Dat 605 HW 6 - PROBABILITY AND COMBINATORICS

HW 6

library(matlib)
library(MASS)
require(plotrix)
require(grid)
library(ggplot2)
require(ggthemes)
require(combinat)

Prob 1

red_marbles <- 54
white_marbles <- 9
blue_marbles <- 75

Tot_marbles <- red_marbles+white_marbles+blue_marbles 

Pred <- round(red_marbles/Tot_marbles,4)
#probability of pulling red marble


Pblue <- round(blue_marbles/Tot_marbles,4)
#probability of pulling blue marble


# P(red or blue) = P(red) + P(blue)
round(Pred + Pblue, 4)

## [1] 0.9348

Prob 2

g <- 19
r <- 20
b <-24
y <- 17

tot <- g+r+b+y

Pr <- round(r/tot,4)

Pr

## [1] 0.25

Prob 3

Prob 4

Going to Gym and Losing Weight Determine if the following events are independent. Going to the gym. Losing weight.

A)Dependent B) Independent

Ans: (A) Dependent

One event affects the other event

Prob 5

C(n,k) = $\frac{n!}{(n-k)!k!}$

Ctot = $\frac{8!}{(8-3)!3!}$ $\frac{7!}{(7-3)!3!}$ $\frac{3!}{(3-1)!1!}$

wraps <- (factorial(8)/(factorial(5)*factorial(3)))*(factorial(7)/(factorial(4)*factorial(3)))*(factorial(3)/(factorial(2)*factorial(1)))
wraps

## [1] 5880

Prob 6

Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news.

Ans: B)Independent

One event does NOT affect another event

Prob 7

P(n,k) = $\frac{n!}{(n-k)!}$

P(14,8) = $\frac{14!}{(14-8)!}$

#Permutation problem
cabinet <- factorial(14)/factorial(6)
cabinet

## [1] 121080960

Prob 8

|S| = $\binom{22}{4}$, All possible combo in the sample space

|R| = $\binom{9}{0}$, Red sample space and its combo

|O| = $\binom{4}{1}$, Orange sample space and its combo

|G| = $\binom{9}{3}$, Greens sample space and its combo

choose(9,0)*choose(4,1)*choose(9,3)/choose(22,4)

## [1] 0.04593301

Prob 9

 factorial(11)/factorial(7)

## [1] 7920

Prob 10

ans:

33% of suscribers to a fitness magazine are 34 or younger.

Prob 11, step 1

Binomial Distribution

Generic Binomial formula:

P(k heads and n-k tails) = $\binom{n}{k}$ * $p^{k}$*$(1-p)^{k}$

whereby p(heads) = 0.5 and p(tail) = 1- 0.5 = 0.5

P(Sucess = 3 heads) = $\binom{4}{3}$ * $0.5^{3}$*$(1-0.5)^{1}$

$\ E$(x) = 0.25$97 - 0.75$30

# Prob of success
ps <- 4*0.5^3*0.5^1


#Prob of failure
qf <- 1-ps

#Expected value of the proposition

E <- 0.25*97 -  0.75*30

E

## [1] 1.75

Prob 11, step 2

N$\ E$(x) = 559$1.75, where N is no. of times of playing the game

Exp1 <- 559*1.75
Exp1

## [1] 978.25

Prob 12, step 1

Following prob 11, P(Sucess = 4 heads or less) = $\binom{n}{k}$ * $p^{k}$*$(1-p)^{k}$

whereby k goes from 0 to 4 while n = 9

P(sucess) = 0.5 and P(failure) = 1-0.5 = 0.5

comb = function(n, x) {
  factorial(n) / factorial(n-x) / factorial(x)
}
p <-0.5
q <- p
p0 <- comb(9, 0)*p^0*q^9

p1 <- comb(9, 1)*p^1*q^8
 
p2 <- comb(9, 2)*p^2*q^7
 
p3 <- comb(9, 3)*p^3*q^6

p4 <- comb(9, 4)*p^4*q^5

#prob of sucess (winning!)
ps <- p0+p1+p2+p3+p4
 
pfail <- 1-ps # prob of losing


#Expected value of the proposition

E2 <- ps*23-pfail*26
E2

## [1] -1.5

Prob 12, step 2

N∗E(X) = 994*(-$1.5)

E3 <- 994*E2
E3

## [1] -1491