We need to withdraw 5 jellybean total where only 1 or 0 of them are from the 5 green jellybeans:
Choose 1 out of 5 green and then 4 out of 7 red plus 0 out of 5 green and 5 out of 7 red.
ans <-choose(5,1) * choose(7,4) + choose(5,0) * choose(7,5)
Answer: 196 ways to withdraw 5 jellybeans so that there will be less than 2 green ones
We need 4 or more out of 5 to be representatives. so either 4 reps or 5 reps
ans <-choose(13,4) * choose(14,1) + choose(13,5) * choose(14,0)
Answer: 11297 ways to select 5 members of congress with at least 4 representatives
5 outcomes from a 2 sided options and then 2 outcomes from a 6 sided options and then 52 options then 51 options then 50 options
ans <- 2^5 * 6^2 * 52 * 51 * 50
Answer: 152755200 outcomes
there at 4 “3” cards in a deck so we just check the probability that we draw 3 cards of which none are a 3 and then subtract that from 1
not_three <- (48/52)*(47/52)*(46/52)
ans <- round(1-not_three,digits = 4)
Answer: 0.2619 probability there is at least one three in three cards picked from a standard deck without replacement
there are 31 movies and he wants to choose 5 of them
ans <- choose(31,5)
Answer: 169911 ways to rent 5 movies from 31
Step 2. How many different combinations of 5 movies can he rent if he wants at least one mystery?
He needs to pick either 1, 2, 3, 4, or 5 mystery movies to ensure at least one mystery
ans <- 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)*choose(17,0)
Answer: 163723 ways to rent 5 movies from 31 with at least one mystery (out of 14)
9 outcomes choosing 3 from 4, 3 from 104, and 3 from 17, assuming no duplicates would be:
ans <- format(4*3*2*104*103*102*17*16*15,scientific = TRUE)
Answer: 1.069897e+11 ways to organize an equal number of 9 symphonies from Brahs, Haydn, and Mendelsson picking 4,104, and 17 respectively.
pick 13 books with at most 4 out of 5 nonfiction (0-4) and 9-13 out of 19
ans <- format(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), scientific = TRUE)
Answer: 2.420562e+06 ways to pick 13 books with no more than 4 non fiction books
Step 2. If he wants to include all 6 plays, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
pick 6 out of 6 plays and then any combination of the remaining 18
ans <- format(6*5*4*3*2*1*18*17*16*15*14*13*12, scientific = TRUE)
Answer: 1.154829e+11 ways to pick 13 books with 6 plays
There are only two ways for this to occur -> all sycamore then all cypress or all cypress then all sycamore
total <- 10
sycamore <- 5
ans <- round(2/(factorial(total)/(factorial(sycamore)^2)),4)
Answer: 0.0079 chance of planting all the same trees in a row
expected value is probability * gain + probability * loss
pq_l <- 44/52
pk_or_a <- 8/52
win <- 4
lose <- -16
ev <- pq_l * win + pk_or_a * lose
Answer: 0.923076923076923 expected value
Step 2. If you played this game 833 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.
It was be expected value times number of games
ans <- round(833 * ev, digits =2)
Answer: 768.92 expected wins