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Chapter 2: Problem 160

An urn contains six chips, numbered 1 through 6 . Two are chosen at random and their numbers are added together. What is the probability that the resulting sum is equal to \(5 ?\)

Short Answer

Expert verified
The probability that the sum of the numbers of two randomly chosen chips is 5 is \( \frac{4}{15} \).

Step by step solution

01

Total number of Possible Outcomes

We are selecting two chips out of six. This can be done in \(\binom{6}{2} = 15\) ways. So, the total number of possible outcomes is 15.
02

Favorable Outcomes

The combinations of two chips' numbers that sum to 5 are (1,4), (2,3), (3,2), (4,1). This means there are 4 favorable outcomes.
03

Probability Calculation

The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of outcomes. Thus, the required probability = \( \frac{4}{15} \).

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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 methods, arrangements, and combinations. In this exercise, we use combinatorics to determine the number of ways to select two chips from an urn containing six chips numbered 1 through 6.
Instead of examining every possible pair, combinatorics facilitates finding the total number of possible selections using combinations.
  • Combinations: This refers to selecting items from a group where the order doesn't matter. In our exercise, we're interested in selecting 2 chips from 6, irrelevant to the order of selection.
  • The formula for combinations is represented by \( \binom{n}{k} \), where **n** is the total number of items, and **k** is the number of items to choose. For our exercise, we calculate \( \binom{6}{2} \), representing the ways to choose 2 chips out of 6.
Evaluating \( \binom{6}{2} \) involves calculating \( \frac{6!}{2!(6-2)!} = 15 \). This means there are 15 potential outcomes when picking any 2 of the 6 chips without regard to the order they are selected.
Random Selection
Random selection in probability theory implies that every outcome has an equal chance of occurring. In this exercise, each pair of chips has an equal probability of being selected from the urn, ensuring a fair and unbiased process.
  • Understanding Randomness: Each chip from 1 to 6 holds the same likelihood of being drawn since selection is random. There is no influence or pattern determining which numbers appear in our selection, making it an unbiased process.
  • Simplifying Assumptions: We assume the chips are perfectly mixed, and each draw is independent of previous draws. These assumptions help maintain the integrity of randomness in the process.
This balanced approach allows us to fairly consider all potential pairs and solely focus on the combinations where their sum equals 5, without worrying about any intervention or pattern in selection.
Probability Calculation
Probability is a measure of how likely an event is to happen; it is the ratio of favorable outcomes to the total outcomes. In this exercise, our task is to calculate the probability of the sum of numbers on two chosen chips equaling 5.
  • Enumerating Favorable Outcomes: First, identify which pairs of numbers between 1 and 6 sum up to 5. The pairs are (1,4), (2,3), (3,2), and (4,1), accounting for each possible way to achieve the sum of 5.
  • Calculating Probability: We've established there are 4 favorable outcomes and 15 possible total outcomes from our combination analysis. Therefore, the probability of drawing two chips whose numbers sum to 5 is \( \frac{4}{15} \).
Understanding and calculating such probabilities is an essential skill in probability theory, allowing for the prediction and understanding of random events in a structured manner.

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Most popular questions from this chapter

Lucy is currently running two dot-com scams out of a bogus chatroom. She estimates that the chances of the first one leading to her arrest are one in ten; the "risk" associated with the second is more on the order of one in thirty. She considers the likelihood that she gets busted for both to be \(0.0025\). What are Lucy's chances of avoiding incarceration?

A dashboard warning light is supposed to flash red if a car's oil pressure is too low. On a certain model, the probability of the light flashing when it should is \(0.99 ; 2 \%\) of the time, though, it flashes for no apparent reason. If there is a \(10 \%\) chance that the oil pressure really is low, what is the probability that a driver needs to be concerned if the warning light goes on?

Your favorite college football team has had a good season so far but they need to win at least two of their last four games to qualify for a New Year's Day bowl bid. Oddsmakers estimate the team's probabilities of winning each of their last four games to be \(0.60,0.50,0.40\), and \(0.70\), respectively. (a) What are the chances that you will get to watch your team play on Jan. 1 ? (b) Is the probability that your team wins all four games given that they have won at least three games equal to the probability that they win the fourth game? Explain. (c) Is the probability that your team wins all four games given that they won the first three games equal to the probability that they win the fourth game?

A man has \(n\) keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourceful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is \(1 / n\). Show that the probability the door opens with the third key he tries is also \(1 / n\). (Hint: What has to happen before he even gets to the third key?)

Suppose that two cards are drawn simultaneously from a standard fifty-two-card poker deck. Let \(A\) be the event that both are either a jack, queen, king, or ace of hearts, and let \(B\) be the event that both are aces Are \(A\) and \(B\) independent? (Note: There are 1326 equally likely ways to draw two cards from a poker deck.)
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