Question

Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be \(\frac{1}{2}\) and probability of occurrence of 0 at the odd place be \(\frac{1}{3}\). Then the probability that '10' is followed by '01' is equal to :

• A. \(\frac{1}{18}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{9}\)
• D. \(\frac{1}{6}\)

Hint:

The required pattern is the 4-digit block 1001. The block can start at an odd place or at an even place; find each probability and add (the two cases are mutually exclusive).

Answer 1

The Correct Option is C

P(0 at even) = \(\frac{1}{2}\), so P(1 at even) = \(\frac{1}{2}\); P(0 at odd) = \(\frac{1}{3}\), so P(1 at odd) = \(\frac{2}{3}\).
The pattern '10' followed by '01' is the block 1, 0, 0, 1 in four consecutive places.
Case 1: block starts at an odd place, so the places are odd, even, odd, even: \(\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{18}\).
Case 2: block starts at an even place, so the places are even, odd, even, odd: \(\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{18}\).
The cases are mutually exclusive, so the required probability = \(\frac{1}{18}+\frac{1}{18}=\frac{1}{9}\).
Answer: (3) \(\frac{1}{9}\)