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2. Two dice are rolled. Find the probabilities of the following as reduced fractions and justify. Use the formula sheet to help you out.  a. Are the events "first die is even" and "sum = 10" independent?

2. Two dice are rolled. Find the probabilities of the following as reduced fractions and justify. Use the formula sheet to help you out.  a. Are the events "first die is even" and "sum = 10" independent?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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2. Two dice are rolled. Find the probabilities of the following as reduced fractions and justify. Use the formula sheet to help you out. 

a. Are the events "first die is even" and "sum = 10" independent?

Basic Probability Principle P(E) = n(s) Product Rule P(EnF) = P(E) · P(FE) = P(F) · P(E|F) Union Rule For sets: n(AUB) = n(A) + n(B) - n(ANB) For probability: P(EUF) = P(E) + P(F) − P(E^F) Complement Rule (Probability) P(E') = 1 - P(E) P(E) = 1 - P(E') Complement Rule (Number of Outcomes) n(E) = n(S) - n(E') n(E') = n(S) — n(E) Permutations P(n, k) Bayes' Theorem P(F;).P(E|F;) P(F;|E) = P(F1).P(E|F1) + P(F2)·P(E|F2) + ... + P(Fn)·P(E|Fn) P(F).P(EF) Special Case: P(F|E) = P(F).P(E|F)+P(F').P(E\F') n! (n-k)! Conditional Probability n(EnF) P(ENF) P(F) n(F) Distinguishable Permutations n! n₁!n₂!...nk! P(E|F) = = Expected Value E(x) = x₁p₁ + x2P2 + ·+XnPn Binomial Probability P(k successes in n trials) = C(n. k)p(1-p)"-h ... Combinations C(n, k) = k!(n-k)! Expected # of Successes of a Binomial or Combination-based Random Variable E(x) = np = 2 Odds Odds in favor of E= If odds in favor of E are m:n, then P(E) m m+n Number of subsets If a set has n elements, then it has 2" subsets. = P(E) P(E') Independent Events P(E|F) = P(E) P(F|E) = P(F) P(EnF) = P(E). P(F) Table of Outcomes for Rolling 2 Dice 1 2 3 6 4 5 1|(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5.3) (5,4) (5,5) (5,6) 6(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Table of Sums when Rolling 2 Dice 2 3 4 5 6 5 6 6 7 8 1 1 2 3 4 2 3 4 5 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Irod
Transcribed Image Text:Basic Probability Principle P(E) = n(s) Product Rule P(EnF) = P(E) · P(FE) = P(F) · P(E|F) Union Rule For sets: n(AUB) = n(A) + n(B) - n(ANB) For probability: P(EUF) = P(E) + P(F) − P(E^F) Complement Rule (Probability) P(E') = 1 - P(E) P(E) = 1 - P(E') Complement Rule (Number of Outcomes) n(E) = n(S) - n(E') n(E') = n(S) — n(E) Permutations P(n, k) Bayes' Theorem P(F;).P(E|F;) P(F;|E) = P(F1).P(E|F1) + P(F2)·P(E|F2) + ... + P(Fn)·P(E|Fn) P(F).P(EF) Special Case: P(F|E) = P(F).P(E|F)+P(F').P(E\F') n! (n-k)! Conditional Probability n(EnF) P(ENF) P(F) n(F) Distinguishable Permutations n! n₁!n₂!...nk! P(E|F) = = Expected Value E(x) = x₁p₁ + x2P2 + ·+XnPn Binomial Probability P(k successes in n trials) = C(n. k)p(1-p)"-h ... Combinations C(n, k) = k!(n-k)! Expected # of Successes of a Binomial or Combination-based Random Variable E(x) = np = 2 Odds Odds in favor of E= If odds in favor of E are m:n, then P(E) m m+n Number of subsets If a set has n elements, then it has 2" subsets. = P(E) P(E') Independent Events P(E|F) = P(E) P(F|E) = P(F) P(EnF) = P(E). P(F) Table of Outcomes for Rolling 2 Dice 1 2 3 6 4 5 1|(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5.3) (5,4) (5,5) (5,6) 6(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Table of Sums when Rolling 2 Dice 2 3 4 5 6 5 6 6 7 8 1 1 2 3 4 2 3 4 5 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Irod
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