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Chapter 4: Problem 84

Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box \(i\) with probability \(p_{i}, \sum_{i=1}^{5} p_{i}=1\) (a) Find the expected number of boxes that do not have any balls. (b) Find the expected number of boxes that have exactly 1 ball.

Short Answer

Expert verified
The expected number of boxes that do not have any balls is given by \(E[\text{empty boxes}] = \sum_{i=1}^{5} (1 - p_{i})^{10}\), and the expected number of boxes with exactly one ball is given by \(E[\text{boxes with 1 ball}] = \sum_{i=1}^{5} \binom{10}{1} \times p_{i}^1 \times (1 - p_{i})^9\).

Step by step solution

01

Part (a) - Probability of having no balls in a box#

First, we need to find the probability of a box having no balls. Since balls are placed independently, we calculate the probability of a ball not being placed in a specific box, which is (1 - \(p_i\)). After placing all 10 balls, the probability that none of them were placed into the box is given by the product of all these probabilities. Thus, the probability of a box having no balls is: \((1 - p_{i})^{10}\)
02

Part (a) - Expected number of empty boxes#

Now we find the expected number of empty boxes. To do this, we need to multiply the probability of a box having no balls by the number of boxes: \(E[\text{empty boxes}] = \sum_{i=1}^{5} (1 - p_{i})^{10}\) This is the expected number of boxes that have no balls.
03

Part (b) - Probability of having exactly one ball in a box#

Now, we need to find the probability of a box having exactly one ball. Using the binomial probability formula, we can calculate the probability of having exactly one ball in a specific box as follows: \(P(\text{exactly 1 ball}) = \binom{10}{1} \times p_{i}^1 \times (1 - p_{i})^9\)
04

Part (b) - Expected number of boxes with exactly one ball#

Finally, to find the expected number of boxes with exactly one ball, we need to multiply the probability of a box having exactly one ball by the number of boxes: \(E[\text{boxes with 1 ball}] = \sum_{i=1}^{5} \binom{10}{1} \times p_{i}^1 \times (1 - p_{i})^9\) This is the expected number of boxes that have exactly one ball.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with the analysis of random events. The fundamental concept here is the 'probability' of an event occurring, which is a measure of how likely it is that the event will happen. This can range from 0 (an event that can never happen) to 1 (an event that is certain to occur). In the context of our exercise, each ball has a probability of being placed into a particular box, summing up to 1 when considering all the boxes. This enforces the concept that a ball must go into one of the boxes, an example of a certainty in probability theory.

When analyzing problems like the one presented, it is essential to grasp the notion that probability theory allows us to make predictions about large groups of events or trials. Even though we cannot predict individual outcomes definitively, we can predict the overall behavior of the system, such as the number of empty boxes or the number with exactly one ball, with a degree of certainty provided by probability calculations.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states and where each state is not equally likely. It is applicable for scenarios with a fixed number of independent experiments (trials), two possible outcomes per trial, and a constant probability of success (or occurrence) per trial.

In our case, each ball represents a trial, and being placed in box i can be seen as 'success'. Thus, when dealing with a fixed number of ten balls and five boxes, with each ball's placement being independent of the others, the binomial distribution is ideal for modeling this situation. Specifically, finding the probability of a box containing exactly one ball involves the binomial probability formula, which factors in the number of trials, the success probability for each trial, and the number of successes out of those trials.
Expected Value
Expected value is a key concept in probability that denotes the long-term average outcome of a random variable after many repetitions of an experiment. It is a way of combining probabilities into a measure of what the average outcome will be and is often referred to as the 'mean' in the probability context.

In our exercise, the expected value helps us determine the average number of empty boxes or boxes with exactly one ball after many repeated experiments of placing the ten balls into the boxes. By calculating the probability of a box being empty or having a single ball and then multiplying by the total number of boxes, we obtain the expected numbers sought in parts (a) and (b) of the question. It's important to recognize that the expected number is not necessarily a value that must occur; it is a statistical measure that guides us towards the central tendency of the data over a long period or a large number of trials.

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Most popular questions from this chapter

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through 80. A player can select from 1 to 15 numbers; a win occurs if some fraction of the player's chosen subset matches any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of \(20,\) and the payoff is \(\$ 2.20\) won for every dollar bet. (As the player's probability of winning in this case is \(\frac{1}{4},\) it is clear that the "fair" payoff should be \(\$ 3\) won for every \(\$ 1\) bet.) When the player selects 2 numbers, a payoff (of odds) of \(\$ 12\) won for every \(\$ 1\) bet is made when both numbers are among the \(20 .\) (a) What would be the fair payoff in this case? Let \(P_{n, k}\) denote the probability that exactly \(k\) of the \(n\) numbers chosen by the player are among the 20 selected by the house. (b) Compute \(P_{n, k}\) (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet, the casino pays off as shown in the following table. Compute the expected payoff:

Each of 500 soldiers in an army company independently has a certain disease with probability \(1 / 10^{3} .\) This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? Now, suppose one of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first \(i-1\) of these tests were negative, and the \(i\) th one-which was on Jones-was positive. (d) Given the preceding scenario, what is the probability, as a function of \(i,\) that any of the remaining people have the disease?

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win \(\$ 1.10 ;\) if they are different colors, then you win \(-\$ 1.00 .\) (That is, you lose \(\$ 1.00 .\) ) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win.

Four independent flips of a fair coin are made. Let \(X\) denote the number of heads obtained. Plot the probability mass function of the random variable \(X-2\).
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