expand.grid(die1=1:6,
die2=1:6,
die3=1:6)## die1 die2 die3
## 1 1 1 1
## 2 2 1 1
## 3 3 1 1
## 4 4 1 1
## 5 5 1 1
## 6 6 1 1
## 7 1 2 1
## 8 2 2 1
## 9 3 2 1
## 10 4 2 1
## 11 5 2 1
## 12 6 2 1
## 13 1 3 1
## 14 2 3 1
## 15 3 3 1
## 16 4 3 1
## 17 5 3 1
## 18 6 3 1
## 19 1 4 1
## 20 2 4 1
## 21 3 4 1
## 22 4 4 1
## 23 5 4 1
## 24 6 4 1
## 25 1 5 1
## 26 2 5 1
## 27 3 5 1
## 28 4 5 1
## 29 5 5 1
## 30 6 5 1
## 31 1 6 1
## 32 2 6 1
## 33 3 6 1
## 34 4 6 1
## 35 5 6 1
## 36 6 6 1
## 37 1 1 2
## 38 2 1 2
## 39 3 1 2
## 40 4 1 2
## 41 5 1 2
## 42 6 1 2
## 43 1 2 2
## 44 2 2 2
## 45 3 2 2
## 46 4 2 2
## 47 5 2 2
## 48 6 2 2
## 49 1 3 2
## 50 2 3 2
## 51 3 3 2
## 52 4 3 2
## 53 5 3 2
## 54 6 3 2
## 55 1 4 2
## 56 2 4 2
## 57 3 4 2
## 58 4 4 2
## 59 5 4 2
## 60 6 4 2
## 61 1 5 2
## 62 2 5 2
## 63 3 5 2
## 64 4 5 2
## 65 5 5 2
## 66 6 5 2
## 67 1 6 2
## 68 2 6 2
## 69 3 6 2
## 70 4 6 2
## 71 5 6 2
## 72 6 6 2
## 73 1 1 3
## 74 2 1 3
## 75 3 1 3
## 76 4 1 3
## 77 5 1 3
## 78 6 1 3
## 79 1 2 3
## 80 2 2 3
## 81 3 2 3
## 82 4 2 3
## 83 5 2 3
## 84 6 2 3
## 85 1 3 3
## 86 2 3 3
## 87 3 3 3
## 88 4 3 3
## 89 5 3 3
## 90 6 3 3
## 91 1 4 3
## 92 2 4 3
## 93 3 4 3
## 94 4 4 3
## 95 5 4 3
## 96 6 4 3
## 97 1 5 3
## 98 2 5 3
## 99 3 5 3
## 100 4 5 3
## 101 5 5 3
## 102 6 5 3
## 103 1 6 3
## 104 2 6 3
## 105 3 6 3
## 106 4 6 3
## 107 5 6 3
## 108 6 6 3
## 109 1 1 4
## 110 2 1 4
## 111 3 1 4
## 112 4 1 4
## 113 5 1 4
## 114 6 1 4
## 115 1 2 4
## 116 2 2 4
## 117 3 2 4
## 118 4 2 4
## 119 5 2 4
## 120 6 2 4
## 121 1 3 4
## 122 2 3 4
## 123 3 3 4
## 124 4 3 4
## 125 5 3 4
## 126 6 3 4
## 127 1 4 4
## 128 2 4 4
## 129 3 4 4
## 130 4 4 4
## 131 5 4 4
## 132 6 4 4
## 133 1 5 4
## 134 2 5 4
## 135 3 5 4
## 136 4 5 4
## 137 5 5 4
## 138 6 5 4
## 139 1 6 4
## 140 2 6 4
## 141 3 6 4
## 142 4 6 4
## 143 5 6 4
## 144 6 6 4
## 145 1 1 5
## 146 2 1 5
## 147 3 1 5
## 148 4 1 5
## 149 5 1 5
## 150 6 1 5
## 151 1 2 5
## 152 2 2 5
## 153 3 2 5
## 154 4 2 5
## 155 5 2 5
## 156 6 2 5
## 157 1 3 5
## 158 2 3 5
## 159 3 3 5
## 160 4 3 5
## 161 5 3 5
## 162 6 3 5
## 163 1 4 5
## 164 2 4 5
## 165 3 4 5
## 166 4 4 5
## 167 5 4 5
## 168 6 4 5
## 169 1 5 5
## 170 2 5 5
## 171 3 5 5
## 172 4 5 5
## 173 5 5 5
## 174 6 5 5
## 175 1 6 5
## 176 2 6 5
## 177 3 6 5
## 178 4 6 5
## 179 5 6 5
## 180 6 6 5
## 181 1 1 6
## 182 2 1 6
## 183 3 1 6
## 184 4 1 6
## 185 5 1 6
## 186 6 1 6
## 187 1 2 6
## 188 2 2 6
## 189 3 2 6
## 190 4 2 6
## 191 5 2 6
## 192 6 2 6
## 193 