/*! 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} Q20SE Suppose that two players A and B... [FREE SOLUTION] | 91影视

91影视

Suppose that two players A and B take turns rolling a pair of balanced dice and that the winner is the first player who obtains the sum of 7 on a given roll of the two dice. If A rolls first, what is the probability that B will win?

Short Answer

Expert verified

The probability that B will win is 0.454545

Step by step solution

01

Given information

There have A and B players, they have two balanced dice.

02

Step 2:Calculation of the probability that B will win

The sample space S is given by,

\(S = \left\{ {\begin{aligned}{{}{}}{\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right)}\\{\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)}\\{\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right)}\\{\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)}\\{\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)}\\{\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)}\end{aligned}} \right\}\)

The probability of sum 7 i.e., the win is,

\(\frac{6}{{36}} = \frac{1}{6}\)

The probability of not sum 7 i.e., the loose is,

\(\frac{{30}}{{36}} = \frac{5}{6}\)

For 1sttrial,

Player A rolled first and lost i.e.,

The probability of player A losing is\(\frac{5}{6}\)

Player B rolled and do 7 and win.

The probability of player B win is\(\frac{1}{6}\)

The probability of player B winning in 1sttrial is,

\(\frac{5}{6} \times \frac{1}{6} = \frac{5}{{36}}\)

The probability of player B winning in 2ndtrial is,

\(\frac{5}{{36}} \times \left( {\frac{5}{6} \times \frac{5}{6}} \right)\)

The probability of player B winning in 3rdtrial is,

\(\frac{5}{{36}} \times {\left( {\frac{5}{6} \times \frac{5}{6}} \right)^2}\)

The probability of no player winning is,

\(\frac{5}{6} \times \frac{5}{6} = \frac{{25}}{{36}}\)

So, the probability of player B winning at the nth round is,

\(\begin{aligned}{}\frac{5}{{36}}{\sum\limits_{n = 0}^\infty {\left( {\frac{{25}}{{36}}} \right)} ^{n - 1}} = \frac{5}{{36}}\left[ {1 + \frac{{25}}{{36}} + {{\left( {\frac{{25}}{{36}}} \right)}^2} + ...} \right]\\ = \frac{5}{{36}} \times \frac{1}{{\left( {1 - \frac{{25}}{{36}}} \right)}}\\ = \frac{5}{{36}} \times \frac{1}{{\left( {\frac{{36 - 25}}{{36}}} \right)}}\\ = \frac{5}{{36}} \times \frac{{36}}{{11}}\\ = \frac{5}{{11}}\\ = 0.454545\end{aligned}\)

Therefore, the probability of player B win is 0.454545

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 91影视!

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

Consider again the three different conditions (a), (b), and (c) given in Exercise 2, but suppose now that p < 1/2. For which of these three conditions is there the greatest probability that gambler A will win the initial fortune of gambler B before he loses his own initial fortune?

Suppose thatkevents\({{\bf{B}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{B}}_{\bf{k}}}\)form a partition of the sample spaceS. For\({\bf{i = 1 \ldots k}}\), let\(\Pr \left( {{{\bf{B}}_{\bf{i}}}} \right)\)denote the prior probability of\({{\bf{B}}_{\bf{i}}}\). Also, for each eventAsuch that\({\bf{Pr}}\left( {\bf{A}} \right){\bf{ > 0}}\), let\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}{\bf{|A}}} \right)\)denote the posterior probability of\({{\bf{B}}_{\bf{i}}}\)given that the eventAhas occurred. Prove that if\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}{\bf{|A}}} \right){\bf{ < Pr}}\left( {{{\bf{B}}_{\bf{1}}}} \right)\)then\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}{\bf{|A}}} \right){\bf{ > Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right)\)for at least one value of i\(\left( {{\bf{i = 2, \ldots ,k}}} \right)\).

Suppose that A and B are independent events such that\({\bf{Pr}}\left( {\bf{A}} \right){\bf{ = }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{3}}}}\right.}\!\lower0.7ex\hbox{\({\bf{3}}\)}}\)and\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ > 0}}\). What is the value of\({\bf{Pr}}\left( {{\bf{A}} \cup {{\bf{B}}{\bf{c}}}{\bf{|B}}} \right)\)?

Bus tickets in a certain city contain four numbers, U, V , W, and X. Each of these numbers is equally likely to be any of the 10 digits , and the four numbers are chosen independently. A bus rider is said to be lucky if\({\bf{U + V = W + X}}\). What proportion of the riders are lucky?

Consider a machine that produces items in sequence. Under normal operating conditions, the items are independent with a probability of 0.01 of being defective. However, it is possible for the machine to develop a 鈥渕emory鈥 in the following sense: After each defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{2}}}{{\bf{5}}}\). After each non-defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{1}}}{{{\bf{165}}}}\).

Please assume that the machine is operating normally for the whole time we observe or has a memory for the whole time we observe. LetBbe the event that the machine is operating normally, and assume that\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ = }}\frac{{\bf{2}}}{{\bf{3}}}\). Let\({{\bf{D}}_{\bf{i}}}\)be the event that theith item inspected is defective. Assume that\({{\bf{D}}_{\bf{1}}}\)is independent ofB.

a. Prove that\({\bf{Pr}}\left( {{{\bf{D}}_{\bf{i}}}} \right){\bf{ = 0}}{\bf{.01}}\) for alli. Hint:Use induction.

b. Assume that we observe the first six items and the event that occurs is\({\bf{E = D}}_{\bf{1}}^{\bf{c}} \cap {\bf{D}}_{\bf{2}}^{\bf{c}} \cap {{\bf{D}}_3} \cap {{\bf{D}}_4} \cap {\bf{D}}_{\bf{5}}^{\bf{c}} \cap {\bf{D}}_{\bf{6}}^{\bf{c}}\). The third and fourth items are defective, but the other four are not. Compute\({\bf{Pr}}\left( {{\bf{B}}\left| {\bf{E}} \right.} \right)\).

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.