If you roll a pair of fair dice, what is the probability of..
Answer: A fair die has 6 sides and the probability of getting a particular side is 1/6. When a pair of fair dice are rolled, we get 36 possible outcome as follows (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)..(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
Therefore, the probility of getting a sum of 1 = 0/36 = 0
Answer: Probability of getting a sum of 5 = \[ \frac {4}{36} = \frac {1}{9}\]
Answer:
Probability of getting a sum of 12 = \[\frac {1}{36}\]
Data collected at elementary schools in DeKalb County, GA suggest that each year roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more days due to sickness.
## Let:
## p1 = Probability of missing one day, pn1 = probability of not missing 1 days
## p2 = probability of missing 2 days, pn2 = probability of not missing 2 days
## p3 = probability of missing 3 days, pn3 = probability of not missing 3 days
p1 <- 25/100
pn1 <- 1 - p1
p2 <- 15/100
pn2 <- 1-p2
p3 <- 28/100
pn3 <- 1-p3
p1## [1] 0.25p2## [1] 0.15p3## [1] 0.28## pn = Let probability of not missing any day
pn = 1 - (p1 + p2 + p3)
pn## [1] 0.32## let pa1 = probability of missing no more than one day => probability of missing at most one day
pa1 <- (pn + p1)
pa1## [1] 0.57## let p1a = probability to miss AT LEAST one day
p1a <- 1 - pn
p1a## [1] 0.68## let p2n = probability neither kid missed any school day. These events are independent of each other therefore, the multiplication ## rule for independent processes applies =>
p2n <- pn * pn
p2n## [1] 0.1024## let pb2 = probability both kids missed at least one school day. These events are still independent of each other =>
pb2 <- p1a * p1a
pb2## [1] 0.4624Answer: Computing the probabilities in (d) and (e) above as independent is reasonable because there is the chance that both kids may not fall sick at the same time and the parent can take the healthy kid to school while the sick one stays at home. Though siblings are likely to fall sick at the same time but it depends on if the sickness is a contagious one, and if the parent opts to take the sick child to a health facility while the healthy one remains at home. This is also a way to separate both kids in this situation.
The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance.
mat=matrix(c(.023, 0.0364, 0.0427, 0.0192, 0.0050,0.2099, 0.3123 ,0.2410 ,0.0817,0.0289), byrow=TRUE, nrow=2)
colnames(mat)=c("Excellent", "Very Good","Good", "Fair","Poor")
rownames(mat)=c("No Coverage","Coverage")
mat## Excellent Very Good Good Fair Poor
## No Coverage 0.0230 0.0364 0.0427 0.0192 0.0050
## Coverage 0.2099 0.3123 0.2410 0.0817 0.0289Answer:
NO. People can be in excellent health while having health coverage. As depicted by the survey data (20.99%), those in excellent health even have substantial amount of health coverage
## Let pe = probability that the individual has excellent health = probability of all individuals with excellent health with/without ## health coverage
pe <- 0.0230 + 0.2099
pe## [1] 0.2329## let pc = probability all with health coverage
## pce = probability that a chosen individual has excellent health given that he has health coverage
pc <- 0.2099+0.3123+0.2410+0.0817+0.0289
pce = 0.2099/pc
pce## [1] 0.2402152## let pnc = probability all without health coverage
## pnce = probability that a chosen individual has excellent health given that he doesn't have health coverage
pnc <- 0.0230+0.0364+0.0427+0.0192+0.0050
pnce = 0.0230/pnc
pnce## [1] 0.1821061## Comparing pce with pnce => probability of being in excellent health with and without coverage
pde = 0.2402152 - 0.1821061
pde## [1] 0.0581091This implies that the probability to be in excellent health varies between having and not having health coverage. Moreover, the amount of data for people in excellent health with coverage is higher than those in excellent health without coverage. Therefore, having excellent health and having health coverage are not independent
Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found that 53% of the respondents voted in favor of Scott Walker. Additionally, they estimated that of those who did vote in favor for Scott Walker, 37% had a college degree, while 44% of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a person who participated in the exit poll and found that he had a college degree. What is the probability that he voted in favor of Scott Walker?
