Problem 165 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

An urn contains nine balls of which three are red, four are blue and two are green. The experiment is to draw three balls at random without replacement from the urn. The probability that the three balls have different color is (A) \(\frac{2}{7}\) (B) \(\frac{1}{21}\) (C) \(\frac{2}{23}\) (D) \(\frac{1}{3}\)

Step-by-Step Solution

Answer

The probability that all three balls are of different colors is \( \frac{2}{7} \).

Step 1: Determine Total Ways to Pick 3 Balls

First, we must determine the total number of ways to draw 3 balls from an urn containing 9 balls. This is a combination problem where we have 9 items and choose 3, calculated as:\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

Step 2: Determine Ways to Pick Different Colors

Next, we find the number of ways to draw one ball of each color: one red, one blue, and one green. There are 3 red balls, 4 blue balls, and 2 green balls, so the number of ways to draw one of each is:\[ 3 \times 4 \times 2 = 24 \]

Step 3: Calculate the Probability

To find the probability that the three balls drawn are all of different colors, we divide the favorable outcomes by the total possible outcomes:\[ P(\text{different colors}) = \frac{24}{84} = \frac{2}{7} \]

Step 4: Match to Options

Finally, compare the calculated probability with the given options. The probability \( \frac{2}{7} \) matches option (A).

Key Concepts

Combinations

Combinations are a fundamental concept in probability, often used to determine how many ways we can select items from a larger group. Unlike permutations, where order matters, combinations are used when the order does not matter. In the context of selecting balls from an urn, we use combinations to calculate the total number of ways to pick a specific number of items, in this case, three balls from nine available balls. To compute this, we use the combination formula:

• General formula: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Where \( n \) is the total number of items, \( r \) is the number of items to choose, and \( n! \) represents the factorial of \( n \).
In our problem, \( n = 9 \) and \( r = 3 \).
By calculating \( \binom{9}{3} \), we find there are 84 different ways to pick any three balls from nine, laying the groundwork for further calculations.

Color Distribution

Understanding color distribution is crucial when calculating probabilities involving different categories. Here, the urn contains balls of three colors: red, blue, and green. Each color has a specific number of balls:

• 3 Red balls
• 4 Blue balls
• 2 Green balls

When drawing three balls, one of each color, we're interested in the number of ways to select one ball per color. We multiply the ways to pick each color of ball:

• Select a red ball: 3 ways
• Select a blue ball: 4 ways
• Select a green ball: 2 ways

Thus, the total number of ways to pick one ball of each color is \( 3 \times 4 \times 2 = 24 \).
This calculation helps us determine the number of successful outcomes where each chosen ball is of a distinct color.

Probability Calculation

Probability quantifies the likelihood of an event occurring and is expressed as a ratio of favorable outcomes to total possible outcomes. When calculating the probability for our scenario of drawing three balls, each a different color, we use our results from the previous sections:

• Total ways to pick any 3 balls from 9: 84
• Ways to pick three balls, one of each color: 24

To find the probability that the three drawn balls are of different colors, divide the successful outcomes by the total outcomes:\[ P(\text{different colors}) = \frac{24}{84} \]Simplifying this fraction, \( 24 \div 12 = 2 \) and \( 84 \div 12 = 7 \), gives us:\[ \frac{2}{7} \]This simplified probability tells us that there is a \(\frac{2}{7}\) chance of picking one red, one blue, and one green ball in a single draw of three balls from the urn without replacement.