All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dic

Q: Wait, why is the sum of all probabilities equal to 1?

That is a fundamental axiom of probability! The set of all possible outcomes (all 120 words) constitutes the sample space. Since one of these words must be chosen, the sum of the probabilities of all individual, mutually exclusive outcomes must equal 1.

Q: How do I systematically count the rank of 'CDBEA'?

Think of it like a tree. First, count all words starting with A (4! = 24) and B (4! = 24). Then, for words starting with C, count those starting with CA (3! = 6) and CB (3! = 6). Then, for words starting with CD, count CDA (2! = 2). Finally, list the words starting with CDB: CDBAE (63rd) and CDBEA (64th).

Q: Why is the common ratio 2 in the geometric progression?

The problem explicitly states $P(W_n) = 2P(W_{n-1})$. This is the definition of a geometric progression where each term is twice the previous one. The common ratio $r$ is simply the multiplier, which is 2.

Question

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1}), n > 1$. If $P(CDBEA) = \frac{2^\alpha}{2^\beta - 1}$, $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to : _______

Sigma Percentile · Accepted Answer

### Analyzing the Sample Space

Imagine a dictionary containing every possible five-letter word formed using the letters $A, B, C, D,$ and $E$ without repetition. Since there are $5$ distinct letters, the total number of unique words is given by the permutation of $5$ items:

$$5! = 120$$

These words are arranged in strict alphabetical order from $W_1$ to $W_{120}$.

### The Hidden Geometric Rhythm

The problem introduces a constraint where the probability of choosing the $n$-th word is exactly twice the probability of choosing the $(n-1)$-th word:

$$P(W_n) = 2P(W_{n-1})$$

If we let the probability of the first word be $x$, the sequence of probabilities forms a geometric progression: $x, 2x, 4x, \dots, x \cdot 2^{n-1}$. Since the sum of all probabilities in a sample space must equal $1$, we have:

$$\sum_{n=1}^{120} P(W_n) = x(1 + 2 + 2^2 + \dots + 2^{119}) = 1$$

Using the sum formula for a geometric series $S = \frac{a(r^n - 1)}{r - 1}$, where $a=x$, $r=2$, and $n=120$, we obtain:

$$x \left( \frac{2^{120} - 1}{2 - 1} \right) = 1 \implies x = \frac{1}{2^{120} - 1}$$

### The Hunt for the Rank

To locate the word $CDBEA$, we count the permutations systematically:

1. Words starting with $A$: $4! = 24$.
2. Words starting with $B$: $4! = 24$.
3. Words starting with $CA$: $3! = 6$.
4. Words starting with $CB$: $3! = 6$.

The total count before reaching words starting with $CD$ is $24 + 24 + 6 + 6 = 60$. Now, we examine words starting with $CD$:

1. Words starting with $CDA$: $2! = 2$.
2. Words starting with $CDB$: The remaining letters are $A$ and $E$. The first is $CDBAE$ ($63$rd word), and the next is $CDBEA$ ($64$th word).

Thus, the rank of our target word is $n = 64$.

### The Grand Synthesis

The probability of choosing $W_{64}$ is given by:

$$P(W_{64}) = x \cdot 2^{64-1} = x \cdot 2^{63}$$

Substituting the value of $x$ derived earlier:

$$P(CDBEA) = \frac{2^{63}}{2^{120} - 1}$$

Given the form $\frac{2^{\alpha}}{2^{\beta} - 1}$, we identify $\alpha = 63$ and $\beta = 120$. The final result is the sum:

$$\alpha + \beta = 63 + 120 = \mathbf{183}$$

Enter Numerical Value:

Visualized Solution

Total Words and Setup

Probability Sequence

Sum of Probabilities

Finding the First Term $x$

Rank Calculation: Words starting with A and B

Rank Calculation: Words starting with C

Rank Calculation: Narrowing down to CD

Reaching the Target Word

Calculating $P (W_{64})$

Finding $α$ and $β$

Final Answer

The Sigma Insight: Classical Definition of Probability

Analyzing the Sample Space

The Hidden Geometric Rhythm

The Hunt for the Rank

The Grand Synthesis

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