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Chapter 3: Problem 43

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the um. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?

Short Answer

Expert verified
The probability that all of the balls selected are white is approximately 0.066 or 6.6%. The conditional probability that the die landed on 3 if all the balls selected are white is approximately 0.154 or 15.4%.

Step by step solution

01

Compute the conditional probabilities for each dice outcome

We need to find the probability that all selected balls are white for each possible outcome of the die. To do this, we will use the formula for conditional probability: P(A|B) = P(A and B) / P(B). In this case, A represents the event that all selected balls are white, and B represents each possible outcome of the die. For each outcome (1 to 6), we will calculate the probability of selecting all white balls as follows: Outcome 1: \(P(\text{All white}|\text{Die}=1)=\frac{\binom{5}{1}}{\binom{15}{1}}=\frac{5}{15}\) Outcome 2: \(P(\text{All white}|\text{Die}=2)=\frac{\binom{5}{2}}{\binom{15}{2}}=\frac{10}{105}\) Outcome 3: \(P(\text{All white}|\text{Die}=3)=\frac{\binom{5}{3}}{\binom{15}{3}}=\frac{10}{455}\) Outcome 4: \(P(\text{All white}|\text{Die}=4)=\frac{\binom{5}{4}}{\binom{15}{4}}=\frac{5}{1365}\) Outcome 5: \(P(\text{All white}|\text{Die}=5)=\frac{\binom{5}{5}}{\binom{15}{5}}=\frac{1}{3003}\) Outcome 6: \(P(\text{All white}|\text{Die}=6)=0\), since there are only 5 white balls.
02

Compute the total probability of selecting all white balls

To find the total probability, we will sum up the conditional probabilities for each outcome. Since the die is fair, each outcome has a probability of 1/6. Thus, the total probability is: \(P(\text{All white})=\sum_{i=1}^{6}P(\text{All white}|\text{Die}=i)P(\text{Die}=i)\) \(P(\text{All white}) = \frac{5}{15}\cdot\frac{1}{6} + \frac{10}{105}\cdot\frac{1}{6} + \frac{10}{455}\cdot\frac{1}{6} + \frac{5}{1365}\cdot\frac{1}{6} + \frac{1}{3003}\cdot\frac{1}{6}\) \(P(\text{All white}) \approx 0.066\)
03

Compute the conditional probability of the die landing on 3 given all selected balls are white

Now, we will calculate the conditional probability that the die landed on 3 given that all the balls selected are white using the formula for conditional probability: P(B|A) = P(A and B) / P(A). \(P(\text{Die}=3|\text{All white})=\frac{P(\text{All white}|\text{Die}=3)P(\text{Die}=3)}{P(\text{All white})}\) \(P(\text{Die}=3|\text{All white}) \approx \frac{\frac{10}{455}\cdot\frac{1}{6}}{0.066} \approx 0.154\) So, the probability that all of the balls selected are white is approximately 0.066 or 6.6%, and the conditional probability that the die landed on 3 if all the balls selected are white is approximately 0.154 or 15.4%.

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Key Concepts

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

Probability Theory
At the heart of many statistical analyses and games of chance, you'll find a branch of mathematics known as probability theory. It is the framework we use to quantify the likelihood of events, ranging from the roll of a die to predictions of market trends. The fundamental premise is that probability is a measure between 0 and 1, where 0 indicates an impossibility and 1 signifies a certainty.

When dealing with probability, the total number of possible outcomes plays a crucial role in determining the likelihood of any specific event. For instance, the flip of a fair coin has two possible outcomes, making the probability of either heads or tails 0.5. In more complex scenarios, like our urn of white and black balls, we have to consider all possible selections and outcomes to determine the probabilities for events of interest, such as drawing white balls.
Combinatorics
Combinatorics is a field of mathematics primarily concerned with counting, but it also delves into the arrangement and combination of objects. It is pivotal in calculating probabilities where the order of objects matters or where we select several items from a larger set.

One key tool from combinatorics is the concept of permutations and combinations. These help us to count how many ways we can select items from a set. The exercise given involves combinations, as the order in which we choose the balls does not matter. In combinatorial terms, we represent this as \( \binom{n}{k} \), which denotes the number of ways to choose k items from a larger set of n. The formulae from combinatorics are what allow us to precisely calculate the probabilities for each possible outcome of the die in the urn example.
Conditional Probability Formula
Conditional probability is the likelihood of an event occurring given that another event has already occurred. Understanding this concept is essential when events are interconnected. The conditional probability formula is expressed as \( P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \) where \(P(A|B)\) is the probability of A occurring given that B has occurred, \(P(A \text{ and } B)\) represents the joint probability of both A and B occurring, and \(P(B)\) is the probability of B.

In our urn example, event A is the balls being all white, and event B could be the die showing a three. We first compute the joint probability of both A and B happening, then divide it by the probability of B. This gives us the conditional probability of A given B, which is a more refined probability measure than considering event A in isolation. Through conditional probability, we can answer questions that take known information into account, providing more context and accuracy to our probability assessments.

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Most popular questions from this chapter

In a certain species of rats, black dominates over brown. Suppose that a black rat with two black parents has a brown sibling. (a) What is the probability that this rat is a pure black rat (as opposed to being a hybrid with one black and one brown gene)? (b) Suppose that when the black rat is mated with a brown rat, all 5 of their offspring are black. Now, what is the probability that the rat is a pure black rat?

Suppose that \(\left\\{E_{n}, n \geq 1\right\\}\) and \(\left\\{F_{n}, n \geq 1\right\\}\) are increasing sequences of events having limits \(E\) and \(F\). Show that if \(E_{n}\) is independent of \(F_{n}\) for all \(n\), then \(E\) is independent of \(F\).

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will. have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person. will have brown eyes. (Because of the latter fact we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue-eyed gene? Suppose that Smith's wife has blue eyes. (b) What is the probability that their first child will have blue eyes? (c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

On the morning of September 31,1982 , the won-lost records of the three leading baseball teams in the western division of the National League of the United States were as follows: Each team had 3 games remaining to be played. All 3 of the Giants games were with the Dodgers, and the 3 remaining games of the Braves were against the San Diego Padres. Suppose that the outcomes of all remaining games are independent and each game is equally likely to be won by either participant. What are the probabilities that each of the teams wins the division? If two teams tie for first place, they have a playoff game, which each team has an equal chance of winning.

\(A\) and \(B\) flip coins. A starts and continues flipping until a tail occurs. At this point \(B\) starts flipping and continues until there is a tail, then \(A\) takes over, and so on. Let \(P_{1}\) be the probability of the coin's landing heads when \(A\) flips, and \(P_{2}\) when \(B\) flips. The winner of the game is the first one to get (a) 2 heads in a row; (b) a total of 2 heads; (c) 3 heads in a row; (d) a total of 3 heads. In each case, find the probability that \(A\) wins.
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