Question

In a chess tournament, assume that your probability of winning a game is 0.3 against level 1 players, 0.4 against level 2 players and 0.5 against level 3 players. It is further assumed that among the players 50% are at level 1, 25% are at level 2 and the remaining are at level 3. Suppose that you win the game. Then the probability that you had played with level 1 player is

• A. \(0.3\)
• B. \(0.4\)
• C. \(0.5\)
• D. \(0.6\)
• E. \(0.2\)

Hint:

In Bayes' theorem problems, always compute total probability first before applying the formula.

Answer 1

The Correct Option is B

Concept: This is an application of Bayes’ Theorem, which allows us to update probabilities based on new information. If \(A_1, A_2, A_3\) are mutually exclusive and exhaustive events and \(B\) is another event, then: \[ P(A_i|B) = \frac{P(A_i)\,P(B|A_i)}{\sum_{j} P(A_j)\,P(B|A_j)} \]

Step 1: Define the events clearly. Let: \[ A_1 = \text{opponent is level 1}, A_2 = \text{opponent is level 2}, A_3 = \text{opponent is level 3} \] \[ B = \text{you win the game} \] From the question: \[ P(A_1)=0.5, P(A_2)=0.25, P(A_3)=0.25 \] Winning probabilities: \[ P(B|A_1)=0.3, P(B|A_2)=0.4, P(B|A_3)=0.5 \]

Step 2: Find the total probability of winning \(P(B)\). Using the law of total probability: \[ P(B) = P(A_1)P(B|A_1) + P(A_2)P(B|A_2) + P(A_3)P(B|A_3) \] Substituting values: \[ P(B) = (0.5)(0.3) + (0.25)(0.4) + (0.25)(0.5) \] Now compute each term separately: \[ (0.5)(0.3) = 0.15 \] \[ (0.25)(0.4) = 0.10 \] \[ (0.25)(0.5) = 0.125 \] Adding: \[ P(B) = 0.15 + 0.10 + 0.125 = 0.375 \]

Step 3: Apply Bayes’ Theorem to find \(P(A_1|B)\). \[ P(A_1|B) = \frac{P(A_1)P(B|A_1)}{P(B)} \] Substitute: \[ P(A_1|B) = \frac{(0.5)(0.3)}{0.375} \] Compute numerator: \[ (0.5)(0.3) = 0.15 \] Thus: \[ P(A_1|B) = \frac{0.15}{0.375} \]

Step 4: Simplify the fraction carefully. Multiply numerator and denominator by 1000 to remove decimals: \[ \frac{150}{375} \] Divide both by 75: \[ = \frac{2}{5} = 0.4 \]

Step 5: Interpret the result. This means that given that you won the game, the probability that your opponent was a level 1 player is \(0.4\).