
4. What if the number of balls in the boxes is different?
5. How would you solve this if there were three boxes instead of two?

**Tip:** When dealing with probability problems, always break down the problem into smaller, manageable parts and calculate the individual probabilities step by step.

Explore a probability problem involving two boxes with different numbers of white and black balls. Understand how to calculate conditional probabilities and solve for the value of n. Suitable for college-level students and enthusiasts of probability theory.

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