## Let v = voters for Scott Walker with/without college degree,
## d = voters for Scott Walker with degree, vd = voters against Scott Walker with degree,
## pvd = probability of a person with degree voting in favor of Scott Walker
## percentage that voted against Scott Walker with/without = 100 - 53 = 47%
pvd <- (0.37 * 0.53)/((0.37 * 0.53) + (.44 * 0.47))
pvd## [1] 0.4867213The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.
mymat2=matrix(c(13,59,15,8),nrow=2,byrow=TRUE)
colnames(mymat2)=c("hard","paper")
rownames(mymat2)=c("fiction","nonfiction")
mymat2## hard paper
## fiction 13 59
## nonfiction 15 8## Let t = total books on the shelf, php = probility of drawing hardcover book first then a paperback fiction without replacement
t <- 13+15+59+8
php <- 28/t * 59/(t-1)
php## [1] 0.1849944## Let pfh = probility of drawing a fiction book first and then a hardcover book second without replacement
pfh <- (13 + 59)/t * (13+15)/(t-1)
pfh## [1] 0.2257559## Let pfpr = drawing as in caseb(b) with replacement
pfhr <- (13 + 59)/t * (13+15)/t
pfhr## [1] 0.2233795Answer:
The two scenarios involved either drawing with replacement or without replacement and only one book (out of a total of 95 books) was removed from the shelf without replacement. This in effect did have a significant impact the outcomes. If like 50 books were removed without replacement, the outcomes would have drastically be different
Andy is always looking for ways to make money fast. Lately, he has been trying to make money by gambling. Here is the game he is considering playing: The game costs 2 dollars to play. He draws a card from a deck. If he gets a number card (2-10), he wins nothing. For any face card (jack, queen or king), he wins 3 dollars. For any ace, he wins 5 dollars and he wins an extra $20 if he draws the ace of clubs.
## Let c = 2 = cost per game,
## t = 52 = total number of cards,
## n = 36 = total number cards (2-10),
## k = 12 = total face cards ('J', 'Q', 'K'),
## a = 4 = total aces,
## p_0 = probability of drawing a number card (probability of loosing $2),
## p_3 = probability of drawing a face card,
## p_5 = probability of drawing an ace but not ace of clubs,
## p_20 = probability of drawing ace of clubs
## x = amount to win per draw (0, 3, 5, 20)
## z = profit per draw
p_0 <- 36/52
p_3 <- 12/52
p_5 <- 3/52
p_20 <- 1/52
x <- c(-2, 3, 5, 20)
y <- c(p_0, p_3, p_5, p_20)
z <- c(p_0 * -2, p_3 * 3, p_5 * 5, p_20 * 20)
df <- data.frame('x' = x, 'y' = y, 'z' = z)
df## x y z
## 1 -2 0.69230769 -1.3846154
## 2 3 0.23076923 0.6923077
## 3 5 0.05769231 0.2884615
## 4 20 0.01923077 0.3846154## let p = expected profit per game = summation of all the profit per chance
p <- z[1] + z[2] + z[3] + z[4]
p## [1] -0.01923077Answer:
I wouldn’t recommend this game as a good way of making money to Andy because from the computations, his chances of loosing $2 per draw (0.69230769 = 69.23%) is higher than every other chances. He will end up loosing 0.02 cents on the average.
Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y . Suppose these random variables have the following means, standard deviations, and variances:
mymat3=matrix(c(48,1,1, 2,.25,.0625), nrow=2, byrow=TRUE)
colnames(mymat3)=c("mean", "SD", "Var")
rownames(mymat3)=c("X, In Box","Y, Scooped")
mymat3## mean SD Var
## X, In Box 48 1.00 1.0000
## Y, Scooped 2 0.25 0.0625Answer:
## Let ti = total ounces of icecream served at the party
ti <- 48 + 3*(2)
ti## [1] 54## Let sd = Var(X + Y1 + Y2 + Y3) = total standard deviations including three independent scoops of 0.0625 each
sd <- 1.0000 + 0.0625 + 0.0625 + 0.0625
sd## [1] 1.1875Answer:
Each box contains 48 ounces of ice cream fluid, one scoop will leave 46 ounces in the box.
## Let sd_1 = standard deviation after one scoop
sd_1 <- 1.0000 + 0.0625
sd_1## [1] 1.0625Answer:
We Add variances even when the random (independent) variables are subtracted from one another because it helps to express the increase in our UNCERTAINTY when the variables (quantity of ice cream in this case) vary.