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Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, three green ones, two white ones, and one purple one. She grabs five of them. Find the probabilities of the following events, expressing each as a fraction in lowest terms. HINT [See Example 1.] She has none of the red ones.

Short Answer

Expert verified
The probability of Suzan grabbing 5 marbles with no red ones is \(\frac{1}{84}\).

Step by step solution

01

Calculate the total marbles in the bag

By adding the number of marbles of each color, we determine there are a total of 4 + 3 + 2 + 1 = 10 marbles in the bag.
02

Calculate the total number of ways to grab 5 marbles

Suzan grabs 5 marbles out of the 10 marbles in the bag. We can determine the total number of ways she can do this by calculating the number of combinations using the binomial coefficient formula, where n is the total number of items and r is the number of items to be chosen: e.g., \(\binom{n}{r}\). In this case, n = 10 and r = 5, so the total number of ways to grab 5 marbles is \(\binom{10}{5} = 252\).
03

Calculate the number of ways to grab 5 marbles without red ones

Since we are trying to find the probability of grabbing 5 marbles with no red marbles, we can only choose the remaining green, white, and purple marbles. There are 3 green ones, 2 white ones, and 1 purple one. So, Suzan needs to grab all 6 non-red marbles out of the 6 available non-red marbles. Since Suzan needs to grab 5 marbles, there are only two possible scenarios in that she grabs all 3 green marbles and the additional 2 marbles are chosen from the remaining 3 non-red ones (1 white and 2 purple). The number of ways to choose 2 marbles from the 3 non-red ones is \(\binom{3}{2} = 3\).
04

Calculate the probability of grabbing 5 marbles with no red ones

Now, we can calculate the probability by dividing the number of ways to grab 5 marbles with no red ones (3 ways) by the total number of ways to grab 5 marbles (252 ways). The probability is given by: \[\frac{3}{252} = \frac{1}{84}\] Hence, the probability of Suzan grabbing 5 marbles with no red ones is \(\frac{1}{84}\).

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Key Concepts

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

Combinatorics
Combinatorics is a field of mathematics that deals with counting, arranging, and finding patterns. It's like a creative math puzzle where you explore how different objects can be combined or switched around. This is not only fun, but it can be extremely useful in solving real-world problems.

When we talk about Suzan grabbing marbles, we're diving into the world of combinatorics. We want to know how many different ways she can grab a set of marbles from the bag. In other words, we're counting how many combinations of marbles are possible when she reaches in and plucks out five marbles at a time. Understanding this helps us calculate probabilities more easily.

The essential thing to grasp here is how to count these combinations efficiently, which leads us to the binomial coefficient.
Binomial Coefficient
The binomial coefficient is a fancy name for a specific way to count combinations. Imagine you have a group of items, and you want to know how many unique ways you can pick a certain number of them. That's what the binomial coefficient tells you.

In Suzan's scenario, we have a total of 10 marbles, and she picks 5. The binomial coefficient, denoted as \(\binom{n}{r}\), represents the number of such combinations possible from \(n\) items when choosing \(r\) at a time. Here, \(\binom{10}{5} = 252\), meaning there are 252 unique combinations of picking 5 marbles from the bag.

Mathematically, the formula for the binomial coefficient is:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

This formula uses the concept of factorial, where \( n! \) (n-factorial) means multiplying all whole numbers from \( n \) down to 1.
Probability Calculation
Probability helps us determine how likely certain events are to happen. In Suzan’s marble adventure, we use probability to find out how likely she is to grab marbles without getting any red ones.

The probability of an event is calculated by dividing the number of successful outcomes by the total number of possible outcomes.

In Suzan's exercise, we determined there are 3 ways she can grab 5 marbles that don't include a red marble. Since we earlier calculated there are 252 different ways to grab any 5 marbles from the bag, the probability of her grabbing no red ones becomes:
  • \( \frac{3}{252} = \frac{1}{84} \)

This means that for every 84 times Suzan grabs 5 marbles from the bag, on average just once will she grab a combination that doesn't include any red marbles. Understanding these calculations is crucial for making sense of how probability informs decisions and predictions in both everyday and scientific contexts.

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Most popular questions from this chapter

\(\nabla\) Two distinguishable dice are rolled. Could there be two mutually exclusive events that both contain outcomes in which the numbers facing up add to 7 ?

Explain how the property \(P\left(A^{\prime}\right)=1-P(A)\) follows directly from the properties of a probability distribution.

Polls According to the New York Times/CBS poll of March 2005, referred to in Exercise \(65,72 \%\) of those polled ranked neither Iraq nor North Korea as the top foreign policy issue. \({ }^{33}\) What is the probability that a randomly selected person polled ranked either Iraq or North Korea as the top foreign policy issue?

\begin{aligned} &\text { 50. I } \nabla \text { \mathrm{\\{} I n c o m e ~ D i s t r i b u t i o n ~ R e p e a t ~ E x e r c i s e ~ } 49 \text { , using the fol- }\\\ &\text { lowing data: }\\\ &\begin{aligned} &\text { Father's } & \multicolumn{4}{c} {\text { Son's income }} \\ &\text { Income } & \text { Bottom } \mathbf{1 0 \%} & \mathbf{1 0 - 5 0 \%} & \mathbf{5 0 - 9 0 \%} & \text { Top 10\% } \\ &\cline { 2 - 5 } \text { Bottom 10\% } & .50 & .32 & .17 & .01 \\ &\cline { 2 - 5 } \mathbf{1 0 - 5 0 \%} & .10 & .48 & .38 & .04 \\ &\cline { 2 - 5 } \mathbf{5 0 - 9 0 \%} & .04 & .38 & .48 & .10 \\ &\cline { 2 - 5 } \text { Top 10\% } & .01 & .17 & .32 & .50 \\ &\cline { 2 - 5 } & & & & \end{aligned} \end{aligned}

Your best friend thinks that it is impossible for two mutually exclusive events with nonzero probabilities to be independent. Establish whether or not he is correct.

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