Problem 53 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/ loss (in Rs.) of the person is : [April 12, 2019 (II)] (a) \(\frac{1}{2}\) gain (b) \(\frac{1}{4}\) loss (c) \(\frac{1}{2} \operatorname{los} \mathrm{s}\) (d) 2 gain

Step-by-Step Solution

Answer

The expected loss is \(\frac{1}{2}\) or Rs. 0.50 per game.

Step 1: Determine Possible Outcomes

Calculate the total number of possible outcomes when two dice are rolled. Each die has 6 sides and the outcome for each roll is independent, so the total number of outcomes is \(6 \times 6 = 36\).

Step 2: Calculate Probability of Winning Rs. 15 (Doublet)

A doublet is a condition when both dice show the same number. Possible doublets are: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Therefore, there are 6 possibilities. The probability of rolling a doublet is \(\frac{6}{36} = \frac{1}{6}\).

Step 3: Calculate Probability of Winning Rs. 12 (Sum of 9)

Calculate the number of ways the sum of the numbers on the two dice can be 9. These combinations are: (3,6), (4,5), (5,4), (6,3). There are 4 possibilities, therefore, the probability is \(\frac{4}{36} = \frac{1}{9}\).

Step 4: Calculate Probability of Losing Rs. 6 (Other Outcomes)

Number of outcomes that neither result in a doublet nor a sum of 9 is the total outcomes minus the favorable outcomes calculated previously. There are \(36 - (6+4) = 26\) such outcomes, so the probability is \(\frac{26}{36} = \frac{13}{18}\).

Step 5: Calculate Expected Value

Expected value \(E\) is calculated by multiplying each outcome by its probability and summing them all up: \[ E = 15 \times \frac{1}{6} + 12 \times \frac{1}{9} - 6 \times \frac{13}{18} \]Breaking calculations down: - \(15 \times \frac{1}{6} = \frac{15}{6} = 2.5\)- \(12 \times \frac{1}{9} = \frac{12}{9} = 1.333\)- \(6 \times \frac{13}{18} = \frac{78}{18} = 4.333\)Now add them: \[ 2.5 + 1.333 - 4.333 = -0.5 \]The expected value is -0.5, which means an expected loss of Rs. 0.50 per game.

Key Concepts

Expected Value

The concept of expected value is a cornerstone of probability theory that helps us determine what we can anticipate as the average result of a certain event or game if it were to be played many times. For this exercise, calculating the expected value involves combining all the possible outcomes of rolling two dice, their associated probabilities, and their specific resulting gains or losses.First, we need to determine the probabilities of each event:

• Gaining Rs. 15 from rolling a doublet.
• Gaining Rs. 12 from a roll that sums to 9.
• Losing Rs. 6 from all other outcomes.

We incorporate these into the expected value formula: \[E = (15 \times \text{Probability of Doublet}) + (12 \times \text{Probability of Sum 9}) - (6 \times \text{Probability of Other Outcomes})\]By calculating each part using the probabilities found, and then summing the results, we arrive at the expected loss of Rs. 0.50, as shown in the original solution. Understanding expected value allows us to grasp not just what can happen, but what's most likely to happen over the long term.

Combinatorics

Combinatorics is the study of counting, arrangement, and combination principles. In the context of this dice rolling exercise, it is used to determine the number of possible outcomes for given events such as rolling the same numbers on both dice or reaching a sum of 9. To understand how we calculate these:

• A doublet occurs when both dice display the same number. Here, each of the six faces can pair with its identical twin, resulting in six doublets: (1,1), (2,2), (3,3), etc.
• To find rolls that total 9, we consider all two-dice pairs adding to that sum: (3,6), (4,5), (5,4), and (6,3).

Combinatorics helps confirm that each calculation is not only accurate but also comprehensive, as it covers every scenario where these conditions are met without repetition or omission. This foundation is crucial in correctly calculating probabilities for the expected value.

Probability Calculations

Probability calculations form the backbone of assessing potential outcomes when two dice are rolled. Each event, such as rolling a doublet, depends on identifying the count of favorable outcomes divided by the total possible outcomes.Here's how these probabilities are calculated:

• For doublets, 6 outcomes out of 36 yield the same number, giving a probability of \( \frac{1}{6} \).
• For a sum of 9, 4 outcomes are possible, providing a probability of \( \frac{1}{9} \).
• The remaining outcomes neither fit these categories, accounting for 26 possibilities, reflecting a probability of \( \frac{13}{18} \).

These precise calculations allow us to multiply probabilities with their respective rewards or penalties in the expected value computation. The accuracy of these probabilities ensures our understanding of potential gains or losses is realistic and repeatable.