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 $ P(W_n) $ 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 :

Hint:

Use the given recursive relation to find the probability of each word. Calculate the rank of the given word to find its serial number.

Answer 1

Correct Answer: 183

Let \( P(W_1) = x \). Given \( P(W_n) = 2P(W_{n-1}) \). Then, \( P(W_2) = 2x \), \( P(W_3) = 2^2x \), ..., \( P(W_n) = 2^{n-1}x \). Since \( \sum_{i=1}^{120} P(W_i) = 1 \), we have \[ x + 2x + 2^2x + \dots + 2^{119}x = 1 \] \[ x(1 + 2 + 2^2 + \dots + 2^{119}) = 1 \] \[ x \cdot \frac{2^{120} - 1}{2 - 1} = 1 \] \[ x(2^{120} - 1) = 1 \] \[ x = \frac{1}{2^{120} - 1} \tag{1} \] Now, let's find the rank of CDBEA. A\_ \_ \_ \_ = \( 4! = 24 \) B\_ \_ \_ \_ = \( 4! = 24 \) CA\_ \_ \_ = \( 3! = 6 \) CB\_ \_ \_ = \( 3! = 6 \) CDA\_ \_ = \( 2! = 2 \) CDBAE = 1 CDBEA = 1 So, the rank of CDBEA is \[ 24 + 24 + 6 + 6 + 2 + 1 = 63 \]
Thus, CDBEA is \( W_{64} \). Therefore, \[ P(CDBEA) = P(W_{64}) = 2^{63} \cdot P(W_1) = 2^{63} \cdot \frac{1}{2^{120} - 1} \] \[ P(CDBEA) = \frac{2^{63}}{2^{120} - 1} \] Comparing with \( P(CDBEA) = \frac{2^\alpha}{2^\beta - 1} \), we have: \[ \alpha = 63,\quad \beta = 120 \] \[ \alpha + \beta = 63 + 120 = 183 \]