Two dice are rolled. Find the probabilities of the following, as reduced fractions, and justify. Use the formula sheet for help if needed.  a. The first die is even, given that the sum is 9.

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Chapter1: Combinatorial Analysis
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Two dice are rolled. Find the probabilities of the following, as reduced fractions, and justify. Use the formula sheet for help if needed. 

a. The first die is even, given that the sum is 9. 

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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