library(graphics)
There are a total of 138 marbles in three colors (red, white and blue). so the probability of picking red or blue marble
is the total number of (red+blue) / total marbles
round((54+75)/(54+75+9), 4)
## [1] 0.9348
So the probablity of picking red or blue is 0.9348.
The probability of having a red ball is: red/(green+ red+blue+yellow)
20/(19+20+24+17)
## [1] 0.25
What is the probability that a customer is not male or does not live with parents? Write your answer as a fraction or a decimal number rounded to four decimal places.
df <- data.frame(male=c(81,116,215,130,129),
female=c(228,79,252,97,72),
row.names = c("apartment","dorm","with parents","Fraternity House","other"))
df
## male female
## apartment 81 228
## dorm 116 79
## with parents 215 252
## Fraternity House 130 97
## other 129 72
p_male_live_with_parents <- 215/(81+116+215+130+129+228+79+252+97+72)
p_male_live_with_parents
## [1] 0.1536812
p_not_male_or_not_with_parents <- round((1-p_male_live_with_parents), 4)
p_not_male_or_not_with_parents
## [1] 0.8463
So the probability of a customer not male and does not live with parents is 0.8463
Answer: A)Dependent B)Independent
The answer is A)Dependent
choose(8, 3) * choose(7, 3) * 3
## [1] 5880
so there are 5880 different veggie wraps can be made.
Answer: A)Dependent B)Independent
The answer is B)Independent
Solution:
14!/(14-8)! = 121,080,960
round((choose(4, 1) * choose(9, 3) / choose((9+4+9), 4)), digits = 4)
## [1] 0.0459
The probability of choosing 0 red, 1 orange and 3 green jelly beans is 0.0459.
Answer: 11!/7! = \(11*10*9*...*1 / 7*6*5*...*1 = 11*10*9*8 = 7920\)
Answer: 33% of subscribers to a fitness magazine are 34 years older or younger.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
p_W <- 0.5^4 * choose(4, 3)
#Step 1.
EV <- p_W * 97 - (1-p_W) * 30
EV
## [1] 1.75
Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)
559 * EV
## [1] 978.25
Win 978.25 dollars
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
p_W_12 <- sum(dbinom(4:0, 9, 0.5))
#Step 1.
EV_12 <- p_W_12 * 23 - (1-p_W_12) * 26
EV_12
## [1] -1.5
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
994 *EV_12
## [1] -1491
Loss 1491 dollars
round(((0.2 * 0.59) / ((0.2 * 0.59) + (0.8 * 0.1))), digits = 4)
## [1] 0.596
round(((0.8 * 0.9) / ((0.8 * 0.9) + (0.2 * 0.41))), digits = 4)
## [1] 0.8978
round((0.2 + 0.8 * 0.1), digits = 4)
## [1] 0.28