Problem 10 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A bag contains 3 red, 4 white and 5 blue balls. All balls are different. Two balls are drawn at random. Find the probability that they are of different colours. [MP-2008]

Step-by-Step Solution

Answer

The probability is \( \frac{47}{66} \).

Step 1: Calculate Total Possible Outcomes

The total number of ways to draw two balls from a set of 12 (3 red + 4 white + 5 blue) is given by the combination formula \( \binom{n}{r} \). Calculate this as \( \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \). These are the total possible outcomes.

Step 2: Calculate Outcomes for Same Colour

First, calculate the number of ways to draw two balls of the same colour. For red balls: \( \binom{3}{2} = 3 \), for white balls: \( \binom{4}{2} = 6 \), and for blue balls: \( \binom{5}{2} = 10 \). Adding these, the total ways to get two balls of the same colour is \( 3 + 6 + 10 = 19 \).

Step 3: Calculate Outcomes for Different Colours

Use the difference between the total possible outcomes and the same colour outcomes to find different colour outcomes. Calculate this as \( 66 - 19 = 47 \). Thus, there are 47 ways to draw two balls of different colours.

Step 4: Calculate Probability of Different Colours

The probability that the two balls drawn are of different colours is the ratio of different colour outcomes to total possible outcomes. So, the probability is \( \frac{47}{66} \).

Key Concepts

Combination Formula

Combination formula is central to understanding how we count possibilities where order does not matter. In this problem, we use it to calculate how many ways we can select two balls from a total. This formula is written as:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, "\( n \)" represents the total number of items to choose from, and "\( r \)" represents the number of items to choose. The exclamation mark denotes a factorial, which is the product of an integer and all the positive integers below it. So, \( n! \) means multiply all whole numbers from 1 to \( n \).

• To solve our exercise, we let \( n = 12 \) (the total number of balls) and \( r = 2 \) (since we are drawing two balls).

Calculate it:

• \( \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \).

Thus, there are 66 different ways to choose 2 balls from 12.

Probability of Events

Probability helps us determine the likelihood of an event occurring. It's calculated by dividing the number of successful outcomes by the total number of possible outcomes. The probability \( P \) of an event is generally written as:\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]In the exercise, our event of interest is drawing two balls of different colours.

• We calculated the number of outcomes where both balls are the same color: 19.
• We also determined the total possible outcomes to be 66.
• Subtracting, we get that there are 47 successful outcomes where the balls are different colors.

Thus, the probability is given by:\[ P = \frac{47}{66} \]

Colour-based Probability Calculations

Calculating probabilities based on colour involves assessing subsets of a total set according to specific attributes (in this case, colour). We want to focus on how color influences probability outcomes for the exercise.The bag contains:

• 3 red balls
• 4 white balls
• 5 blue balls

To find how colour dictates outcomes, one must consider the separate possibilities:

• Draw 2 red balls: \( \binom{3}{2} = 3 \)
• Draw 2 white balls: \( \binom{4}{2} = 6 \)
• Draw 2 blue balls: \( \binom{5}{2} = 10 \)

Adding these, there are 19 ways to draw balls of the same colour. Finally, subtract this from the total outcomes:

• Ways to draw different-colored balls = 66 - 19 = 47

This calculation highlights the impact different colors have on probability: seeking different colours required considering how each color group contributes to overall combinations.