Key Concepts and Formulas

• Sample Space and Probability: For an experiment with a finite number of equally likely outcomes, the total set of all possible outcomes forms the sample space. The probability of an event $E$ is given by $P(E) = \frac{\text{Number of favorable outcomes for } E}{\text{Total number of outcomes in the sample space}}$.
• Expected Value (Expectation) $E(X)$: For a discrete random variable $X$ with possible outcomes $x_1, x_2, \ldots, x_n$ and their corresponding probabilities $P(X=x_1), P(X=x_2), \ldots, P(X=x_n)$, the expected value is the weighted average of these outcomes. It is calculated as: $E(X) = \sum_{i=1}^{n} x_i P(X=x_i)$ Here, $x_i$ represents the numerical value (gain or loss) associated with each outcome. A positive $E(X)$ indicates an expected gain, while a negative $E(X)$ indicates an expected loss.
• Mutually Exclusive Events: Events are mutually exclusive if they cannot occur at the same time. When calculating the probability of "any other outcome," it's often easiest to subtract the probabilities of specific, mutually exclusive events from 1.

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Step-by-Step Solution

Step 1: Determine the Sample Space and Total Possible Outcomes

When two fair dice are thrown, each die has 6 possible outcomes (1, 2, 3, 4, 5, 6). Since the two throws are independent, the total number of distinct ordered pairs (outcomes) in the sample space is the product of the possibilities for each die.

• Why this is important: This gives us the denominator for calculating the probability of any specific event. Total number of outcomes = (Outcomes for 1st die) $\times$ (Outcomes for 2nd die) Total outcomes = $6 \times 6 = 36$. These outcomes range from (1,1) to (6,6).

Step 2: Identify Each Event, its Monetary Outcome (Gain/Loss), and its Probability

We need to categorize all 36 possible outcomes into the three specified events, determine the number of favorable outcomes for each, calculate its probability, and assign the correct monetary value (positive for gain, negative for loss).

• Event 1: Throwing a doublet

  • Description: Both dice show the same number.
  • Favorable Outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
  • Number of Favorable Outcomes: 6.
  • Probability $P(\text{Doublet})$: $P(\text{Doublet}) = \frac{6}{36} = \frac{1}{6}$
  • Monetary Outcome ($X_1$): Wins Rs. 15. So, $X_1 = +15$.

• Event 2: Throwing a sum of 9

  • Description: The numbers on the two dice add up to 9.
  • Favorable Outcomes: (3,6), (4,5), (5,4), (6,3).
    • Note: For distinguishable dice, (3,6) and (6,3) are considered different outcomes.
  • Number of Favorable Outcomes: 4.
  • Probability $P(\text{Sum=9})$: $P(\text{Sum=9}) = \frac{4}{36} = \frac{1}{9}$
  • Monetary Outcome ($X_2$): Wins Rs. 12. So, $X_2 = +12$.

• Event 3: Any other outcome on the throw

  • Description: This includes all outcomes that are neither a doublet nor a sum of 9.
  • Why this is important: These three events (doublet, sum of 9, and "other") are mutually exclusive and exhaustive, meaning they cover all 36 possible outcomes exactly once.
  • Number of Favorable Outcomes: We can find this by subtracting the counts of the first two events from the total: Number of "other" outcomes = Total outcomes - (Number of doublets) - (Number of outcomes with sum 9) Number of "other" outcomes = $36 - 6 - 4 = 26$.
  • Probability $P(\text{Other})$: $P(\text{Other}) = \frac{26}{36} = \frac{13}{18}$
    • Self-Check: Sum of probabilities: $P(\text{Doublet}) + P(\text{Sum=9}) + P(\text{Other}) = \frac{6}{36} + \frac{4}{36} + \frac{26}{36} = \frac{36}{36} = 1$. This confirms all outcomes are correctly accounted for.
  • Monetary Outcome ($X_3$): Loses Rs. 6. So, $X_3 = -6$.

Step 3: Calculate the Expected Gain/Loss using the Expected Value Formula

Now, we apply the formula $E(X) = \sum x_i P(X=x_i)$ using the values derived in Step 2.

• Why this is done: We are calculating the average monetary outcome per throw, weighted by the probability of each outcome. $E(X) = X_1 \cdot P(\text{Doublet}) + X_2 \cdot P(\text{Sum=9}) + X_3 \cdot P(\text{Other})$ Substitute the values: $E(X) = (15) \cdot \left(\frac{1}{6}\right) + (12) \cdot \left(\frac{1}{9}\right) + (-6) \cdot \left(\frac{13}{18}\right)$ Perform the multiplications: $E(X) = \frac{15}{6} + \frac{12}{9} - \frac{6 \times 13}{18}$ Simplify the fractions where possible: $E(X) = \frac{5}{2} + \frac{4}{3} - \frac{13}{3}$ Combine the terms: $E(X) = \frac{5}{2} + \left(\frac{4-13}{3}\right)$ $E(X) = \frac{5}{2} + \left(\frac{-9}{3}\right)$ $E(X) = \frac{5}{2} - 3$ To subtract, find a common denominator (which is 2): $E(X) = \frac{5}{2} - \frac{6}{2}$ $E(X) = \frac{5 - 6}{2}$ $E(X) = -\frac{1}{2}$

Step 4: Interpret the Result

The calculated expected value is $E(X) = -\frac{1}{2}$ Rs.

• Why the sign matters: A negative expected value indicates an average loss over many trials. Therefore, the expected gain/loss for the person is a loss of Rs. $\frac{1}{2}$ (or Rs. 0.50).

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Common Mistakes & Tips

• Sign Convention for Loss: Always represent losses with a negative sign in the expected value formula. Forgetting this is a frequent source of error.
• Listing Outcomes for Dice: When dealing with two dice, remember that (a,b) is a distinct outcome from (b,a) if a $\neq$ b. For example, (3,6) and (6,3) are distinct outcomes for a sum of 9.
• Ensuring Exhaustive Events: Double-check that the sum of probabilities for all defined events equals 1. This confirms you haven't missed any possible outcomes or double-counted any.
• Arithmetic Precision: Be careful with fraction arithmetic and simplification to avoid calculation errors.

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Summary

By systematically identifying all possible outcomes, categorizing them into defined events, assigning monetary values (gains as positive, losses as negative), calculating the probability of each event, and then applying the expected value formula, we found the average outcome per throw. The calculated expected value of $E(X) = -\frac{1}{2}$ Rs. indicates that, on average, the person can expect to lose 50 paise (Rs. 0.50) per game in the long run. This implies the game is not favorable to the player financially.

The final answer is $\boxed{\text{A}}$