\[\begin{equation*} \binom{5}{1} =\frac{5!}{(5-1)!1!} =\frac{120}{24} \\ \binom{7}{4} =\frac{7!}{(7-4)!4!} =\frac{5,040}{144} \\ \binom{7}{5} =\frac{7!}{5!(7-5)!} =\frac{5,040}{240} \\ =\ \binom{5}{1} \times \binom{7}{4} +\binom{7}{5} \\= 175 + 21 = 196\ ways\ \end{equation*}\]
Q1 <- choose(5,1) * choose(7,4) + choose(7,5)
print(paste0("The total ways 5 jellybeans can be withdrawn from the bag so that the number of green ones withdrawn will be less than 2 is: ",Q1))## [1] "The total ways 5 jellybeans can be withdrawn from the bag so that the number of green ones withdrawn will be less than 2 is: 196"\[\begin{equation*} \binom{13}{4}=\frac{13!}{4!(13-4)!} =\frac{6227020800}{8709120} \\ \binom{14}{1}=\frac{14!}{1!(14-1)!} =\frac{87178291200 }{6227020800} \\ \binom{13}{5} =\frac{13!}{5!(13-5)!} =\frac{6227020800}{4838400} \\ = \binom{13}{4} \times \binom{14}{1} +\binom{13}{5} \\= 10,010 + 1,287 \\Answer:\ =\ 11,297\ ways \end{equation*}\]
Q2 <- choose(13,4)*choose(14,1) + choose(13,5)
print(paste0("The total ways a subcommittee of 5 be formed if at least 4 of the members must be representatives is: ",Q2))## [1] "The total ways a subcommittee of 5 be formed if at least 4 of the members must be representatives is: 11297"\[\begin{equation*} =2^{5} \times 6^{2} \times \frac{52!}{3!(52-3)!} \\ = 32 \times 36 \times 22100 \\ = 25,459,200 \ outcomes \end{equation*}\]
Q3 <- 2^5 * 6^2 * choose(52,3)
print(paste0("The total different outcomes are possible is: ",Q3))## [1] "The total different outcomes are possible is: 25459200"\[\begin{equation*} P = 1 - \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \\ P = 0.2173756 \end{equation*}\]
Q4 <- round(((choose(4,1) * choose(48,2)) + (choose(4,2) * choose(48,1)) + choose(4,3)) / choose(52,3),4)
print(paste0("The probability that at least one of the cards drawn is a 3 is: ",Q4))## [1] "The probability that at least one of the cards drawn is a 3 is: 0.2174"\[\begin{equation*} =\ \binom{31}{5} =\frac{31!}{(31-5)!5!} \\ =169,911 \end{equation*}\]
Q5_S1 <- choose(31,5)
print(paste0("The total different combinations of 5 movies can he rent is: ",Q5_S1))## [1] "The total different combinations of 5 movies can he rent is: 169911"\[\begin{equation*} \binom{14}{1} \times \binom{17}{4} = \frac{14!}{1!(14-1)!} \times \frac{17!}{4!(17-4)!} \\ \binom{14}{2} \times \binom{17}{3} = \frac{14!}{2!(14-2)!} \times \frac{17!}{3!(17-3)!} \\ \binom{14}{3} \times \binom{17}{2} = \frac{14!}{3!(14-3)!} \times \frac{17!}{2!(17-2)!} \\ \binom{14}{4} \times \binom{17}{1} = \frac{14!}{4!(14-4)!} \times \frac{17!}{1!(17-1)!} \\ \binom{14}{5} = \frac{14!}{5!(14-5)!} \\ \\ That\ is\ =\ \binom{14}{1} \times \binom{17}{4} + \binom{14}{2} \times \binom{17}{3} + \binom{14}{3} \times \binom{17}{2} + \binom{14}{4} \times \binom{17}{1} + \binom{14}{5} \\ = 33,320 + 61,880 + 49,504 + 17,017 + 2,002 \\ = 163,723 \end{equation*}\]\end{equation*}
Q5_S2 <- (choose(14,1) * choose(17,4)
) + (choose(14,2) * choose(17,3)) + (choose(14,3) * choose(17,2)) + (choose(14,4) * choose(17,1)) + (choose(14,5))
print(paste0("The total different combinations of 5 movies can he rent if he wants at least one mystery is: ",Q5_S2))## [1] "The total different combinations of 5 movies can he rent if he wants at least one mystery is: 163723"\[\begin{equation*} =\ \binom{4}{3} =\frac{4!}{(4-3)!3!} = 4\ ways\ for\ Brahms\ symphonies\ \\ \binom{104}{3} =\frac{104!}{(104-3)!3!} = 182,104\ ways\ for\ Haydn\ symphonies\ \\ \binom{17}{3} = \frac{17!}{(17-3)!3!} = 680\ ways\ for\ symphonies \\ Anwer:\ \binom{4}{3} \binom{104}{3} \binom{17}{3}=4*182,104*680=4.95\cdot10^8 different\ schedules\ possible. \end{equation*}\]dules possible.
