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Two persons are studying for a test.
X’s chances of getting a “B” are 80%. Y’s chances of getting a “B” are 90%.
The probability of at least one of them getting a “B” is 91%.
a. What is the probability that only X gets a “B”?
b. What is the probability that only Y gets a “B”?
c. What is the probability that both won’t get a “B”?
This sounds like your home work... I will give you some hints so you can get started.
Note: if probability of x getting a = some number c
then probability of x not getting a = 1-c
a. = (probability of x getting b)*(probability of y not getting b)
b. = (probability of y getting b)*(probability of x not getting b)
c. = you can figure it out ;)
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I am new to data science and I came across this exercise that I can't figure out.
I have a data set containing binary data, which represents win and loss for the team. I wanna find out the probability for the win and loss depending on the result of the previous game.
Something like this.
win loss
prev win ? ?
prev loss ? ?
I am not asking for the code here. Though it would be helpful if you do. I just want to understand how to go about doing it.
You can generate that prop.table by comparing the result to the lagged (previous) result:
library(dplyr)
results <- data.frame(results = rbinom(100,1, 0.5)) %>%
mutate(prev_result = lag(results))
prop.table(table(results$prev_result, results$results))
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I want to understand the physical significance of n raise to some decimal power.
Like when i say 2^5. I understand that it means 2 multiplied 5 times. But how do i analyse 2^0.1.
Please suggest.
2^0.1 is the tenth root of 2. For rational powers, x^(p/q)=(x^p)^(1/q) is a combination of powers and roots.
For general real numbers,
x^y = exp(log(x)*y).
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In R, how do you generate a vector (data) with outliers? Great if the data is "acceptable" normal distributed.
#DWin is right that this depends on what you mean by "outlier". For the record, I use the same definition that he is using, so I would use (have used) something like the code he, and #Ferdinand.kraft, list. Others sometimes mean a datum more extreme than you might typically find. This is tricky to define for a simulation study, but a common definition is a point more than 1.5 times the interquartile range past the 1st (3rd) quartile. Here is a simple way to find that (I'm sure there will be more efficient ways):
flag <- 0
while(flag==0){
X <- rnorm(N)
bp <- boxplot(X, plot=FALSE)
if(length(bp$out)!=0){
flag <- 1
}
}
This really depends on the definition of "outlier";
c(rnorm(100), 100, -100) # an egregious example
plot(density( c( rnorm(90), rnorm(5, 1) ) ) ) # not as egregious
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For my theory of computation class, we are supposed to do some review/practice problems to work off the rust and make sure we are ready for the course. Some of the problems are induction proofs. I did this at one time, but apparently it has completely escaped me. I've watched a couple tutorials, but still can't do problem 'a'. If anyone can walk me through the first problem I'm pretty sure I could figure out the second one on my own. Any help would be appreciated!
First verify it holds for n = 1.
Then assume it is true for n = x ( the sum of the first x squares ) and then try to compute the sum of the the first x + 1 squares. You know the result for the first x, you just add the last square to that sum. From there it should be easy.
And you posted on the wrong site.
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I'm a student just learning how to use R and thus far I've made a bit of progress. I'm snagging at a question which asks: For what values of i does Y equal 3?
the data set: c(3,5,2,3,5,4,4,2,3,5)
If I understand your question correctly, you want the index, i inside the data set (in this case, a vector) Y such that Y[i]=3?
Then you just need to use the which function. For more information, make sure you try reading the help files, which you can invoke using the command ?which or help(which)
Now, some code:
# Your data
Y <- c(3,5,2,3,5,4,4,2,3,5)
# Find the index where Y is equal to 3
which(Y==3, arr.ind=TRUE)
And welcome to SO. This is a pretty common question for beginners, so next time, make sure you Google or search around for a solution to elementary problems such as these. Have a good day.