Solution:
Total space is 540.
Action of drawing one chip.
\(C_{505}\) Number of chips labeled 505 = 1.
\(P(C_{505})=\) Number of chips labeled 505 divided by the total space.
\[P(C_{505})=\frac{1}{540}\]
round(1/540, 4)## [1] 0.0019Answer: The probability of drawing the chip numbered 505 is 0.19%
Solution:
A = Asparagus
C = Cheese
E = Eggs
T = Turkey
F = French
V = Vinaigrette
S = Sample space will be building a salad.
Answer:
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 VSolution:
S = Space is the total cards in a deck = 52.
H = Number of hearts in a deck = 14.
F = Number of faces in a deck = 12.
\(H^F\) = Number of hearts that are face = 3
\(H^{Fc}\) = Number of hearts that are NOT face = 10
\(F^c\) = Complement of the Number of faces in a deck = 40.
\[P(H^{Fc})=\frac{10}{52}=0.1923\]
round(10/52,4)## [1] 0.1923Answer: The probability that the card will be a heart and not a face card will be 19.23%.
Solution:
Let’s define as follows:
x = number of dots represented on the dice. P(X) = The resulting probability of obtaining x dots in a dice.
Possible Probabilities:
Dice 1
P1(X=1)=1
P1(X=2)=2
P1(X=3)=3
P1(X=4)=4
P1(X=5)=5
P1(X=6)=6
Dice 2
P2(X=1)=1
P2(X=2)=2
P2(X=3)=3
P2(X=4)=4
P2(X=5)=5
P2(X=6)=6
With the above, we can deduct that the total of possible combinations will be a total of 36.
We are being ask for the probability of the sum being less than 6.
Possible combinations:
P1(x=1) + P2(x=1) = 2
P1(x=2) + P2(x=1) = 3
P1(x=3) + P2(x=1) = 4
P1(x=4) + P2(x=1) = 5
P1(x=1) + P2(x=2) = 3
P1(x=2) + P2(x=2) = 4
P1(x=3) + P2(x=2) = 5
P1(x=1) + P2(x=3) = 4
P1(x=2) + P2(x=3) = 5
P1(x=1) + P2(x=4) = 5
By counting the total number of possibilities we find out that there are 10 different options to obtain such results.
With that, in order to find our probability, we can proceed as follows:
\(P(P_1(X)+P_2(X))<6\)
\(P(P_1(X)+P_2(X))=\frac{10}{36}=0.2778\)
round(10/36,4)## [1] 0.2778Answer: The probability of rolling a sum less than 6 will be 27.78%.
What is the probability that a customer is male? Write your answer as a fraction or a decimal number rounded to four decimal places.
Solution:
M = Total number of Males
F = Total number of Females
S = Space total
Males <- c(233, 159, 102, 220, 250)
Females <- c(208, 138, 280, 265, 146)
PM <- round(sum(Males)/(sum(Males) + sum(Females)),4)
cat("Probability of Customer being a Male", PM)## Probability of Customer being a Male 0.4818M = 964
F = 1037
S = 2001
\[P(M)=\frac{964}{2001}=0.4818\]
Answer: The probability that a customer is a male is 48.18%
Solution:
Important: WITH REPLACEMENT!
S = Space is the number of cards in a deck = 52
C = Number of clubs in a deck = 13
B = Number of black cards in a deck = 26
F = Number of faces in a deck = 12
\(P(C)=\frac{13}{52}=0.25\)
\(P(B)=\frac{26}{52}=0.5\)
\(P(F)=\frac{12}{52}=0.2308\)
\[P(C) \cap P(B) \cap P(F) = \frac{13}{52} \cdot \frac{26}{52} \cdot \frac{12}{52} = 0.0288\]
PC <- 13/52
PB <- 26/52
PF <- 12/52
round( PC * PB * PF, 4)## [1] 0.0288Answer: The probability will be 2.88%
Solution:
Important: WITHOUT REPLACEMENT
By using the Theorem of Conditional Probability, we need to do as follows:
\[P(C_2 | C_1) = \frac{P(C_2 \cap C_1)}{P(C_1)}\] We can deduct that \[P(C_2 | C_1) = \frac{P(C_2) P(C_1)}{P(C_1)}=P(C_2)\]
\(S_1\) = Total space for the first card is 52 cards
\(S_2\) = Total space for second card is 51 cards
\(C_1\) = Heart
\(C_2\) = Spade
\(Sp_1\) = Number of spades in the first Space = 13.
