Problem 9 - A Complete Resource Book in Mathematics for JEE Main

Question

\(A\) and \(B\) throw a dice. The probability that \(A\) 's throw is not greater than B's is (A) \(\frac{5}{12}\) (B) \(\frac{7}{12}\) (C) \(\frac{1}{6}\) (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The probability is \(\frac{7}{12}\).

Step 1: Define Possible Outcomes

Each of A and B throw a fair dice, each die showing a number from 1 to 6. Thus, the total number of possible outcomes is the number of pairs (A, B) where A and B are integers from 1 to 6. This gives 6 possible outcomes for A and 6 for B, resulting in a total of 36 possible pairs: 6 x 6 = 36.

Step 2: Define the Event of Interest

We need to find the probability where A's throw is not greater than B's throw. This means that we are interested in the pairs (A, B) where A \( \leq \) B.

Step 3: Count Favorable Outcomes

Count the number of pairs where A's throw is less than or equal to B's throw. The possible pairs are: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,4), (4,5), (4,6), (5,5), (5,6), (6,6). There are a total of 21 such pairs.

Step 4: Calculate Probability

The probability can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This gives us \( P(A \leq B) = \frac{21}{36} = \frac{7}{12} \).

Key Concepts

Combinatorial Probability

Combinatorial probability is a way of calculating the likelihood of an event based on the number of possible outcomes. For example, when rolling a die, each side represents one possible outcome. In problems like these, we use the rules of counting to figure out how likely different scenarios are. When we talk about "combinatorial," we're often referring to counting combinations of different outcomes or arrangements. It's like a puzzle where you figure out all the ways something can happen. In the dice problem, we are dealing with a pair of dice rolls. The main objective is to list or count all possible outcomes, much like when you count all the ways you can wear 3 different colored shirts and 2 types of pants.

• The total number of outcomes when throwing two dice is 36, as each die can show a value between 1 to 6, resulting in 6 x 6 combinations.
• Combinatorial probability involves determining how many of those outcomes are favorable for a certain event, like one player's roll being less than or equal to the other's in this case.

Favorable Outcomes

A favorable outcome is a result that satisfies the conditions of the event you are interested in. In probability, determining favorable outcomes is crucial for calculating how likely an event is to occur. In the dice problem we're discussing, a favorable outcome is when player A's roll is not greater than player B's.

• This means that if A rolls a 1, B's roll must be at least 1. If A rolls a 2, B can roll 2 or more, and so on.
• Tracking these favorable conditions helps us list down all the combinations that meet these criteria. In total, there are 21 such combinations: (1,1), (1,2), (1,3), all the way up to (6,6).

For calculating probability, the number of favorable outcomes (21) is divided by the total possible outcomes (36). This calculation helps determine the likelihood of A's roll not being greater than B's, which in this scenario is \( \frac{7}{12} \).

Independent Events

Independent events refer to events where the outcome of one event does not affect the outcome of another. In probability, understanding this concept helps us analyze situations where various events occur that do not influence each other. Throwing dice is a perfect example of independent events. The number that appears on the die when player A rolls does not change the probabilities of the numbers that appear when player B rolls.

• This means each roll is separate and does not rely on what happened before or after. Each die has the same chances of showing any number from 1 to 6 regardless of previous rolls.
• In this context, the outcomes when both A and B throw their dice are independent of each other, ensuring that the calculation of probabilities for each scenario remains uniform.

Recognizing this independence allows us to multiply probabilities of individual events together to find the probability of compound events if needed. However, in this exercise, the focus stays on counting combinations rather than combining the probabilities of multiple events.