Question

When Viswanath plays chess, he wins with probability 0.80, loses with probability 0.20. Assume independence, find the probability that Viswanath's first win happens when he plays his third game.

• A. 0.8
• B. 0.2
• C. 0.032
• D. 0.512

Hint:

For small values of $x$, write out the sequence of events directly (e.g., Loss-Loss-Win) and multiply their probabilities. This avoids needing to recall formal probability distribution formulas.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem can be solved using the Geometric distribution, which models the number of trials needed to get the first success in a sequence of independent Bernoulli trials.
Key Formula or Approach:
The probability that the first success occurs on the $x^{\text{th}}$ trial is: \[ P(X = x) = q^{x-1} p \] Where: - $p$ is the probability of success on any single trial.
- $q = 1 - p$ is the probability of failure on any single trial.

Step 2: Detailed Explanation:
From the problem parameters:
- Probability of success (winning), $p = 0.80$
- Probability of failure (losing), $q = 0.20$
- We want the first win to occur on the $x = 3^{\text{rd}}$ game.
Using the formula: \[ P(X = 3) = q^{3-1} p \] \[ P(X = 3) = q^2 p \] This is intuitive: he must lose the first game, lose the second game, and win the third game.
Substitute the given probabilities: \[ P(\text{L, L, W}) = (0.20) \times (0.20) \times (0.80) \] \[ P(\text{L, L, W}) = 0.04 \times 0.80 = 0.032 \] Thus, the probability is 0.032.

Step 3: Final Answer:
The probability is 0.032, which matches Option (C).