\(Sp_2\) = Number of spades in the second Space = 13.
\[P(C_2 | C_1)=\frac{13}{51}=0.2549\]
round( 13/51, 4)## [1] 0.2549Answer: The probability of choosing a spade for the second card drawn, if the first card, drawn without replacement, was a heart is 25.49%
Solution:
Important: WITHOUT REPLACEMENT
\(S_1\) = Total space for the first card is 52 cards
\(S_2\) = Total space for second card is 51 cards
\(H_1\) = Number of hearts available in the first space = 13
\(P(H_1)=\frac{13}{52}\) This is the probability of choosing a heart in the first pick.
Since there’s no replacement and since hearts are red, this means that the number of red cards on the second space has decreased by 1.
\(R_2\) = Red card in the second space.
\(R_2\) = 25
\(P(R_2)=\frac{25}{51}\)
In order to get the final probability, we can deduct as follows:
\(P(R_2 | H_1) = P(H_1) \cdot P(R_2)\)
\[P(R_2 | H_1) = \frac{13}{52} \cdot \frac{25}{51} = \frac{325}{2652} = 0.1225\]
# conditional, but not B|A
PH1 <- 13/52
PR2 <- 25/51
round( PH1 * PR2, 4)## [1] 0.1225Answer: The probability of choosing a heart and then, without replacement, a red card is 12.25%.
Males <- c(12, 19, 12, 7)
Females <- c(12, 15, 4, 4)
Grade <- c("Freshmen", "Sophomores", "Juniors", "Seniors")
MyTable <- data.frame("Grade" = Grade, "Males" = Males, "Females" = Females)
MyTable## Grade Males Females
## 1 Freshmen 12 12
## 2 Sophomores 19 15
## 3 Juniors 12 4
## 4 Seniors 7 4Solution:
\(S_1\) = First space total = 85
\(S_2\) = Second space total = 84
Males = 50
Females = 35
\(P(J_F) \cap P(F_M) = P(J_F) \cdot P(F_M)\)
\(P(J_F) \cap P(F_M) = \frac{4}{85} \cdot \frac{12}{84}=0.0067\)
P <- round( 4 / 85 * 12 /84,4)
cat("Probability is", P)## Probability is 0.0067Answer: The probability that a junior female and then a freshmen male are chosen at random is 0.67%
Step 1. What is the probability that a randomly chosen applicant has a graduate degree, given that they are male? Enter your answer as a fraction or a decimal rounded to four decimal places.
Solution:
\[P(G|M)=\frac{P(G \cap M)}{P(M)}\]
\[P(G \cap M)=\frac{52}{300}\]
\[P(M)=\frac{141}{300}\]
Going back to our original equation:
\[P(G|M)=\frac{\frac{52}{300}}{\frac{141}{300}}=0.3688\]
P <- round( (52 / 300) / (141 / 300),4)
cat("Probability is", P)## Probability is 0.3688Answer: The probability that a randomly chosen applicant has a graduate degree, given that they are male is 36.88%.
Step 2. If 102 of the applicants have graduate degrees, what is the probability that a randomly chosen applicant is male, given that the applicant has a graduate degree? Enter your answer as a fraction or a decimal rounded to four decimal places.
Solution:
\[P(M|G)=\frac{P(M \cap G)}{P(G)}\]
\[P(M \cap G)=\frac{52}{300}\]
\[P(G)=\frac{102}{300}\]
Going back to our original equation:
\[P(G|M)=\frac{\frac{52}{300}}{\frac{102}{300}}=0.5098\]
P <- round( (52 / 300) / (102 / 300),4)
cat("Probability is", P)## Probability is 0.5098Answer: The probability that a randomly chosen applicant is male, given that the applicant has a graduate degree is 50.98%.
Solution:
Let’s define as follow:
D = 6 different kind of drinks.
S = 5 different types of sandwiches.
C = 3 different types of chips.
This is basically a combination as follows:
Combination = D * S * C
D <- 6
S <- 5
C <- 3
Combination <- D * S * C
Combination## [1] 90Answer: There are 90 different value meal packages possible.
Solution:
This can be done in 5! ways.
Ways <- factorial(5)
Ways## [1] 120The doctor can visit 5 patients during the morning rounds in 120 different ways.
