Key Concepts and Formulas

• Geometric Series: The sum of a geometric series $a + ar + ar^2 + \dots + ar^{n-1}$ is given by $S_n = a\frac{1-r^n}{1-r}$. If $|r| < 1$ and $n \rightarrow \infty$, $S = \frac{a}{1-r}$.
• Permutations: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times \dots \times 2 \times 1$.
• Probability: The sum of probabilities of all possible outcomes in a sample space is 1.

Step-by-Step Solution

Step 1: Determine the general form of the probability $P(W_n)$

We are given that $P(W_n) = 2P(W_{n-1})$ for $n > 1$. This means the probabilities form a geometric progression. Let $P(W_1) = p$. Then, $P(W_2) = 2p$ $P(W_3) = 2(2p) = 4p = 2^2p$ In general, $P(W_n) = 2^{n-1}p$.

Step 2: Find the value of $p$ using the total probability condition

Since the probabilities of all possible words must sum to 1, we have $\sum_{n=1}^{120} P(W_n) = 1$ Substituting $P(W_n) = 2^{n-1}p$, we get $\sum_{n=1}^{120} 2^{n-1}p = 1$ $p \sum_{n=1}^{120} 2^{n-1} = 1$ The sum is a geometric series with first term 1, common ratio 2, and 120 terms. Therefore, $p \cdot \frac{1(2^{120} - 1)}{2-1} = 1$ $p(2^{120} - 1) = 1$ $p = \frac{1}{2^{120} - 1}$

Step 3: Express the probability of the $n$-th word $W_n$

Substituting the value of $p$ into the expression for $P(W_n)$, we get $P(W_n) = \frac{2^{n-1}}{2^{120} - 1}$

Step 4: Determine the rank of the word "CDBEA" in the dictionary

We need to find how many words come before "CDBEA" alphabetically.

• Words starting with 'A': $4! = 24$
• Words starting with 'B': $4! = 24$
• Words starting with 'CA': $3! = 6$
• Words starting with 'CB': $3! = 6$
• Words starting with 'CDA': $2! = 2$
• Words starting with 'CDBA': $1! = 1$ Adding these up, we get $24 + 24 + 6 + 6 + 2 + 1 = 63$ words before "CDBEA". Therefore, the rank of "CDBEA" is $n = 63 + 1 = 64$.

Step 5: Calculate the probability of the word "CDBEA"

Using the formula for $P(W_n)$ with $n = 64$, we have $P(\text{CDBEA}) = P(W_{64}) = \frac{2^{64-1}}{2^{120} - 1} = \frac{2^{63}}{2^{120} - 1}$

Step 6: Compare with the given form and find $\alpha$ and $\beta$

We are given that $P(\text{CDBEA}) = \frac{2^\alpha}{2^\beta - 1}$. Comparing this with our result, we have $\alpha = 63$ $\beta = 120$

Step 7: Calculate $\alpha + \beta$

$\alpha + \beta = 63 + 120 = 183$

Common Mistakes & Tips

• Be careful with the indexing of the geometric series. The first term corresponds to $n=1$, so the exponent of 2 in $P(W_n)$ is $n-1$.
• Remember to add 1 to the number of words preceding "CDBEA" to find its rank.
• Pay close attention to the order of letters when determining the position of the word in the dictionary.

Summary

We first found the general form of the probability $P(W_n)$ using the given recurrence relation. Then, we used the fact that the sum of all probabilities must equal 1 to find the value of $p$. After finding the rank of the word "CDBEA", we calculated its probability and compared it with the given form to find $\alpha$ and $\beta$, and finally calculated $\alpha + \beta$. The final result differs from the problem's given answer. There must be an error in the question, the options, or the "Correct Answer" provided. Following the logic as presented, the value of $\alpha + \beta$ is 183.

Final Answer

The final answer is \boxed{183}.