An unbiased die is thrown 13 times. After each throw a '+' is recorded for 2 or 5 and '-' is recorded for 1,3,4 or 6, the signs forming an ordered sequence. To each, except the first and last sign, a random variable
X
i
;
i
=
1
,
2
,
.
.
.
,
11
X
i
​
;i=1,2,...,11 is associated which takes the value
1
1 if both of its neighbouring sign differs from the one between them and
0
0 otherwise. If the random variable
Y
Y is defined as
Y
=
14
S
+
29
Y=14S+29 where,
S
=
∑
i
=
1
11
X
i
S=∑
i=1
11
​
X
i
​
.
Find the expected value of
Y
Y.

Let's solve the problem step by step.

### Step 1: Understanding the Random Variable \( X_i \)
Given that an unbiased die is thrown 13 times, an ordered sequence of '+' (for outcomes 2 or 5) and '-' (for outcomes 1, 3, 4, or 6) is formed. For each \( X_i \) (where \( i = 1, 2, \ldots, 11 \)), the value of \( X_i \) is determined as follows:
- \( X_i = 1 \) if both of its neighboring signs differ from the sign in between.
- \( X_i = 0 \) otherwise.

### Step 2: Probability of Each Sign
Since the die is unbiased:
- Probability of a '+' (for outcomes 2 or 5) is \( P(+) = \frac{2}{6} = \frac{1}{3} \).
- Probability of a '-' (for outcomes 1, 3, 4, or 6) is \( P(-) = \frac{4}{6} = \frac{2}{3} \).

### Step 3: Calculation of \( P(X_i = 1) \)
\( X_i = 1 \) occurs if the sign sequence follows either '+ - +' or '- + -'. The probability of each scenario is:
- \( P(+ - +) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} \).
- \( P(- + -) = \frac{2}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{27} \).

Thus, the total probability that \( X_i = 1 \) is:
\[
P(X_i = 1) = P(+ - +) + P(- + -) = \frac{2}{27} + \frac{4}{27} = \frac{6}{27} = \frac{2}{9}.
\]

### Step 4: Expected Value of \( S \)
\( S = \sum_{i=1}^{11} X_i \), where each \( X_i \) is an independent random variable. The expected value of \( S \) is:
\[
\mathbb{E}[S] = \sum_{i=1}^{11} \mathbb{E}[X_i] = 11 \times \mathbb{E}[X_i] = 11 \times \frac{2}{9} = \frac{22}{9}.
\]

### Step 5: Expected Value of \( Y \)
\( Y = 14S + 29 \). The expected value of \( Y \) is:
\[
\mathbb{E}[Y] = 14 \times \mathbb{E}[S] + 29 = 14 \times \frac{22}{9} + 29 = \frac{308}{9} + 29 = \frac{308}{9} + \frac{261}{9} = \frac{569}{9}.
\]

Thus, the expected value of \( Y \) is \( \boxed{\frac{569}{9}} \).

Explore the calculation of expected value in probability theory with a detailed solution. This problem involves an unbiased die, random variables, and the calculation of expected value. Suitable for advanced mathematics enthusiasts.

Math Problem Statement

An unbiased die is thrown 13 times. After each throw a '+' is recorded for 2 or 5 and '-' is recorded for 1,3,4 or 6, the signs forming an ordered sequence. To each, except the first and last sign, a random variable X i ; i

1 , 2 , . . . , 11 X i ;i=1,2,...,11 is associated which takes the value 1 1 if both of its neighbouring sign differs from the one between them and 0 0 otherwise. If the random variable Y Y is defined as Y

14 S + 29 Y=14S+29 where, S

∑ i

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