Solution:
Since we have n=8 songs and want to find the number of ways k=5 songs that can be ordered:
\[P(n,k)= \frac{n!}{(n-k)!}\]
\[P(8,5) = \frac{8!}{(8-5)!}\]
P <-function(n,k){
factorial(n) / (factorial(n-k))
}
Lineup <- P(8,5)
Lineup## [1] 6720Answer: There will be 6720 possible lineups.
Solution:
In this case, we have \(n=9\) ways of obtaining any side, we can deduct as follows:
\[Ways = \frac{9!}{3! \cdot 5! \cdot 1!}\]
Ways <- factorial(9) / (factorial(3) * factorial(5) * factorial(1))
Ways## [1] 504Answer: there are 504 different ways to get get 3 fours, 5 sixes and 1 two
Solution:
Since we have \(n=14\) different toppings and want to chose \(k=6\) toppings without repeating it, we can do as follows:
\[P(n,k)= \frac{n!}{(n-k)!}\]
Then \[P(14,6)=\frac{14!}{(14-6)!}\]
P <-function(n,k){
factorial(n) / (factorial(n-k))
}
Ways <- P(14,6)
Ways## [1] 2162160Answer: Rudy can chose 6 toppings in 2162160 different ways.
Solution:
Important: WITHOUT REPLACEMENT
Since we have \(n=52\) different cards and want to chose \(k=3\) card hands without repeating it, we can do as follows:
\[C(n,k)= \frac{P(n,k)}{k!}\]
\[C(52,3)= \frac{P(52,3)}{3!}\] \[C(52,3)=\frac{52!}{(52-3)! \cdot 3!}\]
C <-function(n,k){
factorial(n) / (factorial(n-k)*factorial(k))
}
Ways <- C(52,3)
Ways## [1] 22100Answer: There are 22100 different 3-card hands possible if the drawing is done without replacement.
Solution:
Let’s take as follows:
TV = 12 different choices for TVs.
SS = 9 different choices for surround sound systems.
DVD = 5 different choices for DVDs.
Ways = TV * SS * DVD
\(Ways = 12 * 9 * 5\)
Ways <- 12 * 9 * 5
Ways## [1] 540Answer: there are 540 different ways to order a new home theater system.
Solution:
Important: CHARACTERS AND DIGITS CAN NOT BE REPEATED.
There are 26 letters in the alphabet.
There are 5 different odd number for 0 - 9 = {1,3,5,7,9}
Length of password = 8
\(Ways = P(26,5) \cdot P(5,3)\)
By using: \[P(n,k)= \frac{n!}{(n-k)!}\]
\[Ways =\frac{26!}{(26-5)!} \cdot \frac{5!}{(5-3)!}\]
P <-function(n,k){
factorial(n) / (factorial(n-k))
}
Ways <- P(26,5) * P(5,3)
Ways## [1] 473616000Answer: There are 473616000 possible choices for the password.
Another option will be to consider UPPER CASE different as a lower case.
For that our answer will be considering 52 letter, 26 lower case and 26 UPPER CASE as follows:
P <-function(n,k){
factorial(n) / (factorial(n-k))
}
Ways <- P(52,5) * P(5,3)
Ways## [1] 18712512000Answer: Considering Upper case as different to lower case we will have 18712512000 possible choices for our password.
Solution:
By interpreting the above notation as the function for calculating the number of r-permutations of n in the form \(_nP_r\).
We have as follows:
nPr <-function(n,k){
factorial(n) / (factorial(n-k))
}
nPr(9,4)## [1] 3024Answer: \(_9P_4=3024\)
Solution:
By interpreting the above notation as the function for calculating the number of r-combination of n in the form \(_nC_r\).
We have as follows:
nCr <-function(n,k){
factorial(n) / (factorial(n-k)*factorial(k))
}
nCr(11,8)## [1] 165Answer: \(_{11}C_8=165\)
Solution:
nPr(12,8)/nCr(12,4)## [1] 40320Answer: \[\frac{_{12}P_8}{_{12}C_4}=40320\]
Solution:
Assuming that one person can hold only one position.
We obtain resolve this, by calculating \(_{13}P_7\)
nPr(13,7)## [1] 8648640Answer: The president can appoint his cabinet in 8648640 different ways.
Solution:
Population has a total of 10 letters.
There are 2 “P” repeating.
There are 2 “o” repeating.
We have to divide in order to avoid the duplicate letters on the same spot.
