/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 Two tennis professionals, \(A\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two tennis professionals, \(A\) and \(B\), are scheduled to play a match; the winner is the first player to win three sets in a total that cannot exceed five sets. The event that \(A\) wins any one set is independent of the event that \(A\) wins any other, and the probability that \(A\) wins any one set is equal to .6. Let \(x\) equal the total number of sets in the match; that is, \(x=3,4,\) or \(5 .\) Find \(p(x)\).

Short Answer

Expert verified
The probability distribution for x is: - p(3) = 0.280 - p(4) = 0.3312 - p(5) = 0.576

Step by step solution

01

Case 1: Total of 3 sets

In this case, either A wins all the sets or B wins all the sets. The probability of A winning all sets is: $$ p_A(3) = \binom{3}{3} (0.6)^3 (1-0.6)^{3-3} = 0.6^3 = 0.216 $$ The probability of B winning all sets is: $$ p_B(3) = \binom{3}{0} (0.6)^0 (1-0.6)^{3-0} = 0.4^3 = 0.064 $$ So, the total probability of having 3 sets is: $$ p(3) = p_A(3) + p_B(3) = 0.216 + 0.064 = 0.280 $$
02

Case 2: Total of 4 sets

In this case, one of them won 3 sets, and the other won 1 set. The probability of A winning 3 sets and B winning 1 set is: $$ p_A(4) = \binom{3}{2} (0.6)^3 (1-0.6)^{3-2} = 3 \times 0.6^3 \times 0.4 = 0.2592 $$ The probability of B winning 3 sets and A winning 1 set is: $$ p_B(4) = \binom{3}{1} (0.6)^1 (1-0.6)^{3-1} = 3 \times 0.6 \times 0.4^2 = 0.072 $$ So, the total probability of having 4 sets is: $$ p(4) = p_A(4) + p_B(4) = 0.2592 + 0.072 = 0.3312 $$
03

Case 3: Total of 5 sets

In this case, one of them won 3 sets, and the other won 2 sets. The probability of A winning 3 sets and B winning 2 sets is: $$ p_A(5) = \binom{4}{2} (0.6)^3 (1-0.6)^{4-2} = 6 \times 0.6^3 \times 0.4^1 = 0.3456 $$ The probability of B winning 3 sets and A winning 2 sets is: $$ p_B(5) = \binom{4}{2} (0.6)^2 (1-0.6)^{4-2} = 6 \times 0.6^2 \times 0.4^2 = 0.2304 $$ So, the total probability of having 5 sets is: $$ p(5) = p_A(5) + p_B(5) = 0.3456 + 0.2304 = 0.576 $$
04

Final Answer

Therefore, the probability distribution for x is: $$ p(3) = 0.280 \\ p(4) = 0.3312 \\ p(5) = 0.576 $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Binomial Coefficients
Binomial coefficients play a crucial role in determining probabilities in situations where there are multiple instances of independent events. They are numbers that help us find out how many different ways we can pick a specific number of successes (e.g., winning sets) from a larger pool (e.g., total sets played). In math, they are represented with a formula and written as \( \binom{n}{k} \), meaning "n choose k". This is calculated using:
  • \( n! \) is the factorial of n, meaning the product of all positive integers up to n.
  • \( k! \) is the factorial of k.
  • \( (n-k)! \) is the factorial of \( n-k \).
So, for example, using binomial coefficients, we found that player A winning 3 sets out of 3 is calculated as \( \binom{3}{3} \). This showcases how this formula helps break down the probability of different scenarios.
Independent Events
Independent events are occasions where one outcome does not affect another. In tennis, this means the result of each set is not influenced by previous ones. So even if player A wins every set, that doesn't raise or lower the chance of them winning the next one. For probability calculations, independence means we simply multiply the probabilities together. For example, if player A's chance to win a set is 0.6, it remains 0.6 for every set, regardless of past set results. Additionally, the probability of multiple independent events happening can also be combined by multiplication, like seeing how many times player A can win in a best-of-five series.
Probability Calculation
Calculating probabilities involves simple math once the groundwork is laid with concepts like binomial coefficients and understanding independent events. To predict how likely it is for player A or player B to win the series (in 3, 4, or 5 sets), we determine the chances of different sequences happening.When we say player A's probability of winning each set is 0.6, this drives the subsequent calculations. For example:
  • Probability of A winning 3 sets and B winning none in 3 total sets: \( 0.6^3 = 0.216 \)
  • This applies similarly across other scenarios with appropriate binomials and powers.
Understanding these calculations helps predict potential match outcomes. By adding up these possibilities, we form a probability distribution like in the step-by-step coding above, where the results such as \( p(3) = 0.280 \) are derived by weighting each possible outcome's chance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particular football team is known to run \(30 \%\) of its plays to the left and \(70 \%\) to the right. A linebacker on an opposing team notes that the right guard shifts his stance most of the time \((80 \%)\) when plays go to the right and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance \(90 \%\) of the time and the shift stance the remaining \(10 \% .\) On a particular play, the linebacker notes that the guard takes a balanced stance. a. What is the probability that the play will go to the left? b. What is the probability that the play will go to the right? c. If you were the linebacker, which direction would you prepare to defend if you saw the balanced stance?

A fire-detection device uses three temperature-sensitive cells acting independently of one another in such a manner that any one or more can activate the alarm. Each cell has a probability \(p=.8\) of activating the alarm when the temperature reaches \(135^{\circ} \mathrm{F}\) or higher. Let \(x\) equal the number of cells activating the alarm when the temperature reaches \(135^{\circ} \mathrm{F}\) a. Find the probability distribution of \(x\). b. Find the probability that the alarm will function when the temperature reaches \(135^{\circ} \mathrm{F}\). c. Find the expected value and the variance for the random variable \(x\)

Two fair dice are tossed. a. What is the probability that the sum of the number of dots shown on the upper faces is equal to 7 ? To \(11 ?\) b. What is the probability that you roll "doubles" that is, both dice have the same number on the upper face? c. What is the probability that both dice show an odd number?

Most coffee drinkers take a little time each day for their favorite beverage, and many take more than one coffee break every day. The table below, adapted from a Snapshot in \(U S A\) Today, shows the probability distribution for \(x\), the number of daily coffee breaks taken per day by coffee drinkers. \({ }^{8}\) $$\begin{array}{l|lllllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .28 & .37 & .17 & .12 & .05 & .01\end{array}$$ a. What is the probability that a randomly selected coffee drinker would take no coffee breaks during the day? b. What is the probability that a randomly selected coffee drinker would take more than two coffee breaks during the day? c. Calculate the mean and standard deviation for the random variable \(x\). d. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

City crime records show that \(20 \%\) of all crimes are violent and \(80 \%\) are nonviolent, involving theft, forgery, and so on. Ninety percent of violent crimes are reported versus \(70 \%\) of nonviolent crimes. a. What is the overall reporting rate for crimes in the city? b. If a crime in progress is reported to the police, what is the probability that the crime is violent? What is the probability that it is nonviolent? c. Refer to part b. If a crime in progress is reported to the police, why is it more likely that it is a nonviolent crime? Wouldn't violent crimes be more likely to be reported? Can you explain these results?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.