Suppose I have an unfair coin that has a 70% chance to land on heads and a 30% chance to land on tails.

Which of the following codes correctly sets up flipping this coin 10 times?

Question

Suppose I have an unfair coin that has a 70% chance to land on heads and a 30% chance to land on tails.

Which of the following codes correctly sets up flipping this coin 10 times?

BlackTom AI · Accepted Answer

sample() in R accepts a vector of outcomes, a sample size, an optional probability vector of the same length, and whether sampling is with replacement — exactly what a sequence of independent biased coin flips requires.

1) Relevant knowledge points and concepts
- sample(x, size, replace, prob) samples from vector x.
- prob must be a numeric vector of the same length as x; its entries give relative probabilities and should sum to 1 (or be proportional).
- replace = TRUE is required to allow repeated outcomes (independent flips).
- For an unfair coin with P(Heads)=0.7 and P(Tails)=0.3, the prob vector must be c(0.7, 0.3) corresponding to the order of the outcome vector.

2) Step-by-step reasoning
- Create the outcome vector: box = c("Heads", "Tails").
- Call sample on that vector with size = 10 to perform 10 flips.
- Provide prob = c(0.7, 0.3) so each draw uses those probabilities.
- Use replace = TRUE so each flip is independent and values can repeat.

3) Intermediate checks and small calculations
- Probabilities sum: 0.7 + 0.3 = 1, so they form a valid probability distribution.
- Expected number of Heads in 10 flips = 10 * 0.7 = 7 (useful check but not required for sampling).
- Without replace = TRUE, sampling 10 items from 2 distinct values without replacement would be impossible/error.

4) Why this answer is correct
- The code exactly matches the required argument structure: the outcome labels, number of trials, the correct probability vector in the same order, and replacement to model independent flips. Each of the 10 draws is performed with P(Heads)=0.7.

5) Why other common options are incorrect
- Omitting prob uses equal probabilities (not 0.7/0.3).
- Omitting replace = TRUE attempts sampling without replacement (invalid for 10 > 2).
- Swapping the prob order would assign 0.7 to Tails, reversing the bias.
- Using rbinom(10,1,0.7) returns 0/1 rather than labeled "Heads"/"Tails" (though convertible).

6) Conclusion
The correct code is:
box = c("Heads", "Tails")
sample(box, 10, prob = c(0.7, 0.3), replace = TRUE)
This matches the provided answer.

Suppose I have an unfair coin that has a 70% chance to land on heads and a 30% chance to land on tails. Which of the following codes correctly sets up flipping this coin 10 times? 单项选择题

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