Total sample space is 540 plastic chips.
Total chips labeled 505 = 1
\(P(C) = \frac{1}{540}\)
P <- round(1/540, 4)
cat("The probability is: ", P)## The probability is: 0.0019When deciding what you want to put into a salad for dinner at a restaurant, you will choose one of the following extra toppings: asparagus, cheese. Also, you will add one of following meats: eggs, turkey. Lastly, you will decide on one of the following dressings: French, vinaigrette. (Note: Use the following letters to indicate each choice: A for asparagus, C for cheese, E for eggs, T for turkey, F for French, and V for vinaigrette.)toppings <- c("A","A","A","A","C","C","C","C")
meats <- c("E","E","T","T","E","E","T","T")
dressings <- c("F","V","F","V","F","V","F","V")
salad <- data.frame(toppings, meats, dressings)
salad## toppings meats dressings
## 1 A E F
## 2 A E V
## 3 A T F
## 4 A T V
## 5 C E F
## 6 C E V
## 7 C T F
## 8 C T VTotal # of hearts: 14
Total # of Face cards: 12
Total # of hearts that are a face card: 3
Total # of hearts that aren’t a face card: 10
Sample size: 52
P <- round(10/52, 4)
cat("The probability is: ", P)## The probability is: 0.1923TOtal possibilities per die is 6. 6*6 = 36 total outcomes.
10 out of the possible 36 would equal less than a sum of 6. \({\frac {10}{36}} = 0.2778\) or 27.78%
P <- round(10/36, 4)
cat("The probability is: ", P)## The probability is: 0.2778males <- 233 + 159 + 102 + 220 + 250
females <- 208 + 138+ 280 + 265 +147
males## [1] 964females## [1] 1038total <- males + females
total## [1] 2002P <- round(males/total, 4)
cat("The probability is: ", P)## The probability is: 0.4815\(P(M) = \frac {964}{2001}\) = 0.4818
\(P (A \cap B) = P (A) P (B)\)
S = Sample Space = 52 cards in a deck
C = Total # of cubs in a deck = 13
B = Total # of black cards in a deck = 26
F= Total # of face cards in a deck = 12
\(P(C)\cap P(B) \cap P(F) = \frac{13}{52} * \frac{26}{52} * \frac {12}{52}\) = 2.88%
PC <- 13/52
PB <- 26/52
PF <- 12/52
P <- round(PC * PB * PF, 4)
cat("The probability is: ", P)## The probability is: 0.0288
Sample space for 1st card drawn = 52 cards
Sample space for 2nd card drawn = 51 cards
Total # of spades for 1st drawing = 13
Total # of spades for 2nd drawing = 13
P <- round(13/51, 4)
cat("The probability is: ", P)## The probability is: 0.2549Sample for 1st card: 52 cards
Total hearts: 13
\(P(H): \frac {13}{52}\)
Sample for 2nd card: 51 cards
Red card for 2nd sample: 25
\(P(R)=\frac {25}{51}\)
PH <- 13/52
PR <- 25/51
P <- round(PH*PR,4)
cat("The probability is: ", P)## The probability is: 0.1225males <- c(12, 19, 12, 7)
females <- c(12, 15, 4, 4)
total <- sum(males + females)
total## [1] 85Total students for 1st random choosing: 85
Total students for 2nd random choosing: 84
TOtal junior females: 4
Total freshmen males: 12
JF <- 4/85
FM <- 12/84
P <- round(JF * FM, 4)
cat("The probability is: ", P)## The probability is: 0.0067Total sample: 300 applicants
Graduate degree male applicants: 52 \(P(G) = \frac{52}{300}\)
Total male applicants: 141 \(P(M) = \frac{141}{300}\)
G <- 52/300
M <- 141/300
P <- round(G/M, 4)
cat("The probability is: ", P)## The probability is: 0.3688Total Sample: 300 applicants
Graduate degree applicants: 102
Graduate degree male applicants: 52
M <- 52/300
G <- 102/300
P <- round(M/G, 4)
cat("The probability is: ", P)## The probability is: 0.5098drinks <- 6
sandwiches <- 5
chips <- 3
total <- drinks * sandwiches * chips
cat("There are", total, "value meal combos")## There are 90 value meal combosvisit <- factorial(5)
cat("Total number of ways is:", visit)## Total number of ways is: 120\[_nP_r = \frac {n!}{(n-r)!} \ = \ _8P_5 = \frac {8!}{(8-5)!}\]
P <- function(n, r){
factorial(n)/factorial(n-r)
}
permutation <- P(8, 5)
cat("Total number of ways is:", permutation)## Total number of ways is: 6720$ P = $
P <- factorial(9)/(factorial(3)*factorial(5)*factorial(1))
P## [1] 504n = 52 playing cards r = 3 card hands
\[_nC_r = \frac {n!}{(n-r)!} \ = \ _{52}C_3 = \frac {52!}{(52-3)!*3!}\]
C <- function(n, r){
factorial(n)/(factorial(n-r)*factorial(r))
}
combination <- C(52, 3)
cat("Total number of ways is:", combination)## Total number of ways is: 22100TV = 12
Surround_sound = 9
DVD_player = 5
home_theater <- TV * Surround_sound * DVD_player
cat("The total # of ways to build a home theater system is: ", home_theater)## The total # of ways to build a home theater system is: 54026 letters in alphabet