1 3 6
## 194 2 3 6
## 195 3 3 6
## 196 4 3 6
## 197 5 3 6
## 198 6 3 6
## 199 1 4 6
## 200 2 4 6
## 201 3 4 6
## 202 4 4 6
## 203 5 4 6
## 204 6 4 6
## 205 1 5 6
## 206 2 5 6
## 207 3 5 6
## 208 4 5 6
## 209 5 5 6
## 210 6 5 6
## 211 1 6 6
## 212 2 6 6
## 213 3 6 6
## 214 4 6 6
## 215 5 6 6
## 216 6 6 6total <- rowSums(expand.grid(die1 = 1:6, die2 = 1:6, die3 = 1:6))
total ## [1] 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7
## [26] 8 9 10 11 12 8 9 10 11 12 13 4 5 6 7 8 9 5 6 7 8 9 10 6 7
## [51] 8 9 10 11 7 8 9 10 11 12 8 9 10 11 12 13 9 10 11 12 13 14 5 6 7
## [76] 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12 8 9 10 11 12 13 9 10 11 12
## [101] 13 14 10 11 12 13 14 15 6 7 8 9 10 11 7 8 9 10 11 12 8 9 10 11 12
## [126] 13 9 10 11 12 13 14 10 11 12 13 14 15 11 12 13 14 15 16 7 8 9 10 11 12
## [151] 8 9 10 11 12 13 9 10 11 12 13 14 10 11 12 13 14 15 11 12 13 14 15 16 12
## [176] 13 14 15 16 17 8 9 10 11 12 13 9 10 11 12 13 14 10 11 12 13 14 15 11 12
## [201] 13 14 15 16 12 13 14 15 16 17 13 14 15 16 17 18?length## starting httpd help server ... donelength(total[total==12])## [1] 25## P=N/D
x<-25/216
x## [1] 0.1157407round(x, digits = 3)## [1] 0.116install.packages(“Matrix”)
install.packages(“concatenate”)
mymat = matrix(data = c(200,200,100,200,200, 300,100,200,100,100), nrow = 5, byrow = FALSE # implies sequence is filled into columns )
cbind(mymat, rowSums(mymat))
mymat = rbind(mymat, colSums(mymat) )
#Q3
#Two cards are drawn without replacement from a standard deck of 52 playing cards. #What is the probability of choosing a diamond for the second card drawn, if the first #card, drawn without replacement, was a diamond?
#A= First card drawn #B= Second card drawn
#Conditional Probability Formula will be applied.
x<-(13/52) * (12/51) / (13/52)
x
#Answer 0.2352941
round(x, digits = 3)
#Answer: 0.235
#Q4
#Permutation Problem
Formula
#(20!)/(20!-10!)
factorial(20)/factorial(10)
#Answer: 670442572800
#Q5
#Answer can be fetched by Multiplication rule of counting
20 * 20* 18
#7200 is the answer
#10!/10!-10! #10!
1098765432*1
#Answer 3628800
#Q7
#‘Multiplication rule of Counting’
2^76^3choose(52,4)
#Q8
factorial(3)*factorial(4)
#Q9
#Using Bayes Formula
#A=user, A’=Not a user, B=Positive , B’=Negative
.95 .03 / (.95 .03 + .01*.97 )
#Answer 0.7460733
#10
#Solved using Bayes Probability theorem
#g1g2, g2g1, b1b2, b2b1, g1b2, b2g1
x <- ((3/6) * (2/3) ) / (3/6)
#Answer 0.6666667
round(x,digits=3)
0.667