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STATS 550 SDSU

HW 4 Page 33. #s 1.41, 1.43, 1.44

Last edited 152 days ago by Eddie Coda

1.41 Modify the code in the R script CoinFlip.R to simulate the probability of getting exactly one head in four coin tosses.

1.43 Use R to simulate the probability of getting at least one 8 in the sum of two dice rolls.

Two dice are rolled ➡️ sample (_ , 2, _______)

Values of the dice are 1 to 6 (inclusive) ➡️ sample (1:6 , _ , _______)

trial <- sample(1:6, 2, replace=TRUE)

OK so the probability of getting exactly one 8 in the sum of two dice rolls would be different from checking at least 8, where (𝑋 = 8)->>>>> `success <- if (sum(trial) == 8) 1 else 0`

would be ≅ 0.1442❌❌❌

P(𝑋 ≥ 8) ->>>> `success <- if (sum(trial) >= 8) 1 else 0`

would be ≅ 0.4119✅✅✅

I think the trick to understanding this question would be to recognize that, “…getting at least one 8 in the sum of two dice rolls.” Is the same as saying “… sum of the two rolls is greater than or equal to 8.” P(𝑋 ≥ 8) we want to calculate the sum of exactly 8, as well as accounting for getting the sum of 9, 10, 11, 12.

1.44 Use R to simulate the probability in Exercise 1.30. (1.30 A tetrahedron dice is four-sided and labeled with 1, 2, 3, and 4. When rolled it lands on the base of a pyramid and the number rolled is the number on the base. In five rolls, what is the probability of rolling at least one 2?) See image regarding problem 2 (1.43)

mean( simlist )

Please copy and paste your answers (output) under each code chunk.

n <- 10000 # Number of repetitions

simlist <- numeric(n) # Initialize vector

for (i in 1:n) {

trial <- sample(0:1, 4, replace=TRUE)

success <- if (sum(trial)==1) 1 else 0

simlist[i] <- success }

mean(simlist)

[1] 0.2528

#1.41

Modify the code in the R script CoinFlip.R to simulate the probability of getting exactly one head in four coin tosses.

Let’s Try Mine :

Sure! Here is the reformatted code:

# We can simulate the flips using the sample line, where heads=0 and tails=1, and the number of flips is 4.

trial <- sample(0:1, 4, replace = TRUE)

# Routine to decide if we get one-and-only one head in a toss.

# We dump this into the value 'success' and replicate it 4 times.

#if (sum(trial) == 1) 1 else 0

# Using the code above, we can simulate replication over 10,000 trials.

n <- 10000 # Number of repetitions

simlist <- numeric(n) # Initialize vector "simlist"

for (i in 1:n) {

trial <- sample(0:1, 4, replace = TRUE)

success <- if (sum(trial) == 1) 1 else 0

simlist[i] <- success

}

mean(simlist) # Proportion of trials with 3 heads in 10,000 samples

Let me know if there's anything else I can help you with!

[1] 0.2478

We can see that the resulting estimate is fairly close to the exact solution 1/4=0.25

Increase n to 100,000 or even a million to get more precise estimates.

n <- 10000 # Number of repetitions

simlist <- numeric(n) # Initialize vector

for (i in 1:n) # Loop through 10000 trails

{

trial <- sample(0:1, 4, replace=TRUE) # new trail of four flips

success <- if (sum(trial)==1) 1 else 0 # 1 if exactly one head, 0 otherwise

simlist[i] <- success # keep record of successes

}

mean(simlist) # Proportion of trials with one head

[1] 0.2491

#1.43

Use R to simulate the probability of getting at least one 8 in the sum of two dice rolls.

We want to estimate the probability of getting a sum of 8 when rolling two dice. Instead of taking the basic "for-loop approach, we can make things functionally better:

we can create 10000 columns with 2 rows, simply put-> we run 10,000 trials of rolling 2 dice

then we can sum the columns, metaphorically adding the two dice together

and the last like finds the average of sums that are equal to 8

ntrials = 10000

x = matrix(sample(1:6, 2*ntrials, replace=TRUE), nrow=2, ncol=ntrials)

y = colSums(x)

mean(y >= 8)

[1] 0.4201

We can see that the resulting estimate is fairly close to the exact solution 5/36= 0.1389

n <- 10000 # Number of repetitions

simlist <- numeric(n) # Initialize vector

for (i in 1:n) # Loop through 10000 trails

{

trial <- sample(1:6, 2, replace=TRUE) # new trail of two rolls

success <- if (sum(trial)>=8) 1 else 0 # 1 if sum >= 8, 0 otherwise

simlist[i] <- success # keep record of successes

}

mean(simlist) # Proportion of trials with one head

[1] 0.4215

#1.44

Use R to simulate the probability in Exercise 1.30:

1.30) A tetrahedron dice is four-sided and labeled with 1, 2, 3, and 4. When rolled it lands on the base of a pyramid and the number rolled is the number on the base. In five rolls, what is the probability of rolling at least one 2?

Sure! Here is the reformatted code:

# Let's create a matrix map of a 5x10 array.

# Each column represents 5 rolls.

# The columns represent the trials.

# We will simulate rolling a 4-sided die.

sides <- 1:4 # Sides of the die

ntrials <- 10000 # Number of trials

x <- matrix(sample(sides, 4 * ntrials, replace = TRUE), nrow = 5, ncol = ntrials)

y <- (x == 2) # Check which rolls are 2's. The results in y are TRUE/FALSE

z <- colSums(y) # Sum the columns to get the number of 2's in each trial

mean(z > 0) # Proportion of trials with at least one 2

Let me know if there's anything else I can help you with!

we can see that the resulting estimate is fairly close to the exact solution = .763

Sure! Here is the reformatted code:

n <- 10000 # Number of repetitions

simlistp <- numeric(n) # Initialize vector

for (i in 1:n) { # Loop through 10000 trails

trial <- sample(1:4, 5, replace=TRUE) # new trail of 5 rolls

success <- if (sum(trial==2) >= 1) 1 else 0 # 1 if at least one 2, 0 otherwise

simlistp[i] <- success # keep record of successes

}

mean(simlistp) # Proportion of trials with one head

Let me know if there's anything else I can help you with!

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