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Chapter 9: Problem 67

American roulette is a game in which a wheel turns on a spindle and is divided into \( 38 \) pockets.Thirty-six of the pockets are numbered \( 1-36 \), of which half are red and half are black. Two of the pockets are green and are numbered \( 0 \) and \( 00 \) (see figure). The dealer spins the wheel and a small ball in opposite directions. As the ball slows to a stop, it has an equal probability of landing in any of the numbered pockets. (a) Find the probability of landing in the number \( 00 \) pocket. (b) Find the probability of landing in a red pocket. (c) Find the probability of landing in a green pocket or a black pocket. (d) Find the probability of landing in the number \( 14 \) pocket on two consecutive spins. (e) Find the probability of landing in a red pocket on three consecutive spins.

Short Answer

Expert verified
(a) The probability of landing in the number 00 pocket is \(1/38\). (b) The probability of landing in a red pocket is \(9/19\). (c) The probability of landing in a green pocket or a black pocket is \(10/19\). (d) The probability of landing in the number 14 pocket on two consecutive spins is \(1/1444\). (e) The probability of landing in a red pocket on three consecutive spins is \(729/6859\).

Step by step solution

01

Understand the question

The game of American Roulette has 38 equally likely outcomes. Therefore, the probability of any single outcome is \(1/38\) because each pocket has an equal chance be selected.
02

Find the probability of landing in the number 00 pocket

There is only one '00' pocket, out of the total 38 pockets. Therefore, the probability is \(1/38\).
03

Find the probability of landing in a red pocket

There are 18 red pockets out of a total 38 pockets. So, the probability of landing in a red pocket is \(18/38\) or \(9/19\).
04

Find the probability of landing in a green pocket or a black pocket

A green pocket or a black pocket outcome means the ball could land in either of the two, so we add their individual probabilities. There are 2 green pockets and 18 black pockets, so the probability is \((2 + 18)/38\) or \(20/38 = 10/19\).
05

Find the probability of landing in the number 14 pocket on two consecutive spins

The events of spinning the roulette wheel are independent, so the probability of landing in pocket 14 twice in a row is the product of their individual probabilities, which is \((1/38) * (1/38)\) or \(1/1444\).
06

Find the probability of landing in a red pocket on three consecutive spins

Similarly, the probability of landing in a red pocket thrice in a row is the product of their individual probabilities, which is \((9/19) * (9/19) * (9/19)\) or \(729/6859\).

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Most popular questions from this chapter

You and a friend agree to meet at your favorite fast - food restaurant between \( 5:00 \) and \( 6:00 \) P.M.The one who arrives first will wait \( 15 \) minutes for the other, and then will leave (see figure). What is the probability that the two of you will actually meet,assuming that your arrival times are random within the hour?

In Exercises 21 - 24, find the probability for the experiment of selecting one card from a standard deck of \( 52 \) playing cards. The card is not a face card.

In Exercises 53 - 60, the sample spaces are large and you should use the counting principles discussed in Section 9.6. A class is given a list of \( 20 \) study problems, from which \( 10 \) will be part of an upcoming exam. A student knows how to solve \( 15 \) of the problems. Find the probabilities that the student will be able to answer (a) all \( 10 \) questions on the exam, (b)exactly eight questions on the exam, and (c) at least nine questions on the exam.

Consider a group of people. (a) Explain why the following pattern gives the probabilities that the people have distinct birthdays. \( n = 2: \dfrac{365}{365} \cdot \dfrac{364}{365} = \dfrac{365 \cdot 364}{365^2} \) \( n = 3: \dfrac{365}{365} \cdot \dfrac{364}{365} \cdot \dfrac{363}{365} = \dfrac{365 \cdot 364 \cdot 363}{365^3} \) (b) Use the pattern in part (a) to write an expression for the probability that \( n = 4 \) people have distinct birthdays. (c) Let \( P_n \) be the probability that the \( n \) people have distinct birthdays. Verify that this probability can be obtained recursively by \( P_1 = 1 \) and \( P_n = \dfrac{365 - (n - 1)}{365} P_{n - 1} \). (d) Explain why \( Q_n = 1 - P_n \) gives the probability that at least two people in a group of \( n \) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \( \dfrac{1}{2} \)? Explain.

In Exercises 1 - 7, fill in the blanks. If two events from the same sample space have no outcomes in common, then the two events are ________ ________.
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