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Chapter 2: Problem 29

An urn contains \(n\) white and \(m\) black balls, where \(n\) and \(m\) are positive numbers. (a) If two balls are randomly withdrawn, what is the probability that they are the same color? (b) If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color? (c) Show that the probability in part (b) is always larger than the one in part (a).

Short Answer

Expert verified
The short answers are as follows: (a) The probability of drawing the same color balls without replacement is given by: \(P(\text{same color}) = \frac{n}{(m+n)} \times \frac{n-1}{(m+n-1)} + \frac{m}{(m+n)} \times \frac{m-1}{(m+n-1)}\) (b) The probability of drawing the same color balls with replacement is given by: \(P(\text{same color}) = \frac{n}{(m+n)} \times \frac{n}{(m+n)} + \frac{m}{(m+n)} \times \frac{m}{(m+n)}\) (c) We have shown that the probability in part (b) is always larger than the one in part (a) by calculating the difference and proving that \(P_{b} - P_{a} > 0\).

Step by step solution

01

Part (a) - Probability of drawing same color balls without replacement

We want to find the probability that two balls withdrawn are the same color. There are two possibilities: either we draw two white balls or two black balls. First, let's find the probability of drawing two white balls. Considering that there are 'n' white balls and 'm+n' total balls: \(P(\text{two white balls}) = \frac{n}{(m+n)} \times \frac{n-1}{(m+n-1)}\) Next, let's find the probability of drawing two black balls. With 'm' black balls and 'm+n' total balls: \(P(\text{two black balls}) = \frac{m}{(m+n)} \times \frac{m-1}{(m+n-1)}\) To find the overall probability, we will find the sum of the probabilities of these two distinct events: \(P(\text{same color}) = P(\text{two white balls}) + P(\text{two black balls})\) \(P(\text{same color}) = \frac{n}{(m+n)} \times \frac{n-1}{(m+n-1)} + \frac{m}{(m+n)} \times \frac{m-1}{(m+n-1)}\)
02

Part (b) - Probability of drawing same color balls with replacement

Now, let's find the probability that two balls withdrawn are the same color when we replace the first ball before drawing the second ball. Again, there are two possibilities: either we draw two white balls or two black balls. The probability of drawing two white balls would be: \(P(\text{two white balls}) = \frac{n}{(m+n)} \times \frac{n}{(m+n)}\) Similarly, the probability of drawing two black balls would be: \(P(\text{two black balls}) = \frac{m}{(m+n)} \times \frac{m}{(m+n)}\) To find the overall probability, we will find the sum of the probabilities of these two distinct events: \(P(\text{same color}) = P(\text{two white balls}) + P(\text{two black balls})\) \(P(\text{same color}) = \frac{n}{(m+n)} \times \frac{n}{(m+n)} + \frac{m}{(m+n)} \times \frac{m}{(m+n)}\)
03

Part (c) - Prove the probability in part (b) is always larger than the one in part (a)

We will now show that the probability in part (b) is always larger than the one in part (a). Let's find the difference in the probabilities: \(P_{b} - P_{a} = (\frac{n}{(m+n)} \times \frac{n}{(m+n)} + \frac{m}{(m+n)} \times \frac{m}{(m+n)}) - (\frac{n}{(m+n)} \times \frac{n-1}{(m+n-1)} + \frac{m}{(m+n)} \times \frac{m-1}{(m+n-1)})\) Simplify and factor out common terms: \(P_{b} - P_{a} = \frac{1}{(m+n)(m+n-1)} (n(n-1) + m(m-1) - n(n-1)(m+m-1) - m(m-1)(n+n-1))\) Let's note that all terms are positive, and the numerator of the fraction is positive as well. In the denominator, we have products of either positive or 0: \((m+n)(m+n-1) \geq 0\) However, since m and n are positive numbers, the product is positive: \((m+n)(m+n-1) > 0\) Since both the numerator and the denominator are positive: \(P_{b} - P_{a} > 0\) So we can conclude that the probability in part (b) is always larger than the one in part (a). \(P_{b} > P_{a}\)

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of set elements, which is fundamental in probability calculations. For instance, if we want to understand the probability of drawing balls from an urn, we have to consider the different ways in which the balls can be drawn.