3 odd digits between 0-9 (1,3,5,7,9) - 5 total
password = 8 (5 letters, 3 odd digits)
choice_password = P(26,5) * P(5,3)
\[_nP_r = \frac {n!}{(n-r)!} \ =\ \ \ \frac {26!}{(26-5)!} * \frac {5!}{(5-3)!} \]
P <- function(n, r){
factorial(n)/(factorial(n-r))
}
permutation <- P(26, 5) * P(5, 3)
cat("Total number of ways is:", permutation)## Total number of ways is: 473616000 $$_9 P_4$$P <- function(n, r){
factorial(n)/factorial(n-r)
}
probability <- P(9, 4)
cat("Total number of ways is:", probability)## Total number of ways is: 3024$$_{11} C_8$$C <- function(n, r){
factorial(n)/(factorial(n-r)*factorial(r))
}
combination <- C(11, 8)
cat("Total number of ways is:", combination)## Total number of ways is: 165$$( _{12} P_8)/( _{12} C_4 )$$P <- function(n, r){
factorial(n)/factorial(n-r)
}
C <- function(n, r){
factorial(n)/(factorial(n-r)*factorial(r))
}
permutation <- P(12, 8)
combination <- C(12, 4)
permutation/combination## [1] 40320\[_{13} P_7\]
P <- function(n, r){
factorial(n)/(factorial(n-r))
}
permutation <- P(13, 7)
cat("Total number of ways is:", permutation)## Total number of ways is: 8648640The word ‘Population’ has a total of 10 letters with 2 sets of duplicates - 2 P’s and 2 O’s. We divide to avoid having duplicate letters in the same spot
\[total\ ways = \frac{10!}{2!*2!}\]
total_ways <- factorial(10)/(factorial(2)*factorial(2))
cat("The total ways the word population can be arranged is:", total_ways, "ways.")## The total ways the word population can be arranged is: 907200 ways.x <- c(5, 6, 7, 8, 9)
px <- c(0.1, 0.2, 0.3, 0.2, 0.2)
df <- data.frame(x, px)
names(df) <- c("x", "P(x)")
df## x P(x)
## 1 5 0.1
## 2 6 0.2
## 3 7 0.3
## 4 8 0.2
## 5 9 0.2Step 1. Find the expected value E( X ). Round your answer to one decimal place.
ex <- round(sum(x*px), 1)
ex## [1] 7.2Step 2. Find the variance. Round your answer to one decimal place.
variance <- sum((x - ex)^2 *px)
round(variance, 1)## [1] 1.6Step 3. Find the standard deviation. Round your answer to one decimal place.
sd <- round(sqrt(variance), 1)
sd## [1] 1.2Step 4. Find the value of P(X >= 9). Round your answer to one decimal place.
step4 <- round(sum((x >= 9) * px), 1)
step4## [1] 0.2Step 5. Find the value of P(X <= 7). Round your answer to one decimal place.
step5 <- round(sum((x <= 7) * px), 1)
step5## [1] 0.6Step 1. Find the expected value of the proposition. Round your answer to two decimal places. Probability of the player making the first shot is 188/376. The probability of the player making the first 2 shots is (188/376)^2. The probability of the first 3 shots is (188/376)^3 or 1/8.
The probability of the player not making the first 3 shots is 1 - 1/8 = 7/8
Expected value of the proposition is: (P all 3 shots * $23 + P not making all 3 shots * (-4)) or (1/8) * 23 + 7/8 * -4
round((1/8)*(23) + (7/8)*(-4), 2)## [1] -0.62Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
(-0.62 * 994)## [1] -616.28Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
n1 <- 11
k1 <- 8
temp <- 0
while (k1 >= 0){
temp <- temp + (factorial(n1) / ((factorial(k1) * factorial(n1-k1))))
k1 <- k1 - 1
}
p1 <- temp/(2^11)
p1## [1] 0.9672852pf <- 1 - (p1)
pf## [1] 0.03271484E1 <- round(1 * (p1) + (-7) * pf, 2)
E1## [1] 0.74Step 2. If you played this game 615 times how much would you expect to win or lose? (Losses must be entered as negative.)
round(E1 * 615, 2)## [1] 455.1Step 1. Find the expected value of the proposition. Round your answer to two decimal places. # of cards in deck: 52 # of clubs: 13 \(P(E)= \frac{13}{52} *\frac{12}{52}\)
P <- 13/52 * 12/51
PF <- 1 - (P)
E <- round(583 * (P) + (-35) * PF, 2)
E## [1] 1.35Step 2. If you played this game 632 times how much would you expect to win or lose? (Losses must be entered as negative.)
E2 <- 632 * E
E2## [1] 853.2n = 10, k = 2, p = 0.3
P <- round(pbinom(2, size = 10, prob = .3), 3)
P## [1] 0.3835*.3## [1] 1.5The mean is E = 5.5 P of special orders will be more than 5?
\[p(k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}\]
round(ppois(5, 5.5, lower.tail = FALSE), 4)## [1] 0.4711round(ppois(4, 5.7, lower.tail = FALSE), 4)## [1] 0.6728round(ppois(1, 0.4*7, lower.tail = TRUE), 4)## [1] 0.2311q <- 1
m <- 6
n <- 19
k <- 8
p <- round(phyper(q, m, n, k, lower.tail = F),3)
p## [1] 0.651# phyper(q, m, n, k, lower.tail = TRUE, log.p = FALSE)
q <- 6
m <- 10
n <- 15
k <- 8
p <- round(phyper(q, m, n, k, lower.tail = T),3)
p## [1] 0.998