Question

An urn contains 6 blue and 'a' green balls. If the probability of drawing a green ball is double that of drawing a blue ball, then 'a' is equal to (1) 6 (2) 18 (3) 24 (4) 12

Short answer

Answer: 12

Step-by-step solution

Setup the probability equation

Let P(B) be the probability of drawing a blue ball, and P(G) be the probability of drawing a green ball. According to the problem, P(G) = 2 * P(B). We know that P(B) = (Number of blue balls) / (Total number of balls) and P(G) = (Number of green balls) / (Total number of balls).

Plug in known values and create an equation

We have 6 blue balls, and 'a' green balls. So, P(B) = 6 / (6 + a) and P(G) = a / (6 + a). Now, we plug these values into the equation P(G) = 2 * P(B), which gives us: a / (6 + a) = 2 * (6 / (6 + a))

Simplify the equation

Next, we will clear the fraction by multiplying both sides of the equation by (6 + a). a = 2 * (6 / (6 + a)) * (6 + a) This simplifies to: a = 12

Identify the correct option

Thus, the correct value for 'a' is 12. So, the urn contains 12 green balls. The correct option is: (4) 12

Key concepts

Probability Equations

Understanding probability equations is like deciphering a code that predicts the likelihood of an event. These equations underpin how we evaluate uncertainty and make calculated guesses in everyday decisions.

Probability is calculated using a simple equation: \( P(E) = \frac{n(E)}{n(S)} \), where \( P(E) \) is the probability of an event E, \( n(E) \) is the number of ways that event can occur, and \( n(S) \) is the total number of possible outcomes. This becomes particularly useful when the events are about drawing balls from an urn, as in the problem we're exploring.

When given specific conditions, such as the probability of drawing a green ball being double that of a blue ball, we use this knowledge to construct a relationship between the two probabilities. Mathematically, this involves setting up a ratio and solving for unknowns, which leads us towards a well-formed probability equation that we can solve.

Mathematical Problem Solving

The heart of mathematical problem solving lies in identifying patterns, establishing relationships, and applying logic. To solve the given problem, we must break down the task into manageable steps, which helps streamline the thinking process and leads to a successful solution.

Initially, the problem may seem complex, but by translating the words into a set of mathematical expressions, it becomes much clearer. For instance, by setting up the correct probability equation that reflects the given conditions and logically solving for the unknown 'a', we draw closer to the answer.

The aforementioned exercise with balls being drawn from an urn serves to demonstrate the technique of translating a real-world scenario into a mathematical framework. It encapsulates not just computing numbers but also understanding the concepts, like relative probability and algebraic manipulation, to solve for an unknown variable.

Probability for Drawing Balls

Let's dive deeper into the concept of 'Probability for Drawing Balls', which is a common illustrative example in probability exercises. It's a type of problem that helps explain complex theoretical ideas through a tangible scenario.

In such problems, each ball in the urn represents a potential outcome. When the balls are of different colors, they symbolize distinct events with their respective probabilities. Our task is to calculate the likelihood of drawing one ball over another, considering the total number of balls and their distribution.

Here's how it goes: to determine the exact probability, we first count the number of balls of each color, which represent the favorable outcomes for that event. We then divide that by the total number of balls to get the probability of drawing a ball of that specific color from the urn. As the problem often introduces certain conditions, like in our exercise where the probability of drawing a green ball is double that of a blue ball, we use this information to set a proportional relationship between the two occurrences.