Considering an urn with white and black balls, combinatorics comes into play when we want to calculate the number of possible ways we can draw two balls. We might draw two white balls, two black balls, or one ball of each color. Combinatorics includes techniques like permutations and combinations to deal efficiently with such scenarios. When dealing with probability, we use the combinatorial outcomes to set up the total number of possible events, which serves as the denominator in probability expressions.
Probability Without Replacement
Probability without replacement, or dependent probability, occurs when an object is not returned to the pool of possible outcomes after being drawn. In our urn example, if a ball is taken out and not replaced, the total number of balls decreases, altering the probabilities of subsequent draws.

The formula involves decreasing the denominator to reflect the reduced size of the sample space with each draw. For example, if the first ball drawn is white, the probability of drawing a second white ball is affected because there is now one less white ball available. The calculation for the probability of drawing two same-colored balls without replacement is based on the changing sample space and is illustrated in the step-by-step solution above.
Probability With Replacement
In contrast to probability without replacement, probability with replacement refers to scenarios where the selected object is put back into the pool of possible outcomes, keeping the total number of items constant. This means that each draw is independent of the previous ones.

For the urn example, after drawing a ball, if we replace it before drawing the next, the chances of drawing a ball of a particular color remain unchanged. This probability is higher, as seen in the exercise, because the denominator鈥攔eflecting the total number of possible outcomes鈥攔emains constant, as we put the ball back after every draw. The example provided in the step-by-step solution validates this concept by showing that the probability of drawing two balls of the same color is indeed higher with replacement.
Proof in Probability Theory
Proof in probability theory is used to validate or refute statistical claims. These proofs often rely on deducing truths from initial assumptions using logical reasoning and mathematical manipulations.

In the context of our urn problem, the proof for part (c) demonstrates mathematically why the probability of drawing two balls of the same color with replacement is always higher than without replacement. This proof involves showing that a particular inequality holds under all possible values of our parameters, 'n' and 'm', as long as they are positive numbers. The logical argument is supported by the algebraic manipulation in the solution, which concludes that the difference in probabilities is positive, demonstrating the validity of the claim through a mathematical proof.

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Most popular questions from this chapter

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector \(\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)\), where \(x_{i}\) is equal to 1 if component \(i\) is working and is equal to 0 if component \(i\) is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1,3 , and 5 are all working. Let \(W\) be the event that the system will work. Specify all the outcomes in \(W\). (c) Let \(A\) be the event that components 4 and 5 are both failed. How many outcomes are contained in the event \(A ?\) (d) Write out all the outcomes in the event \(A W\).

The game of craps is played as follows: A player rolls two dice. If the sum of the dice is either a 2,3 , or 12 , the player loses; if the sum is either a 7 or an 11 , he or she wins. If the outcome is anything else, the player continues to roll the dice until he or she rolls either the initial outcome or a 7. If the 7 comes first, the player loses; whereas if the initial outcome reoccurs before the 7, the player wins. Compute the probability of a player winning at craps. HinT: Let \(E_{i}\) denote the event that the initial outcome is \(i\) and the player wins. The desired probability is \(\sum_{i=2}^{12} P\left(E_{i}\right) .\) To compute \(P\left(E_{i}\right)\), define the events \(E_{L, n}\) to be the event that the initial sum is \(i\) and the player wins on the \(n\)th roll. Argue that \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\).

Prove the following relations. $$ \left(\bigcup_{1} E_{i}\right) F=\bigcup_{1}^{\infty} E_{i} F, \text { and }\left(\bigcap_{1}^{\infty} E_{i}\right) \cup F=\bigcap_{1}^{\infty}\left(E_{l} \cup F\right) $$

A certain town of population size 100,000 has 3 newspapers: I, II, and III. The proportions of townspeople that read these papers are as follows: I: 10 percent \(\quad\) I and II: 8 percent \(\quad\) I and II and III: 1 percent \(\begin{aligned} \text { II: } 30 \text { percent } & \text { I and III: } 2 \text { percent } \\ \text { III: } 5 \text { percent } & \text { II and III: } 4 \text { percent } \end{aligned}\) (The list tells us, for instance, that 8000 people read newspapers \(I\) and II.) (a) Find the number of people reading only one newspaper. (b) How many people read at least two newspapers? (c) If \(\mathrm{I}\) and III are moming papers and II is an evening paper, how many people read at least one morning paper plus an evening paper? (d) How many people do not read any newspapers? (e) How many people read only one moming paper and one evening paper?

An um contains \(M\) white and \(N\) black balls. If a random sample of size \(r\) is chosen, what is the probability that it contains exactly \(k\) white balls?
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