Question

A box has 2 red, 3 white and 6 black balls. In how many ways can 3 balls be taken out from the box if at least one of the balls is white. All given balls are considered to be distinct. (1) 135 (2) 61 (3) 84 (4) 109

Short answer

There are 109 ways to choose 3 balls with at least one white ball.

Step-by-step solution

- Calculate Total Ways to Choose 3 Balls

First, calculate the total number of ways to choose 3 balls from the box. The box contains 2 red, 3 white, and 6 black balls, a total of 11 balls. Use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, \( n = 11 \) and \( r = 3 \). Thus, \[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \]

- Calculate Ways with No White Balls

Next, calculate the number of ways to choose 3 balls from only the red and black balls, as they are the selections where no white ball is included. There are 2 red and 6 black balls, giving us a total of 8 balls. Use the combination formula again: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

- Calculate Ways with At Least One White Ball

To find the number of ways to choose 3 balls with at least one white ball, subtract the number of ways with no white balls (from Step 2) from the total number of ways (from Step 1): \[ 165 - 56 = 109 \]

Key concepts

Combinations

When we talk about combinations, we're dealing with a way to select items from a group, where the order does not matter. It’s different from permutations where the order does matter. To calculate combinations, we use the formula:
\( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

Here, \( n \) is the total number of items to pick from, and \( r \) is the number of items we want to pick. In our exercise, we have 11 balls in total (2 red, 3 white, and 6 black), and we need to choose 3. Using the combination formula, we can find that there are 165 ways to choose 3 balls from these 11.

Remember, combinations are used when the way we order our selections doesn’t matter. Whether we pick the first ball as red, second as white, and third as black, it doesn’t change if we pick them in the reverse order. So, the combination formula simplifies our calculation by ignoring the order.

Binomial Coefficient

The binomial coefficient is represented by \( \binom{n}{r} \), and it plays a critical role in calculating combinations. It tells us how many ways we can choose \( r \) items out of \( n \) without worrying about the order. This coefficient is derived from the formula:
\( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

In simpler terms:

• \( n! \) (n factorial) means we multiply all whole numbers from 1 up to n.
• \( r! \) is the factorial of the number of items we want to choose.
• \( (n-r)! \) is the factorial of the difference between the total items and the items we want to choose.

This might look complex, but it’s just about multiplying and dividing numbers to find out how many ways we can create groups. Use the binomial coefficient when solving combination problems to simplify calculations.

In our specific exercise, we used the binomial coefficient twice: first to find the total ways to choose any 3 balls from 11, and then to find the ways to do it without picking any white balls.

Probability with Distinct Objects

When dealing with probability, the distinction between objects plays a significant role. If all objects were indistinguishable, we wouldn't need combinations. But since each ball in our box is distinct, we have to consider this in our calculations.

In our exercise, each ball color (red, white, black) represents distinct objects. This affects our probability calculations because:

• We need to account for different ways we can choose balls while making sure the specific count of white balls is considered.
• We calculated the overall possibility and then excluded the scenarios without any white balls to ensure that at least one ball is white.
• This helps us filter out the unwanted selections, focusing only on the desired combinations.

By removing scenarios where no white ball is chosen, we ensure that every combination meets the condition of having at least one white ball.

Finally, subtracting these unwanted cases (ways with no white balls) from the total helps us find our desired probability. This method simplifies complex problems by breaking them down step-by-step, ensuring accurate results.