Virtual Laboratories > Bernoulli Trials > 1 2 3 4 5 6 [7]

7. Notes

Simulating Bernoulli Trials

It is very easy to simulate a Bernoulli trials process with random numbers.

1. Let p be in [0, 1] and let U₁, U₂, U₃, ... be a sequence of independent random variables, each uniformly distributed on (0, 1). Show that the following sequence is a Bernoulli trials process with parameter p:

I_j = 1 if U_j p, I_j = 0 if U_j > p

The binomial and negative binomial experiments can now be simulated directly from the Bernoulli trials sequence, since these variables are functions of the Bernoulli trials sequence.

Related Topics

Bernoulli trials appear in many chapters in this project, further evidence of the importance of the model.

• Sampling with replacement from a dichotomous population produces Bernoulli trials. The chapter on Finite Sampling Models includes a number of topics based on this type of sampling.
• Many games are based on Bernoulli trials. The chapter on Games includes a number of these.
• The chapter on Red and Black is more advanced and includes a study of competing strategies for a game based on Bernoulli trials.
• The simple random walk, discussed in the chapter on Random Walk is based on Bernoulli trials.
• Estimation of p is discussed in the chapters on Point Estimation and Interval Estimation.
• Hypothesis tests of p are discussed in the chapter on Hypothesis Testing.

Books

The Bernoulli trials model is discussed in virtually every book on probability. In particular see

• A First Course in Probability, 5th edition, by Sheldon Ross
• An Introduction to Probability Theory and Its Applications, (Vol 1) 3rd edition, by William Feller

Selected Answers for Section 1

1.8. Yes, probably so. The outcomes are correct and incorrect and p = 1 / 4.

1.9. Yes, approximately. The outcomes are prefer A and do not prefer A; p is the proportion of voters in the entire district who prefer A.

1.10. Yes, the outcomes are red and black, and p = 18 / 38.

1.11. No, probably not. The games are almost certainly dependent, and the win probably depends on who is serving and thus is not constant from game to game.

1.17.

1. k = 10, E(Y_k) = 19.56
2. k = 5, E(Y_k) = 426.22
3. k = 32, E(Y_k) = 62.76

Selected Answers for Section 2

2.5. f(0) = 0.4019, f(1) = 0.4019, f(2) = 0.1608, f(3) = 0.0322, f(4) = 0.0032, f(5) = 0.0001.

2.6. 0.07813

2.11.

1. P(at least 1 ace in 6 rolls) = 0.6551
2. P(at least 2 aces in 12 rolls) = 0.6187

2.12.

1. P(at least 1 ace in 4 rolls of 1 die) = 0.5177
2. P(at least 2 aces in 24 rolls of 2 dice) = 0.4914.

2.23. X = Number of failures. E(X) = 1, sd(X) = 0.9899

2.24. X = Number of aces. E(X) = 166.67, sd(X) = 11.79

2.31. X_n = Number of heads in the first n tosses. P(X₂₀ = j | X₁₀₀ = 30) = C(20, j) C(80, 30 - j) / C(100, 30).

2.37. X = Number who prefer A

1. E(X) = 20, sd(X) = 3.464.
2. P(X < 19) = 0.3356.
3. P(X < 19) ~ 0.3335

2.44.

1. R_3,2(p) = 3p² - 2p³.
2. R_5,3(p) = 10p³ - 15p⁴ + 6p⁵.
3. 3 out of 5 is better for p 1 / 2.

Selected Answers for Section 3

3.13. R: Reject null hypothesis that the coin is fair.

1. P(R) = 0.180, P(R^c) = 0.820
2. P(R) = 0.384, P(R^c) = 0.616
3. P(R) = 0.678, P(R^c) = 0.322
4. P(R) = 0.930, P(R^c) = 0.070

3.15. No: 0.0262

Selected Answers for Section 4

4.5. 0.482

4.10. X = # launches. E(X) = 50, sd(X) = 49.497.

4.12. 0.4

4.18. Geometric with p = 18 / 38.

4.22. $1000.

4.27.

1. P(W = 1) = 2/3, P(W = 2) = 1/3
2. P(W = 1) = 4/7, P(W = 2) = 2/7, P(W = 3) = 1/7.
3. P(W = i) = 2^{n - i} / (2ⁿ - 1) for i = 1, 2, ..., n.

Selected Answers for Section 5

5.6. 0.579

5.13. X = launch number of 4'th failure. E(X) = 200, sd(X) = 98.995

5.17. X = number of tosses needed to get 50 heads.

1. 0.0072
2. No.

5.30.

1. 0.6825.
2. 0.7102

5.36. A gets $72.56, B gets $27.44

Selected Answers for Section 6

6.11.

1. 0.0075
2. 0.0178
3. 0.205
4. 0.123

6.12. f(u, v, w, x, y, z) = C(4; u, v, w, x, y, z) (1/4)^{u + z} (1/8)^{v + w + x+ y} for u, v, w, x, y, z nonnegative integers that sum to 4

6.14.

1. -0.625
2. -0.0386

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