Problem 4 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

There are three bags which are known to contain 2 white and 3 black; 4 white and 1 black and 3 white and 7 black balls, respectively. A ball is drawn at random from one of the bags and found to be a black ball. The probability that it was drawn from the bag containing the most black balls is (a) \(7 / 15\) (b) \(5 / 19\) (c) \(3 / 4\) (d) None of these

Step-by-Step Solution

Answer

The probability that the ball was drawn from the bag with the most black balls is \(\frac{7}{15}\).

Step 1: Define the Problem

We have three bags each with a different composition of white and black balls. We're told a black ball is drawn, and we need to find the probability that it was drawn from the bag with the most black balls (third bag). We label the bags for clarity: - Bag 1: 2 white, 3 black. - Bag 2: 4 white, 1 black. - Bag 3: 3 white, 7 black.

Step 2: Calculate Probabilities of Drawing a Black Ball

Calculate the probability of drawing a black ball from each bag:- Bag 1: Probability = \(\frac{3}{5}\)- Bag 2: Probability = \(\frac{1}{5}\)- Bag 3: Probability = \(\frac{7}{10}\)

Step 3: Apply Bayes' Theorem

Let A be the event of drawing a black ball, and B1, B2, B3 be the events of choosing Bag 1, 2, and 3 respectively.Using Bayes' theorem, calculate the probability of drawing from Bag 3 (the bag with the most black balls) given that the ball is black:\[ P(B3 | A) = \frac{P(A | B3) \cdot P(B3)}{P(A | B1) \cdot P(B1) + P(A | B2) \cdot P(B2) + P(A | B3) \cdot P(B3)} \]Given equal selection probability for each bag, - \(P(B1) = P(B2) = P(B3) = \frac{1}{3}\).- \(P(A | B1) = \frac{3}{5}\), \(P(A | B2) = \frac{1}{5}\), \(P(A | B3) = \frac{7}{10}\).

Step 4: Compute Results

Substitute the values into Bayes' theorem:\(P(B3 | A) = \frac{\frac{7}{10} \cdot \frac{1}{3}}{\frac{3}{5} \cdot \frac{1}{3} + \frac{1}{5} \cdot \frac{1}{3} + \frac{7}{10} \cdot \frac{1}{3}}\)Simplify these terms:\[P(B3 | A) = \frac{7/30}{1/5}\]\[P(B3 | A) = \frac{7}{30} \div \frac{6}{30} = \frac{7}{6} \times \frac{5}{6} \]Thus,\[P(B3 | A) = \frac{7}{15}\].

Key Concepts

Probability Theory

Probability theory is the mathematical framework for understanding random events and uncertainty. It allows us to quantify the chance of different outcomes, making it crucial for fields ranging from statistics to everyday decision-making processes.
At its core, probability theory involves assigning a likelihood, typically between 0 and 1, to different possible events. An event with a probability of 0 is impossible, while an event with a probability of 1 is certain to occur.
With any random event, the sum of all probabilities for all potential outcomes must equal 1.

• Imagine you're rolling a six-sided die. Each face (or number) has an equal chance, or probability, of landing face up, namely \(\frac{1}{6}\).
• If someone asks the probability of rolling either a 1 or 2, you just add the individual probabilities: \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).

Probability provides a way to calculate and understand the likelihood of different events, playing a crucial role in analyzing scenarios involving random events.

Conditional Probability

Conditional probability is a refinement in probability theory that looks at how probabilities change in the light of new information or existing conditions. It's the probability of an event occurring, given that another event has already occurred.
This is symbolized as \(P(A|B)\), which reads as the probability of \(A\) given \(B\).
Here's how to think about it:

• If you know it's cloudy (event B), the probability of rain (event A) might increase, compared to a sunny day.
• Mathematically, it can be expressed as:
  
  \[P(A|B) = \frac{P(A \cap B)}{P(B)}\] Where \(P(A \cap B)\) is the probability of both events occurring together, and \(P(B)\) is the probability of the condition already known.

This concept is vital in fields like risk assessment, where we need to reevaluate probabilities based on new data.

Random Events

Random events are occurrences that are not certain and can't be exactly predicted. They're the foundation of probability theory and insist on chance and uncertainty. Every time an event is random, multiple outcomes are possible, but they follow a certain probability distribution.
Consider the random event of tossing a coin. Before it lands, you'd say the outcome (heads or tails) is uncertain and random. However, each outcome has a known probability of 0.5 because there's no reason for the coin to favor one side over another.
Here are some critical points to understand about random events:

• Despite the name, randomness doesn't imply complete chaos. Instead, it means that the outcomes follow a pattern when repeated many times.
• In any random experiment, every possible outcome should belong to a sample space—a set of all possible outcomes.
• Random events are essential not just in games of chance, but in scientific experiments and surveys, where they help model realities and anticipate outcomes.

This comprehension of randomness allows statisticians and data analysts to make informed predictions based on observed data trends.