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Tossing 3 fair coins in R (2)

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2

1

If X = # of heads showing when three coins are tossed, find P(X=1), and E(X).

The experiment has the following distribution:

According to my calculation,
P(X=1) = 0.375 ≈ 0.40
E(X) = 1.50

And,

where n = # of repetitions of the experiment.

Source Code

noOfExperiments = 10000;
mySample <- sample(c(0,1,2,3), noOfExperiments, replace = T)
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount / noOfExperiments
eX <- sum(mySample)/noOfExperiments 

According to the above code, I obtained
P(X=1) = 0.2518
E(X) = 1.4917 ≈ 1.50

It seems like the value of E(X) is coming well, but the value of P(X=1) is not coming correctly.

What is wrong with my code?

Edit:

I have written the following code on the basis of Edward Carney's comment.

noOfExperiments = 1000;
mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) )
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount/noOfExperiments
eX <- sum(mySample)/noOfExperiments

---

Note: related question.

r statistics probability

share|improve this question

edited Mar 8 at 3:13

user366312

asked Mar 8 at 1:52

user366312user366312

3,88046159319

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  An experiment would not involve inserting the probabilities ahead of time. A more realistic simulation can be obtained using mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) followed by table(mySample)/noOfExperiments `.
  
  – Edward Carney
  Mar 8 at 2:32
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In your original code, you're not using the given probabilities at all - you could apply them using sample(c(0,1,2,3), noOfExperiments, replace = T, prob = c(0.125, 0.375, 0.375, 0.125))
  
  – Marius
  Mar 8 at 3:18
  

  
  
  
  

add a comment | 

2

1

If X = # of heads showing when three coins are tossed, find P(X=1), and E(X).

The experiment has the following distribution:

According to my calculation,
P(X=1) = 0.375 ≈ 0.40
E(X) = 1.50

And,

where n = # of repetitions of the experiment.

Source Code

noOfExperiments = 10000;
mySample <- sample(c(0,1,2,3), noOfExperiments, replace = T)
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount / noOfExperiments
eX <- sum(mySample)/noOfExperiments 

According to the above code, I obtained
P(X=1) = 0.2518
E(X) = 1.4917 ≈ 1.50

It seems like the value of E(X) is coming well, but the value of P(X=1) is not coming correctly.

What is wrong with my code?

Edit:

I have written the following code on the basis of Edward Carney's comment.

noOfExperiments = 1000;
mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) )
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount/noOfExperiments
eX <- sum(mySample)/noOfExperiments

---

Note: related question.

r statistics probability

share|improve this question

edited Mar 8 at 3:13

user366312

asked Mar 8 at 1:52

user366312user366312

3,88046159319

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  An experiment would not involve inserting the probabilities ahead of time. A more realistic simulation can be obtained using mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) followed by table(mySample)/noOfExperiments `.
  
  – Edward Carney
  Mar 8 at 2:32
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In your original code, you're not using the given probabilities at all - you could apply them using sample(c(0,1,2,3), noOfExperiments, replace = T, prob = c(0.125, 0.375, 0.375, 0.125))
  
  – Marius
  Mar 8 at 3:18
  

  
  
  
  

add a comment | 

If X = # of heads showing when three coins are tossed, find P(X=1), and E(X).

The experiment has the following distribution:

According to my calculation,
P(X=1) = 0.375 ≈ 0.40
E(X) = 1.50

And,

where n = # of repetitions of the experiment.

Source Code

noOfExperiments = 10000;
mySample <- sample(c(0,1,2,3), noOfExperiments, replace = T)
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount / noOfExperiments
eX <- sum(mySample)/noOfExperiments 

According to the above code, I obtained
P(X=1) = 0.2518
E(X) = 1.4917 ≈ 1.50

It seems like the value of E(X) is coming well, but the value of P(X=1) is not coming correctly.

What is wrong with my code?

Edit:

I have written the following code on the basis of Edward Carney's comment.

noOfExperiments = 1000;
mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) )
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount/noOfExperiments
eX <- sum(mySample)/noOfExperiments

---

Note: related question.

r statistics probability

share|improve this question

edited Mar 8 at 3:13

user366312

asked Mar 8 at 1:52

user366312user366312

3,88046159319

noOfExperiments = 10000;
mySample <- sample(c(0,1,2,3), noOfExperiments, replace = T)
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount / noOfExperiments
eX <- sum(mySample)/noOfExperiments 

noOfExperiments = 1000;
mySample <- replicate(noOfExperiments, sum(sample(c(0, 1), 3, replace=T)) )
outcomeCount <- length(which(mySample==1))
prob <- outcomeCount/noOfExperiments
eX <- sum(mySample)/noOfExperiments

share|improve this question

edited Mar 8 at 3:13

user366312

asked Mar 8 at 1:52

user366312user366312

3,88046159319

share|improve this question

edited Mar 8 at 3:13

user366312

asked Mar 8 at 1:52

user366312user366312

3,88046159319

asked Mar 8 at 1:52

user366312user366312

3,88046159319

asked Mar 8 at 1:52

user366312user366312

3,88046159319