Problem 13 - A Complete Resource Book in Mathematics for JEE Main

Question

A box contains two white balls, three black balls and four red balls. The number of ways in which three balls can be drawn from the box if atleast one black ball is to be included in the draw, is (A) 32 (B) 64 (C) 128 (D) None of these

Step-by-Step Solution

Answer

64 ways.

Step 1: Calculate Total Ways to Draw 3 Balls

First, calculate the total number of ways to draw 3 balls from the 9 balls in the box. Use the combination formula \( \binom{n}{r} \) where \( n \) is the total number of items, and \( r \) is the number of items to choose. Here, \( \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \).

Step 2: Calculate Ways to Draw 3 Balls with No Black Ball

Next, calculate the number of ways to draw 3 balls such that no black balls are included. There are 6 non-black balls (2 white + 4 red), so calculate \( \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).

Step 3: Calculate Ways to Draw 3 Balls with At Least One Black Ball

To find the number of ways to draw at least one black ball, subtract the number of ways to draw no black balls from the total number of ways to draw any 3 balls. Thus, \( 84 - 20 = 64 \).

Key Concepts

Combinations and Permutations

In probability theory, combinations and permutations are fundamental concepts that help us understand how to count and arrange different sets of items:

• Combinations refer to the selection of items where the order does not matter. For example, choosing 2 fruits from a basket of 5 is a combination.


• Permutations, on the other hand, refer to arrangements where the order does matter. So, arranging 2 fruits out of 5 would be a permutation.

In our exercise, we are concerned with combinations since we are selecting balls and the order of selection does not change the outcome.
Using the concept of combinations, we can determine how many ways we can select a group of items from a larger pool without considering the sequence of selection. This is essential when dealing with problems like drawing balls from a box.

Binomial Coefficient Calculation

The binomial coefficient, denoted as \( \binom{n}{r} \), is a way of expressing the number of possible combinations of \( r \) objects from a set of \( n \) objects. It is central in calculating probabilities and is derived using the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In the exercise, we used this formula to first calculate the total number of ways to draw 3 balls from a total of 9 balls in the box:
* There are 9 balls, and we need to select 3, so it becomes \( \binom{9}{3} \).
* We find that there are 84 different ways to draw 3 balls when considering all possibilities without restrictions.
* This calculation is crucial as the starting point when we have further conditions to consider, like drawing at least one black ball.

Constraints in Probability

Constraints in probability refer to specific conditions or restrictions placed on an event that must be met. In this exercise, our constraint is that at least one black ball must be included in the draw.
When dealing with constraints in probability, it's often beneficial to use complementary counting techniques:

• First, calculate the total possible outcomes by including all conditions.


• Then, determine the outcomes that do not meet the constraint (e.g., drawing no black balls).


• Finally, subtract the non-compliant outcomes from the total to find the number of ways the constraint is satisfied.

Applying this to our problem: While there are 84 total ways to choose 3 balls, by calculating \( \binom{6}{3} \), we find that there are 20 ways to choose balls without any black ones. By subtracting these from the total, we find 64 ways to meet our constraint of drawing at least one black ball.