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If 16 chocolate-chip cookies are distributed among four children, what is the probability that every child gets (a) at least one cookie? (b) at least two cookies?

Short Answer

Expert verified
(a) The probability for each child to get at least one cookie is \(\frac{C(15,3)}{C(19,3)}\). (b) The probability for each child to get at least two cookies is \(\frac{C(11,3)}{C(19,3)}\).

Step by step solution

01

Total Ways Of Distribution

First, calculate the total number of ways 16 cookies can be distributed among four children without any restrictions. Here, the concept of 'stars and bars' or the 'balls and urns' method is appropriate. The number of ways can be calculated using the formula \( C(n+k-1,k-1) \) where n is the number of cookies and k is the number of children. Therefore, the total number of distributions is \( C(16+4-1,4-1) = C(19,3) \).
02

Distribution So That Each Child Gets At Least One Cookie

Now, we need to calculate the number of ways the cookies can be distributed so that each child gets at least one cookie. The situation is the same as the previous one with the conditional that each child has already received one cookie. Hence, we distribute the remaining 16-4 cookies in any way. The number of ways can be calculated as \( C((16-4)+4-1,4-1) = C(15,3) \). The required probability is then \(\frac{C(15,3)}{C(19,3)}\).
03

Distribution So That Each Child Gets At Least Two Cookies

For each child to get at least two cookies, each child must initially be given two cookies, leaving us with 16-2*4 = 8 cookies. Now the cookies can be distributed in any way among the four children. The number of ways can be calculated as \( C(8+4-1,4-1) = C(11,3) \). The required probability is then \(\frac{C(11,3)}{C(19,3)}\).

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Key Concepts

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

Understanding Probability Distribution
Probability distribution is a mathematical concept that describes how probabilities are spread across possible outcomes in a random experiment. In the context of distributing cookies among children, it represents the likelihood that a specific distribution occurs under certain conditions. For example, calculating the probability that each child receives at least one cookie involves considering all possible allocations and then focusing on those that meet the criteria.

To compute such probabilities, we look at the fraction of favorable outcomes over possible outcomes. In our problem, we're first interested in finding out how many ways 16 cookies can be distributed among four kids with no restrictions, and then how that changes if conditions like 'at least one cookie per child' are applied.
  • The probability of a desired event is calculated as: \ \[ \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
  • Total possible outcomes are determined without restrictions.
  • Favorable outcomes meet the specific problem conditions.
The knowledge of probability distribution helps us assess how likely various outcomes are, given certain parameters or constraints.
Applying the Stars and Bars Method
The stars and bars method is a clever combinatorial technique used to solve problems of distributing indistinguishable objects (like cookies) into distinguishable boxes (children). It's often called the 'balls and urns' method too.

The main idea is to visualize the distribution as arranging a sequence of stars (cookies) and bars (dividers between children). The formula used is based on the idea of choosing positions for the bars among the stars:
  • Formula: The number of ways to distribute n cookies among k children is \ \[C(n+k-1, k-1)\]
  • 'n' is the total number of cookies.
  • 'k' is the number of children.
In our exercise:
  • With no constraints, distribute 16 cookies among 4 children, calculated as \( C(19, 3) \) .
  • For at least one cookie per child, subtracting 4 as every child gets one first, calculated as \( C(15, 3) \) .
  • For at least two cookies per child, first allocate 8 cookies to ensure each has two. Then distribute the remaining as \(C(11, 3)\) .
This method is crucial for breaking down complex distributions into manageable problems.
Explaining the Binomial Coefficient
The binomial coefficient is a central element in combinatorics used to calculate the number of ways to choose a subset of a certain size from a larger set. It's frequently represented as \( C(n,k)\), where it quantifies the number of ways to pick \( k\) elements from \( n\) total items.

In simpler terms, if you have a collection of items, the binomial coefficient tells you how many different combinations you can form by selecting a specific number of them.
  • The formula for the binomial coefficient: \ \[ C(n,k) = \frac{n!}{k!(n-k)!} \]
  • 'n!' (n factorial) is the product of all positive integers up to n.
  • 'k!' is the factorial of the subset size, and '(n-k)!' factors for the exclusion of those not picked.
Referencing our problem:
  • We calculate \( C(19, 3)\) for unrestricted distribution.
  • We calculate \( C(15, 3)\) to ensure all children get at least one.
  • And \( C(11, 3)\) for each to get at least two.
Being versed with binomial coefficients is essential for tackling various combinatorial tasks, making seemingly complex distribution problems approachable and straightforward.

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Most popular questions from this chapter

Let ? be a given universe with \(A, B \subseteq \mathscr{,} A \cap B=\) \(\theta,|A|=12\), and \(|B|=10\). If seven elements are selected from \(A \cup B\), what is the probability the selection contains four elements from \(A\) and three from \(B\) ?

a) For positive integers \(m, n, r\), with \(r \leq \min \\{m, n\\}\), show that $$ \begin{gathered} \left(\begin{array}{c} m+n \\ r \end{array}\right)=\left(\begin{array}{c} m \\ 0 \end{array}\right)\left(\begin{array}{l} n \\ r \end{array}\right)+\left(\begin{array}{c} m \\ 1 \end{array}\right)\left(\begin{array}{c} n \\ r-1 \end{array}\right)+\left(\begin{array}{c} m \\ 2 \end{array}\right)\left(\begin{array}{c} n \\ r-2 \end{array}\right) \\ +\cdots+\left(\begin{array}{c} m \\ r \end{array}\right)\left(\begin{array}{l} n \\ 0 \end{array}\right)=\sum_{k=0}^{r}\left(\begin{array}{c} m \\ k \end{array}\right)\left(\begin{array}{c} n \\ r-k \end{array}\right) . \end{gathered} $$ b) For \(n\) a positive integer, show that $$ \left(\begin{array}{l} 2 n \\ n \end{array}\right)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)^{2} . $$

Write a computer program (or develop an algorithm) that prints out all of the subsets of \(\\{1,2,3, \ldots, n\\}\), where \(1 \leq n \leq 10\). (The value of \(n\) should be supplied during program execution.)

a) How many subsets of \(\\{1,2,3, \ldots, 11\\}\) contain at least one even integer? b) How many subsets of \(\\{1,2,3, \ldots, 12\\}\) contain at least one even integer? c) Generalize the results of parts (a) and (b).

At a high-school science fair, 34 students received awards for scientific projects. Fourteen awards were given for projects in biology, 13 in chemistry, and 21 in physics. If three students received awards in all three subject areas, how many received awards for exactly (a) one subject area? (b) two subject areas?

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