Our formula will be \[Ways=\frac{10!}{2! \cdot 2!}\]
factorial(10)/(factorial(2) * factorial(2))## [1] 907200Answer: The word “population”" can be arranged in 907200 different ways.
x <- c(5, 6, 7, 8, 9)
Px <- c(0.1, 0.2, 0.3, 0.2, 0.2)
MyTable <- data.frame( x, Px)
names(MyTable) <- c("x", "P(x)")
MyTable## 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.
Preamble:
Expected Value
The expected value of a random variable indicates its weighted average.
Definition: Let \(X\) be a random variable assuming the values \(x_1, x_2, x_3\), … with corresponding probabilities \(p(x_1), p(x_2), p(x_3)\),….. The mean or expected value of \(X\) is defined by \[E(X) = \sum_{k=0}^n(x_k \cdot p(x_k))\]
Solution:
xPx <- x * Px
xPx## [1] 0.5 1.2 2.1 1.6 1.8Ex <- round(sum(x * Px), 1)
cat("E(x) = ", Ex)## E(x) = 7.2Answer: \(E(x) = 7.2\)
Step 2. Find the variance. Round your answer to one decimal place.
Preamble:
By definition \[Var(x) = \sum_{i} ^{n} \left[ (x_i)^2 \cdot p(x_i) \right] - \left[ E(x) \right]^2\]
xx <- x^2
xxPx <- xx * Px
xxPx## [1] 2.5 7.2 14.7 12.8 16.2Vx <- round(sum(xxPx) - Ex^2, 1)
cat("V(x) = ", Vx)## V(x) = 1.6Answer: The variance is \(V(x) = 1.6\)
Step 3. Find the standard deviation. Round your answer to one decimal place.
Preamble:
The Standard deviation of a random variable is defined:
\[SD(X) = \sqrt{V(X)} \]
Solution:
SDx <- round((Vx)^(1/2),1)
cat("SD(x) = ", SDx)## SD(x) = 1.3The standard deviation is \(SD(X) = 1.3\)
Step 4. Find the value of \(P(X \ge 9)\). Round your answer to one decimal place.
Solution:
\[E(x \ge 9) = \sum(x \ge 9) \cdot p(x \ge 9))\]
By following the values from our table we have as follows:
NEx <- round(sum((x >= 9) * Px), 1)
cat("E(x) = ", NEx)## E(x) = 0.2Answer: \(P(X \ge 9)=0.2\)
Step 5. Find the value of \(P(X \le 7)\). Round your answer to one decimal place.
By following the values from our table we have as follows:
NEx <- round(sum((x <= 7) * Px), 1)
cat("E(x) = ", NEx)## E(x) = 0.6Answer: \(P(X \le 7)=0.6\)
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Solution:
Since the Probability of an Event E is represented as follows:
\[P(E) = \frac{number \: of \: ways \: an \: event \: can \: occur \: (dessired)}{total \: number \: of \: possible \: outcomes \:(total)}\] Where \(0 \le P(E) \le 1\)
Since the probability for a single shot, defined \(p_1\) is as follows:
\[p_1 = \frac{188}{376}=0.5\]
p1 <- 188/376
cat("p1 = ", p1)## p1 = 0.5Since losses need to be entered as a negative value and putting our problems in terms of the player; the expected gain for the basketball player turns out to be:
E = (Probability of making three shots) * $23 + (Probability of NOT making three shots) * (-$4)
We can deduct that the probability for three shots will be the probability of one shot multiplied by itself three times:
Probability of three shots: \(P_1*P_1*P1=(p_1)^3\)
p3 <- (p1) ^ 3
cat("p3 = ", p3)## p3 = 0.125Since the probability of NOT making the three shots is the complement, we can deduct as follows:
Probability of not making three shots: \(p_{fail} = 1 - p_3\)
pf <- 1 - (p3)
cat("pf = ", pf)## pf = 0.875By Solving our Expected value of the proposition, we have as follows:
\[E = (23 \cdot p_3) + (-4 \cdot p_{fail})\]
E <- round(23 * (p3) + (-4) * pf,2)
cat("E = ", E)## E = -0.62Answer: The expected value will be $ -$0.62$
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
Solution:
\[E_{994}= 994 \cdot E\]
E994 <- 994 * E
cat("E = ", E994)## E = -616.28Answer: If I played this game 994 times I would lose $616.28.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Preamble:
The binomial distribution consists of the probabilities of each of the possible numbers of successes on N trials for independent events that each have a probability of \(\pi\) (the Greek letter pi) of occurring. For the coin flip example, \(N = 2\) and \(\pi = 0.5\). The formula for the binomial distribution is shown below:
\[P(x)=\frac{N!}{x!(N-x)!} \pi^x (1 - \pi )^{N-x}\] where P(x) is the probability of x successes out of N trials, N is the number of trials, and \(\pi\) is the probability of success on a given trial.
