Problem 116 - A Complete Resource Book in Mathematics for JEE Main

Question

A person throws two dice, one the common cube and the other a regular tetrahedron, the number on the lowest face being taken in the case of tetrahedron. The probability that the sum of the numbers appearing on the dice is 6 is (A) \(\frac{1}{3}\) (B) \(\frac{1}{4}\) (C) \(\frac{1}{6}\) (D) none of these

Step-by-Step Solution

Answer

The probability is \( \frac{1}{6} \) (option C).

Step 1: Understand the Problem

We have two dice: a common cube with numbers 1 to 6 and a tetrahedron with numbers 1 to 4. We need to find the probability that the sum of the numbers on the two dice equals 6.

Step 2: Determine Possible Outcomes

The cube die has 6 faces, and the tetrahedron die has 4 faces, giving us a total of 24 possible outcomes (6 \( \times \) 4 = 24).

Step 3: Find Combinations for Sum of 6

List all number pairs from the cube and tetrahedron dice that sum to 6: (2, 4), (3, 3), (4, 2), and (5, 1). That gives us 4 favorable outcomes.

Step 4: Calculate Probability

The probability is the number of favorable outcomes divided by the total number of possible outcomes: \( \frac{4}{24} \). Simplify this fraction to \( \frac{1}{6} \).

Key Concepts

Combinatorics

Combinatorics is the area of mathematics that deals with counting, arranging, and finding patterns. It helps us determine how many different ways we can combine objects or events. In our dice-related problem, combinatorics allows us to systematically analyze which dice rolls can result in a specific sum. Here’s how we applied combinatorics in our solution:

• First, we identified the number of faces on each die, which are the cube die (6 faces) and the tetrahedron die (4 faces).
• Then, we calculated all possible outcomes by multiplying the number of faces: 6 on the cube and 4 on the tetrahedron. So, there are 24 possible outcomes.
• Finally, we found all combinations where the sum of the two dice equals 6. These possible combinations are (2, 4), (3, 3), (4, 2), and (5, 1).

These steps show how combinatorics helps us efficiently list possible outcomes and combinations, leading to totally 4 favorable outcomes for the event we're interested in.

Dice Probability

Dice probability involves calculating the chances of certain outcomes when rolling dice. Since dice rolls are random, the actual result can vary each time, but probability helps us estimate the likelihood of different results.For the problem, we need to find out how likely it is to get a sum of 6 with two dice. Here’s the breakdown:

• The number of favorable outcomes that result in a sum of 6 is 4, obtained as combinations like (2, 4), (3, 3), (4, 2), and (5, 1).
• Total possible outcomes in this scenario are 24, combining all face values from both dice, a product of their faces: 6 faces on the cube times 4 on the tetrahedron.
• Using these, the probability calculation becomes the ratio of favorable outcomes to total outcomes: \( \frac{4}{24} \).
• We simplify this fraction, as both the numerator and the denominator can be divided by 4, resulting in \( \frac{1}{6} \).

Ultimately, the probability of rolling a sum of 6 is \( \frac{1}{6} \), which means in an ideal setting, you might see this outcome once in every 6 attempts.

Sample Space

In probability, the sample space is the set of all possible outcomes of an experiment. It forms the foundation for calculating probabilities.For our dice problem, the sample space includes every possible roll combination that can occur with our two dice:

• The cube die, having 6 faces, can result in numbers from 1 to 6.
• The tetrahedron has 4 faces, labeled from 1 to 4.
• When combining these, the sample space is the set of all 24 ordered pairs \( (a, b) \) where "a" is from the cube, and "b" is from the tetrahedron.

Therefore, listing out the sample space would show scenarios like (1,1), (1,2), (1,3), (1,4), (2,1), and so forth, all the way up to (6,4). This list helps us visualize all possible outcomes, which is crucial for identifying favorable outcomes.