Problem 33 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A random variable \(X\) has the probability distribution: \(\begin{array}{lcccccccc}X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ P(X) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05\end{array}\) For the events \(E=\\{X\) is a prime number \(\\}\) and \(F=\\{X<4\\}\), then \(P(E \cup F)\) is: (a) \(0.77\) (b) \(0.87\) (c) \(0.35\) (d) \(0.50\)

Step-by-Step Solution

Answer

(a) 0.77

Step 1: Identify prime numbers and construct event E

To find event \(E\), list the prime numbers among \(1, 2, 3, 4, 5, 6, 7, 8\). The prime numbers are \(2, 3, 5,\) and \(7\). Hence, \(E = \{2, 3, 5, 7\}\).

Step 2: Identify numbers less than 4 and construct event F

Event \(F\) consists of numbers less than 4. These numbers are \(1, 2,\) and \(3\). Hence, \(F = \{1, 2, 3\}\).

Step 3: Apply the formula for \(P(E \cup F)\)

According to the formula, \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). We need to find \(P(E \cap F)\) first.

Step 4: Find the intersection \(E \cap F\) and calculate its probability

The intersection \(E \cap F\) includes numbers common to both events, which are \(2\) and \(3\). Thus, \(E \cap F = \{2, 3\}\). Compute \(P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35\).

Step 5: Calculate \(P(E)\)

Compute the probability of event \(E\): \(P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\).

Step 6: Calculate \(P(F)\)

Compute the probability of event \(F\): \(P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50\).

Step 7: Calculate \(P(E \cup F)\)

Substitute the calculated probabilities into the formula: \(P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77\).

Step 8: Select the correct answer

Compare the value with the provided options. The answer is (a) 0.77.

Key Concepts

Random Variables

A random variable is a foundational concept in probability theory. It is essentially a function that assigns a numerical value to each outcome in a sample space of a random experiment. Imagine rolling a die; each face represents an outcome, and we can assign numerical values based on a particular rule or interest. In our scenario, the random variable \(X\) is mapped to outcomes between 1 and 8, reflecting all possible results of an event, like flipping several coins or drawing cards from a deck. The values \( X = 1, 2, 3, \) etc., correspond to the different outcomes of the random experiment defined by your interest or study.

Random variables can be classified into two types: discrete and continuous. Discrete random variables, like our example with a finite number of potential outcomes, have distinct, separate values. These can be counted, such as the sum of dice rolls, making it easy to list and handle them in calculations. Continuous random variables have outcomes that form a continuous range, like measuring the time or distance, where values are not countable but can cover virtually any range within defined limits.

Understanding random variables is key to interpreting and calculating probabilities correctly as they lay the groundwork for probability distribution, discussed next.

Probability Distribution

Probability distribution describes how the probabilities of the different outcomes of a random variable are spread out. It's like a blueprint that tells us the likelihood of each outcome occurring.

For a discrete random variable as in our case, the probability distribution is a simple list or function depicting the probability for each potential outcome. The table given in the exercise outlines this idea: it pairs each outcome (\(X = 1, 2, 3,\) etc.) with a corresponding probability \( P(X) \), showing exactly the chance each outcome happens during the event. The sum of all probabilities in a complete distribution should add up to one, ensuring that one of the possible outcomes will certainly occur.

A critical application of probability distributions is using them to calculate probabilities of more complex events. For example, for combined probabilities of events like \(E\) (being a prime number) or \(F\) (less than 4), we consider the distribution to derive results like \(P(E \cup F)\). This employs sum and difference rules within probabilities, reflecting the logical operators you apply when interpreting events in a probability model.

Events and Outcomes

Events in probability theory refer to a set of outcomes to which a probability is assigned. They are considered subsets of the sample space of a random variable. In our example, event \(E\) is defined as the occurrence of prime numbers, while event \(F\) includes numbers less than 4. Each event is tied to certain outcomes within the broader context of the exercise's random variable.

Understanding how events intersect and combine is crucial. Two events intersecting, as with \(E \cap F\), means outcomes shared by both, which we calculated to find mutual probabilities. Conversely, unions, such as \(E \cup F\), represent any outcome in either set. Calculating probabilities for such combinations underpins many applications in statistics, allowing you to predict and infer from data effectively.

The use of operations like union and intersection complements event understanding, enhancing how outcomes manifest in probability models and harmonizing randomness with data-driven insights. Probability theory enables interpreting and navigating through possible event outcomes by methodically structuring these logical patterns.