Now, applying this to the coin flip example we have as follows:
\[P_{win} = \frac{11!}{8!(11-8)!} (0.5)^8 (1 - 0.5 )^{11-8} \]
Probability of winning:
#Expected probability of winning for multiple cases
spw <- function(n,k,p)
{
spx <-0
for (i in 1:k ) {
px <- factorial(n)/(factorial(i)*factorial(n-i))*(p^i)*(1-p)^(n-i)
spx <- spx + px
}
return(spx)
}
#Expected probability of winning for a single case, comparable to dbinom(k,n,p)
pw <- function(n,k,p)
{
px <- factorial(n)/(factorial(k)*factorial(n-k))*(p^k)*(1-p)^(n-k)
return(px)
}
#pw1 <- pb(11,1,0.5)
#pw2 <- pb(11,2,0.5)
#pw3 <- pb(11,3,0.5)
#pw4 <- pb(11,4,0.5)
#pw5 <- pb(11,5,0.5)
#pw6 <- pb(11,6,0.5)
#pw7 <- pb(11,7,0.5)
#pw8 <- pb(11,8,0.5)
#p <- pw1 + pw2 + pw3 + pw4 + pw5 + pw6 + pw7 + pw8
p <- spw(11,8,0.5)
p## [1] 0.9667969# Another alternative will be to use dbinom(8,11,0.5) instead of programing the formulaProbability of failure:
pf <- 1 - (p)
cat("pf = ", pf)## pf = 0.03320312Expected value of the proposition:
E <- round(1 * (p) + (-7) * pf,2)
cat("E = ", E)## E = 0.73Answer: The expected value will be \(E=0.73\)
Step 2. If you played this game 615 times how much would you expect to win or lose? (Losses must be entered as negative.)
Solution:
E615 <- 615 * E
cat("E = ", E615)## E = 448.95Answer: If I play 615 times I will expect to win $448.95.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Solution:
Number of cards in a standard deck = 52.
Number of clubs in a standard deck = 13.
\(P(E) = \frac{13}{52} \cdot \frac{12}{52}\)
PWin <- 13/52 * 12/51
cat("P(E) = ", PWin)## P(E) = 0.05882353Probability of failure:
pf <- 1 - (PWin)
cat("pf = ", pf)## pf = 0.9411765Expected value of the proposition:
E <- round(583 * (PWin) + (-35) * pf,2)
cat("E = ", E)## E = 1.35Answer: The expected value of the proposition is \(E=1.35\)
Step 2. If you played this game 632 times how much would you expect to win or lose? (Losses must be entered as negative.)
Solution:
E632 <- 632 * E
cat("E = ", E632)## E = 853.2Answer: If I play 615 times I will expect to win $853.20.
Solution:
Sample size = 10 bulbs.
If less than 2 defectives the lot passes inspection.
30 % in the lot are defective.
Probability of passing inspection?
This is very similar to the binomial distribution question #26 for which I will use the same function for \(x \le 2\).
\[P(x)=\sum \frac{N!}{x!(N-x)!} \pi^x (1 - \pi )^{N-x}\]
We will take as follows:
\(n = 10\)
\(k = 2\)
\(p = 0.3\)
p <- round(spw(10,2,0.3),3)
p## [1] 0.355Answer: The probability that the lot will pass inspection is 35.5%
Solution:
n = 5 Number of bulbs.
p = 30% are defective.
\(EV = n \cdot p\)
n <-5
p <- 0.3
EV <- n * p
EV## [1] 1.5Answer: The expected value of defective bulbs is 1.5.
Solution:
Since we know the mean E = 5.5.
We can employ the Poisson distribution as follows:
\[p(k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}\]
p <- function(k, l)
{
spoisson <-0
for (i in 0:k ) {
poisson <- (l^(i) * exp(-l))/(factorial(i) )
spoisson <- spoisson + poisson
}
return(spoisson)
}
p(5,5.5)## [1] 0.5289187# Another way of solving this will be employing the ppois(5, 5.5, lower=FALSE) in r.That result will represent the lower tail, since we are being asked for more than 5, what we need to do is as follows: \(Upper = 1-poisson\).
ut <- round(1-p(5, 5.5),4)
cat("Upper Tail", ut)## Upper Tail 0.4711Answer: The probability that, for any day, the number of special orders sent out will be more than 5 is 47.11%.
