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A bowl contains 10 red balls and 10 blue balls. A woman selects balls at random without looking at them. a) How many balls must she select to be sure of having at least three balls of the same color? b) How many balls must she select to be sure of having at least three blue balls?

Short Answer

Expert verified
a) 5 balls, b) 13 balls

Step by step solution

01

- Understanding Part (a)

To ensure having at least three balls of the same color, we need to consider the worst-case scenario. The worst-case scenario is drawing two red balls and two blue balls alternately.
02

- Calculating for Part (a)

In the worst case, she could draw two red balls and two blue balls, but not yet have three balls of the same color. This means she has to draw one more ball to guarantee that she will have at least three balls of the same color. Thus, she must draw 5 balls.
03

- Understanding Part (b)

To ensure having at least three blue balls specifically, we again consider the worst-case scenario. The worst-case scenario would be drawing all the red balls before getting three blue balls.
04

- Calculating for Part (b)

In the worst case, the woman could draw all ten red balls first. After that, she would need to draw another three balls to make sure she has at least three blue balls. Thus, she must draw 13 balls.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

worst-case scenario
When we talk about a worst-case scenario in problems involving drawing balls from a bowl, we mean the most unfavorable or extreme circumstance that can occur. This helps us determine the maximum number of attempts needed to achieve a certain outcome.

In the given problem, for both parts (a) and (b), we start by assuming the worst-case scenario.
- For part (a), it involves alternating drawing two red balls and two blue balls, resulting in not reaching three balls of the same color until a fifth ball is drawn.
- For part (b), the worst-case scenario involves drawing all red balls before any blue ones, ensuring that we must draw a substantial number of balls to guarantee three blue ones.
probability
Probability is the measure of the likelihood that an event will occur. It ranges from 0 (impossible) to 1 (certain). Although the given problem doesn't directly ask for probabilities, understanding them can still help.

When the woman draws balls without looking, each draw's outcome is uncertain. The probability that she draws a red or a blue ball changes with each draw, especially as she keeps drawing more balls.
- If she starts with 10 red and 10 blue balls, the probability of drawing a specific color changes as balls are drawn. For instance, after two red balls are drawn, the probability of drawing another red ball diminishes, while the probability for blue balls increases.
- This dynamic nature of probability underscores why considering the worst-case scenario is crucial for guarantees.
drawing balls
Drawing balls is a common problem context in combinatorics and probability exercises. It involves selecting items from a set, often without replacement, which makes each selection impact the subsequent ones.

In our problem:
- The woman draws balls from a bowl containing 10 red and 10 blue balls. Without replacement, each draw changes the composition of the balls remaining in the bowl.
- For instance, if she draws two red balls first, only 8 red and 10 blue balls remain, affecting the likelihood of each color being drawn next. This ongoing change influences the strategy used to solve the problem.
- Understanding the nature of drawing balls helps grasp why some results (like needing to draw up to 13 balls in part b) might seem large but are necessary to cover worst-case scenarios.
combinatorial analysis
Combinatorial analysis involves counting, arranging, and analyzing ways to choose items from sets. It's essential for problems like ours where we need to ensure a certain outcome by considering various arrangements.

Let's break it down:
- In part (a): We're considering combinations of drawing red and blue balls. By examining the different ways balls can be selected, we find that drawing five balls covers all possible arrangements, ensuring at least three balls of one color.
- In part (b): We analyze the combinations where red balls are drawn first. To ensure at least three blue balls, we calculate the number drawn by starting with the total red balls and then drawing additional ones.
- This systematic approach lets us cover all potential cases and find guarantees under worst-case scenarios.

Combinatorial analysis ensures that we don't miss any possibilities and helps in providing a methodical solution to such problems.

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Most popular questions from this chapter

This procedure is used to break ties in games in the championship round of the World Cup soccer tournament. Each team selects five players in a prescribed order. Each of these players takes a penalty kick, with a player from the first team followed by a player from the second team and so on, following the order of players specified. If the score is still tied at the end of the 10 penalty kicks, this procedure is repeated. If the score is still tied after 20 penalty kicks, a sudden-death shootout occurs, with the first team scoring an unanswered goal victorious. a) How many different scoring scenarios are possible if the game is settled in the first round of 10 penalty kicks, where the round ends once it is impossible for a team to equal the number of goals scored by the other team? b) How many different scoring scenarios for the first and second groups of penalty kicks are possible if the game is settled in the second round of 10 penalty kicks? c) How many scoring scenarios are possible for the full set of penalty kicks if the game is settled with no more than 10 total additional kicks after the two rounds of five kicks for each team?

How many permutations of {a, b, c, d, e, f, g} end with a?

A circular \(r\) -permutation of \(n\) people is a seating of \(r\) of these \(n\) people around a circular table, where seatings are considered to be the same if they can be obtained from each other by rotating the table. Find the number of circular 3 -permutations of 5 people.

Suppose that a weapons inspector must inspect each of five different sites twice, visiting one site per day. The inspector is free to select the order in which to visit these sites, but cannot visit site X, the most suspicious site, on two consecutive days. In how many different orders can the inspector visit these sites?

Prove the Multinomial Theorem: If \(n\) is a positive integer, then $$\left(x_{1}+x_{2}+\cdots+x_{m}\right)^{n}$$ $$=\sum_{n_{1}+n_{2}+\cdots+n_{m}=n} C\left(n ; n_{1}, n_{2}, \ldots, n_{m}\right) x_{1}^{n_{1}} x_{2}^{n_{2}} \cdots x_{m}^{n_{m}}$$ where $$ C\left(n ; n_{1}, n_{2}, \ldots, n_{m}\right)=\frac{n !}{n_{1} ! n_{2} ! \cdots n_{m} !} $$ is a multinomial coefficient.

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