Problem 171 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let two fair six-faced dice \(\mathrm{A}\) and \(\mathrm{B}\) be thrown simultaneously If \(\mathrm{E}\), is the event that die A shows up four, \(\mathrm{E} 2\) is the event that die \(\mathrm{B}\) shows up two and \(\mathrm{E} 3\) is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true? (A) E, E2 and E3 are independent. (B) E, and E2 are independent. (C) E2 and E3 are independent. (D) E, and E3 are independent.

Step-by-Step Solution

Answer

Statement (A) is NOT true. E, E2, and E3 are not mutually independent.

Step 1: Identify Probability of Individual Events

To solve this, we first calculate the probability of each event. For \(E\), the event that die A shows 4, the probability is \(\frac{1}{6}\), since there is one face with 4 out of 6 total faces. For \(E_2\), the event that die B shows 2, the probability is also \(\frac{1}{6}\) for the same reason. For \(E_3\), the event that the sum of the numbers on both dice is odd, we calculate this as the sum of the events where one die shows odd and the other even, which is \(\frac{3}{6} \times \frac{3}{6} + \frac{3}{6} \times \frac{3}{6} = \frac{1}{2}\).

Step 2: Determine Independence of E and E2

Events E (die A shows 4) and E2 (die B shows 2) are independent if the probability of both occurring together is the product of their probabilities. The joint probability \( P(E \cap E_2) \) is \(\frac{1}{36}\) since both are single face events on different dice, matching the product \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\). Hence, E and E2 are independent.

Step 3: Determine Independence of E and E3

Events E (die A shows 4) and E3 (sum of numbers is odd) are independent if \(P(E \cap E_3) = P(E) \times P(E_3)\). E gives a 4 which requires the second die to show odd (1, 3, or 5) for a sum being odd. Therefore, \(P(E \cap E_3) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\), where \(P(E_3) = \frac{1}{2}\) matches the product \(\frac{1}{6} \times \frac{1}{2}\). Thus, E and E3 are independent.

Step 4: Determine Independence of E2 and E3

Events E2 (die B shows 2) and E3 (sum of numbers is odd) are independent if \(P(E_2 \cap E_3) = P(E_2) \times P(E_3)\). For E3, die B's 2 requires die A to be odd (1, 3, or 5). Thus, \(P(E_2 \cap E_3) = \frac{3}{36} = \frac{1}{12}\), matching \(\frac{1}{6} \times \frac{1}{2}\). Therefore, E2 and E3 are independent.

Step 5: Confirm Which Statement is Not True

Statements (B) and (D) have been confirmed as true. Since statements (A), (B), (C), and (D) except one need to be true, and we proved independence throughout, (A), suggesting dependence, is actually false based on given options since all events should be additive, whereas only three of four can be suggested true to maintain characteristic constraints.

Key Concepts

Independent Events

In probability, two events are described as independent if the occurrence of one event does not affect the probability of the other. This means the outcome of one event has no impact on the outcome of another.

The basic rule for determining if two events, say event A and event B, are independent is by checking the multiplication property:

• Their joint probability is equal to the product of their individual probabilities.
• Mathematically, this is expressed as: \( P(A \cap B) = P(A) \times P(B) \)

For example, when rolling two separate dice, the result of one die does not affect the result of the other, making each roll independent. This assumption helps in calculating the combined outcomes of two independent events.

Dice

A six-faced die, or commonly known as a standard die, is a cube with numbers from 1 to 6 on its faces. When you roll a die, you have an equal chance of landing on any one of those numbers.

Each number thus has a probability of occurring equal to \( \frac{1}{6} \). When two dice are rolled simultaneously, like in this particular exercise, you're dealing with what is known as compound experiments.

The total number of outcomes is determined by multiplying the outcomes of each individual dice. Therefore, with two six-faced dice, there are \( 6 \times 6 = 36 \) possible outcomes. Understanding the basic principles of rolling dice provides a foundation for exploring more complex probability scenarios.

Event Probability

Event probability is the measure of the likelihood that a specific event will occur when a random experiment is performed. It is calculated as the number of favorable outcomes divided by the total number of possible outcomes. Formulaically, it is expressed as:

\[P(Event) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\]In our exercise, to calculate the probability of die A showing a specific number, say 4, the probability is \( \frac{1}{6} \).

Similarly, for die B showing any specific number, like 2, the probability is also \( \frac{1}{6} \).

• This idea extends to any single event on a fair dice.
Understanding this can help analyze situations involving dice games and other scenarios where discrete probability comes into play.

Joint Probability

Joint probability refers to the probability of two events occurring simultaneously. In situations involving two simultaneous actions, like throwing two dice, it's essential to calculate the likelihood of both events happening together.

The joint probability of two independent events, A and B, can be calculated with the formula: \( P(A \cap B) = P(A) \times P(B) \).

For instance, the probability that die A shows a 4 (event E) and die B shows a 2 (event E2) is given by:

• The product of both probabilities: \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \).

Joint probability is crucial in understanding complex systems of events where multiple things happen together, such as multiple dice being thrown and their outcomes considered at once.