Solution:
Since we are being asked for the Upper Tail, by employing the above function; we obtain as follows:
ut <- round(1-p(4, 5.7),4)
cat("Upper Tail", ut)## Upper Tail 0.6728Answer: The probability that, in any hour, more than 4 customers will arrive is 67.28%.
Solution:
Since we have a daily mean of 0.4 per day and we are looking to obtain a weekly probability, we need to do as follows:
daily lambda = 0.4
Weekly lambda = 0.4 * 7
With this transformation we need to take our k = 1 week as well.
Since we are being asked for the lower Tail, by employing the above function; we obtain as follows:
lt <- round(p(1, 0.4 * 7),4)
cat("Lower Tail", lt)## Lower Tail 0.2311Answer: The probability that, in any 7-day week, the computer will crash no more than 1 time is 23.11%
Preamble:
Since the above example explain a distributions related to the number of successes in a sequence of draws with no replacement, I will use the hyper geometric distribution as follows:
Definition: The following conditions characterize the hyper geometric distribution:
The result of each draw (the elements of the population being sampled) can be classified into one of two mutually exclusive categories (e.g. Pass/Fail or Employed/Unemployed).
The probability of a success changes on each draw, as each draw decreases the population (sampling without replacement from a finite population).
A random variable \({\displaystyle X}\) follows the hyper geometric distribution if its probability mass function (pmf) is given by:
\[P(X=x)=\frac{{{\displaystyle K} \choose k} {{\displaystyle N-K} \choose {n-k}}}{{{\displaystyle N} \choose n}}\] where
\({\displaystyle N}\) is the population size,
\({\displaystyle K}\) is the number of success states in the population,
\({\displaystyle n}\) is the number of draws,
\({\displaystyle k}\) is the number of observed successes
Solution:
From the above formula, we can rewrite it as follows:
\[P(X=x)=\frac{(_{K}C_{k}) \cdot (_{N-K}C_{n-k})}{(_{N}C_{n})}\]
In order for us to resolve the above question, we will take our values as follows:
N = 25
K = 6
n = 8
k = {0, 1, 2, 3, 4, 5, 6, 7, 8} will be dismissed.
By using our previous defined combinatory function, we can do as follows:
nCr <-function(n,k){
factorial(n) / (factorial(n-k)*factorial(k))
}In this case we need to add all the probabilities for the one given that is for more than 1 employee was over 50.
I am defining k1 as my starting point of observed successes > 1; k1 =2 and k2 as my final number of observed successes; k2 = 6.
Px <- function(N, K, n, k1, k2)
{
sPx <- 0
for (i in k1:k2 ) {
sPx <- sPx + nCr(K,i) * nCr(N-K, n-i) / nCr(N,n)
}
return(sPx)
}
#Probability function for more than 1 employee who is over 50, staring with 2 ending wit 6
N <- 25
K <- 6
n <- 8
k1 <- 2
k2 <- 6
# Adding all the possibilities > 1 up to 6
round(Px(N, K, n, k1, k2),3)## [1] 0.651# Or by addig the possibilities < 1 and taking the complement.
k1 <- 0
k2 <- 1
1- round(Px(N, K, n, k1, k2),3)## [1] 0.651Answer: The probability of dismissed employees in which more than 1 employee was over 50 is 65.1%
Solution:
Let’s take as follows:
\({\displaystyle N}\) is the population size,
\({\displaystyle K}\) is the number of success states in the population,
\({\displaystyle n}\) is the number of draws,
\({\displaystyle k}\) is the number of observed successes
N = 25
K = 15
n = 8
k = {0, 1, 2, 3, 4, 5, 6, 7, 8} will receive the drug.
#Probability function for less than 7 people; I will define k1 = 0 and k2 = 6
N <- 25
K <- 10
n <- 8
k1 <- 0
k2 <- 6
# Adding all the possibilities > 1 up to 6
round(Px(N, K, n, k1, k2),3)## [1] 0.998# Or by addig the possibilities < 1 and taking the complement.
#k1 <- 0
#k2 <- 1
#1- round(Px(N, K, n, k1, k2),3)Answer: The probability that less than 7 patients will die is 99.8%.