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`.
The probability that the referee observes a plus sign on the slip if it is known that `A` wrote a plus sign is

Options

A. `14//27`
B. `16//27`
C. `13//27`
D. `17//27`

Explanation

To solve the problem, we need to find the probability that the referee observes a plus sign on the slip given that A wrote a plus sign. Let's denote the events as follows:

- Let \( A^+ \) be the event that A writes a plus sign.
- Let \( R^+ \) be the event that the referee observes a plus sign.

We are given:
- \( P(A^+) = \frac{1}{3} \)
- The probability that B, C, or D changes the sign is \( \frac{2}{3} \).
- The probability that B, C, or D does not change the sign is \( \frac{1}{3} \).

### Step 1: Calculate the probability that the referee sees a plus sign given that A wrote a plus sign.

When A writes a plus sign, we need to consider the actions of B, C, and D:

1. **No one changes the sign**:
- The probability that B does not change the sign is \( \frac{1}{3} \).
- The probability that C does not change the sign is \( \frac{1}{3} \).
- The probability that D does not change the sign is \( \frac{1}{3} \).
- Thus, the probability that none of them changes the sign is:
\[ P(\text{No change}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

2. **Exactly one person changes the sign**:
- There are 3 ways to choose which one of B, C, or D changes the sign. The probability that one specific person changes the sign and the other two do not is:
\[ P(\text{One change}) = 3 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right)^2 = 3 \cdot \frac{2}{3} \cdot \frac{1}{9} = \frac{6}{27} = \frac{2}{9} \]

3. **Exactly two people change the sign**:
- There are 3 ways to choose which two of B, C, or D change the sign. The probability that two specific people change the sign and one does not is:
\[ P(\text{Two change}) = 3 \cdot \left(\frac{2}{3}\right)^2 \cdot \left(\frac{1}{3}\right) = 3 \cdot \frac{4}{9} \cdot \frac{1}{3} = \frac{12}{27} = \frac{4}{9} \]

4. **All three change the sign**:
- The probability that all three change the sign is:
\[ P(\text{All change}) = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]

### Step 2: Combine these probabilities to find \( P(R^+ | A^+) \)

The referee sees a plus sign if:
- No one changes the sign (probability \( \frac{1}{27} \)).
- Exactly two change the sign (probability \( \frac{4}{9} \)).

Thus, we can sum these probabilities:
\[ P(R^+ | A^+) = P(\text{No change}) + P(\text{Two change}) = \frac{1}{27} + \frac{4}{9} \]

To add these fractions, we convert \( \frac{4}{9} \) to a fraction with a denominator of 27:
\[ \frac{4}{9} = \frac{4 \cdot 3}{9 \cdot 3} = \frac{12}{27} \]
So,
\[ P(R^+ | A^+) = \frac{1}{27} + \frac{12}{27} = \frac{13}{27} \]

### Final Answer
The probability that the referee observes a plus sign on the slip if it is known that A wrote a plus sign is:
\[ \boxed{\frac{13}{27}} \]