a)Probability would be zero as there is no possible outcome for two fair dice to get a sum of 1 b)There are four possible combinations that could give a sum of 5: (2,3) , (3.2), (4,1) , (1,4). The maximum amount of outcomes is 36, as per 6 possible outcomes per dice.
4/36
## [1] 0.1111111
c)To get a sum of two you must throw two sixes which means there are only two possible combinations.
2/36
## [1] 0.05555556
# Let A = event that kid misses one day
# Let B = event that kid misses two days
# Let C = event that kid misses three or more days
pa <- .25
pna <- 1-pa
pb <- .15
pnb <- 1-pb
pc <- .28
pnc <- 1-pc
1- (pa+pb+pc)
## [1] 0.32
1- (pb+pc)
## [1] 0.57
pa+pb+pc
## [1] 0.68
{1- (pa+pb+pc)}*{1- (pa+pb+pc)}
## [1] 0.1024
(pa+pb+pc)*(pa+pb+pc)
## [1] 0.4624
I believe it was reasonable because it is safe to assume that when one child gets sick in the house he/she is kept separate from the healthy child as to avoid spreading the sickness. We also did not have numerical data as to make an assumption including the probability of the contagion of disease.
Are being in excellent health and having health coverage mutually exclusive? No they are not because the intersection between both events does not equal to zero
What is the probability that a randomly chosen individual has excellent health?
# Let E = excellent health
pe <- 0.2327
pne <- 1-pe
The probability is 0.2327
# Let H = health insurance
ph <- 0.8738
pnh <- 1-ph
pewh <- 0.2099
pegh <- pewh/ph
pewh/ph
## [1] 0.2402152
pewnh <- 0.0230
pegnh <- pewnh/pnh
pewnh/pnh
## [1] 0.1822504
## Let S = votes in favor of Scott
## Let G = voters for Scott graduated from college
## Let F = voters against Scott that graduated from college
ps <- 0.53
pns <- 1-ps #voters against Scott
pg <- 0.37 #voters in favor of Scott with college education
png <- 1-pg
pf <- 0.44 #voters against Scott with college education
pnf <- 1-pf
(ps*pg)+(pns*pf) #Voters with a college degree with or against Scott
## [1] 0.4029
pce <- 0.4029
psgg <-(ps*pg) #Given that voted for Scott voter is college educated
psgce <- {(ps*pg)/pce}
{(ps*pg)/pce}
## [1] 0.4867213
The probability that any college educated voter voted for Scott was 0.4867
# Let HB = Hardcover book
# Let PF = Paperback Fiction No Replacement
phb <- 28/95
ppf <- 59/94
phbapf <- phb*ppf
phb*ppf
## [1] 0.1849944
# Let FB = Fiction Book
# Let HBNR = Hardcover Book No Replacement
pfb <- 72/95
phbnr <- 28/94
pfbahbnr <- pfb*phbnr
pfb*phbnr
## [1] 0.2257559
pfb*phb
## [1] 0.2233795
pnc <- (52-12-4)/52 #Number card
pfc <- (12/52) #Face card
pac <- (4/52) #Ace card
pacc <- (1/52) #Ace of Clubs card
-2+(pnc*0)+(pfc*3)+(pac*5)+(pacc*20)
## [1] -0.5384615
Completely lost on this one I tried to look online for help but answers and methods were very very different.
An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served?
How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of X Y. What is the standard deviation of the amount left in the box?