Question

A company is migrating its database, and two software engineers, Ishaan and Kavya, take turns running a data-sync script that has a constant success rate of \(\dfrac38\) per attempt. If Ishaan initiates the first attempt and they persist until the migration is successful, what is the probability that Kavya is the one who initiates the successful sync?

• A. \(\dfrac{8}{13}\)
• B. \(\dfrac{3}{11}\)
• C. \(\dfrac{8}{11}\)
• D. \(\dfrac{5}{13}\)

Hint:

Infinite geometric series formula: \[ S=\frac{a}{1-r} \] is valid only when: \[ |r|<1 \]

Answer 1

The Correct Option is D

Concept: This problem is based on infinite geometric probability. The process continues until success occurs.

Step 1: Finding success and failure probabilities. Probability of success: \[ p=\frac38 \] Probability of failure: \[ q=1-\frac38=\frac58 \]

Step 2: Determining Kavya's winning cases. Ishaan starts first. Thus the turns are: \[ I,\ K,\ I,\ K,\ I,\ K,\dots \] Kavya succeeds if success occurs on: \[ 2^\text{nd},4^\text{th},6^\text{th},\dots \] Probability that Kavya succeeds on second turn: \[ qp \] Probability that Kavya succeeds on fourth turn: \[ q^3p \] Probability that Kavya succeeds on sixth turn: \[ q^5p \] Hence: \[ P=qp+q^3p+q^5p+\dots \]

Step 3: Forming the geometric series. Factor \(qp\): \[ P=qp(1+q^2+q^4+\dots) \] This is an infinite geometric series with: \[ a=1,\quad r=q^2 \] Therefore: \[ P=\frac{qp}{1-q^2} \] Substituting: \[ P=\frac{\left(\frac58\right)\left(\frac38\right)}{1-\left(\frac58\right)^2} \] \[ P=\frac{\frac{15}{64}}{1-\frac{25}{64}} \] \[ P=\frac{\frac{15}{64}}{\frac{39}{64}} \] \[ P=\frac{15}{39} \] \[ P=\frac{5}{13} \]