A slip of paper is given to a person `A` who marks it either with a plus sign or a minus sign. The probability of his writing a plus sign is `1//3`. A passes the slip to `B`, who may either leave it alone or change the sign before passing it to `C`. Next `C` passes the slip to `D` after perhaps changing the sign. Finally `D` passes it to a refere after perhaps changing the sign. `B,C,D each change the sign with probability `2//3`.
If the refree observes a plus sign on the slip then the probability that A originally wrote a plus sign is

Options

A. `13//41`
B. `19//27`
C. `17//25`
D. `21//37`

Explanation

To solve the problem step by step, we will use the concept of conditional probability and Bayes' theorem.

### Step 1: Define the Events
Let:
- \( A^+ \): Event that A writes a plus sign.
- \( A^- \): Event that A writes a minus sign.
- \( R^+ \): Event that the referee observes a plus sign.

Given:
- \( P(A^+) = \frac{1}{3} \)
- \( P(A^-) = \frac{2}{3} \)

### Step 2: Calculate the Probability of Referee Observing a Plus Sign Given A's Sign
We need to calculate \( P(R^+ | A^+) \) and \( P(R^+ | A^-) \).

#### Case 1: A writes a plus sign (\( A^+ \))
If A writes a plus sign, the probabilities for B, C, and D changing the sign are:
- Each of B, C, D changes the sign with probability \( \frac{2}{3} \) and does not change it with probability \( \frac{1}{3} \).

The possible scenarios for \( R^+ \) given \( A^+ \):
1. None of B, C, D change the sign:
\[ P(B^-, C^-, D^-) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
2. Exactly one of B, C, D changes the sign:
- Choose 1 from 3 to change (3 ways), and the others do not change:
\[ P(B^+, C^-, D^-) + P(B^-, C^+, D^-) + P(B^-, C^-, D^+) = 3 \cdot \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^2 = 3 \cdot \frac{2}{27} = \frac{6}{27} \]
3. Exactly two of B, C, D change the sign:
- Choose 2 from 3 to change (3 ways), and the other does not change:
\[ P(B^+, C^+, D^-) + P(B^+, C^-, D^+) + P(B^-, C^+, D^+) = 3 \cdot \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right) = 3 \cdot \frac{4}{27} = \frac{12}{27} \]
4. All three change the sign:
\[ P(B^+, C^+, D^+) = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]

Adding these probabilities gives:
\[ P(R^+ | A^+) = \frac{1}{27} + \frac{6}{27} + \frac{12}{27} + \frac{8}{27} = \frac{27}{27} = 1 \]

#### Case 2: A writes a minus sign (\( A^- \))
If A writes a minus sign, we analyze similarly:
1. None of B, C, D change the sign:
\[ P(B^-, C^-, D^-) = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]
2. Exactly one of B, C, D changes the sign:
\[ 3 \cdot \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right) = 3 \cdot \frac{4}{27} = \frac{12}{27} \]
3. Exactly two of B, C, D change the sign:
\[ 3 \cdot \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^2 = 3 \cdot \frac{2}{27} = \frac{6}{27} \]
4. All three change the sign:
\[ P(B^+, C^+, D^+) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

Adding these probabilities gives:
\[ P(R^+ | A^-) = \frac{8}{27} + \frac{12}{27} + \frac{6}{27} + \frac{1}{27} = \frac{27}{27} = 1 \]

### Step 3: Apply Bayes' Theorem
Now we apply Bayes' theorem to find \( P(A^+ | R^+) \):
\[ P(A^+ | R^+) = \frac{P(R^+ | A^+) P(A^+)}{P(R^+ | A^+) P(A^+) + P(R^+ | A^-) P(A^-)} \]
Substituting the values:
\[ P(A^+ | R^+) = \frac{1 \cdot \frac{1}{3}}{1 \cdot \frac{1}{3} + 1 \cdot \frac{2}{3}} = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{2}{3}} = \frac{\frac{1}{3}}{1} = \frac{1}{3} \]

### Final Answer
Thus, the probability that A originally wrote a plus sign given that the referee observes a plus sign is:
\[ \boxed{\frac{1}{3}} \]