\end{equation*}
Q6 <- formatC(round(choose(4,3) * choose(104,3) * choose(17,3), -2), format="e")
print(paste0("The different schedules are possible are: ",Q6))## [1] "The different schedules are possible are: 4.9532e+08"Q7_S1 <- formatC(round((choose(5,0)*choose(19,13)) + (choose(5,1)*choose(19,12)) + (choose(5,2)*choose(19,11)) + (choose(5,3)*choose(19,10)) + (choose(5,4)*choose(19,9)), -2), format="e")
print(paste0("The total different reading schedules possible if he wants to include no more than 4 nonfiction books are : ",Q7_S1))## [1] "The total different reading schedules possible if he wants to include no more than 4 nonfiction books are : 2.4206e+06"If the teacher wants to include all six plays, then he has to pick 7 more books out of the 6 novels, 7 poetry books, and 5 nonfiction books. The total number of ways to choose 7 books out of 18 is:
\[\begin{equation*} \binom{18}{7} =\frac{18!}{(18-7)!7!} \\ =\frac{18!}{7!11!} =31824 =\ 3.1824e+04 \end{equation*}\]
Expressing this number in scientific notation and rounding to the hundredths place gives: \[3.18\times 10^{4}\] Therefore, there are approximately 3.1824e+04 different reading schedules possible.
Q7_S2 <- formatC(round(factorial(6) * (factorial(6 + 7 + 5)/factorial(6 + 7 + 5-7)), -2), format="e")
print(paste0("The total different reading schedules possible if he wants to include all 6 plays are : ",Q7_S2))## [1] "The total different reading schedules possible if he wants to include all 6 plays are : 1.1548e+11"There are a total of 10 trees to plant, and we want to calculate the probability that all 5 sycamores are together and all 5 cypress trees are together.
First, we need to calculate the total number of ways to arrange the 10 trees in a row, which is \[10! = 3,628,800\]
We can count the total number of ways to arrange the 10 trees and then count the number of arrangements where all 5 sycamores and all 5 cypress trees are together.
We can treat the 5 sycamores as a single entity and the 5 cypress trees as another single entity. This reduces the problem to arranging 2 entities, which can be done in 2! = 2 ways (either sycamores first or cypress trees first).
Within each group of trees, there are 5! = 120 ways to arrange the individual trees. So the total number of arrangements where all 5 sycamores and all 5 cypress trees are together is:
\[ 2 x 5! x 5! = 2 x 120 x 120 = 28,800 \]
To count the total number of arrangements of 10 trees, we can simply calculate \[10! = 3,628,800\].
P = number of favorable outcomes / total number of outcomes P = 2,880 / 3,628,800
Therefore, the probability of randomly planting the trees so that all 5 sycamores are next to each other and all 5 cypress trees are next to each other is: \[28,800 / 3,628,800 = 0.0079\] (rounded to four decimal places) or \[ 79/10000\].
# each block of five can be arranged in two ways so total is multiplied by 2. The total permutations of all 10 tress is calculated.
prob <- round((2*factorial(5)*factorial(5))/factorial(10),4)
print(paste0('The probability that he randomly plants the trees so that all 5 sycamores are next to each other and all 5 cypress trees are next to each other are ', prob,''))## [1] "The probability that he randomly plants the trees so that all 5 sycamores are next to each other and all 5 cypress trees are next to each other are 0.0079"Q9_S1 <- round((4* (44/52)) + (-16 * (8/52)),2)
print(paste0('The expected value of the proposition is: ', Q9_S1))## [1] "The expected value of the proposition is: 0.92"Q9_S2 <- round(((4* (44/52)) + (-16 * (8/52))) * 833,2)
print(paste0('If I played this game 833 times I would you expect to win or lose $', Q9_S2))## [1] "If I played this game 833 times I would you expect to win or lose $768.92"