On A Rainy Day, An Unnamed Statistics Professor Decides To P

On A Rainy Day An Unnamed Statistics Professor Decides To Play With a

On A Rainy Day An Unnamed Statistics Professor Decides To Play With a

On a rainy day, an unnamed statistics professor decides to play with an urn containing bottle caps. The urn contains caps from three brands: Guinness, Bass, and Smithwicks. Specifically, there are 5 Guinness caps, 10 Bass caps, and 10 Smithwicks caps. The professor randomly picks a cap from the urn, observes its brand, and then replaces it back into the urn. This process is repeated eight times, as he is very bored.

The assignment involves analyzing the probabilities associated with this experiment, which follows a binomial distribution due to the Bernoulli process of each independent trial with replacement.

Paper For Above instruction

To analyze the probability and statistical measures associated with the professor’s selection process, we consider the probabilities involved in each trial and the characteristics of binomial distribution. The key probability is the chance of selecting a Guinness cap during one draw, which is determined by the proportion of Guinness caps relative to the total caps in the urn.

Calculation of the probability of drawing a Guinness cap in a single trial:

Number of Guinness caps = 5

Total number of caps = 5 (Guinness) + 10 (Bass) + 10 (Smithwicks) = 25

Probability of selecting a Guinness cap (p) = 5/25 = 1/5 = 0.20

Since each draw is independent and with replacement, the number of Guinness caps drawn in 8 trials follows a binomial distribution with parameters n=8 and p=0.20. The binomial probability mass function (PMF) is given by:

P(X = k) = C(n, k) p^k (1 - p)^{n - k}

where C(n, k) is the binomial coefficient, "n choose k".

1. Probability that he will pick a Guinness every time

This is the probability of obtaining Guinness in all 8 trials, i.e., k=8.

P(X=8) = C(8,8) (0.20)^8 (0.80)^0 = 1 (0.20)^8 1 = (0.20)^8

Calculating: (0.20)^8 = 2.56 × 10^{-6}

2. Probability that he will pick a Guinness exactly seven times

P(X=7) = C(8,7) (0.20)^7 (0.80)^1 = 8 (0.20)^7 0.80

Calculating: (0.20)^7 = 1.28 × 10^{-5}

Thus, P(X=7) = 8 1.28 × 10^{-5} 0.80 ≈ 8 * 1.024 × 10^{-5} ≈ 8.192 × 10^{-5}

3. Probability that he will pick something other than a Guinness exactly once

Complementary probability: probability of not picking a Guinness is 0.80 in each trial.

Number of non-Guinness caps in 8 trials, with exactly 1 Guinness, meaning 7 non-Guinness caps.

P(X=1) = C(8,1) (0.20)^1 (0.80)^7 = 8 0.20 (0.80)^7

Calculating: (0.80)^7 ≈ 0.2097

Therefore, P(X=1) = 8 0.20 0.2097 ≈ 8 * 0.04194 ≈ 0.3355

4. Probability that he will pick a Guinness at most six times

This is the cumulative probability P(X ≤ 6). It can be calculated as:

P(X ≤ 6) = 1 - P(X=7) - P(X=8)

Using previous results: P(X=7) ≈ 8.192 × 10^{-5}, P(X=8) ≈ 2.56 × 10^{-6}

P(X ≤ 6) ≈ 1 - (8.192 × 10^{-5} + 2.56 × 10^{-6}) ≈ 1 - 8.448 × 10^{-5} ≈ 0.99991552

5. Probability that he will pick something other than a Guinness at least twice

Number of non-Guinness caps at least twice means X ≤ 6 (Guinness caps) ≤ 6, so non-Guinness caps are at least 2.

This is the same as the probability that the number of Guinness caps is at most 6, which we computed as P(X ≤ 6) ≈ 0.99991552.

Therefore, the probability of drawing non-Guinness caps at least twice = 1 - P(X=0) - P(X=1) - ... - P(X=6).

Alternatively, since total probability is 1, and P(X ≥ 2) = 1 - P(X=0) - P(X=1).

Calculate P(X=0): C(8,0) (0.20)^0 (0.80)^8 = 1 1 (0.80)^8 ≈ 0.1678

Recall P(X=1) ≈ 0.3355, thus:

P(X ≥ 2) = 1 - 0.1678 - 0.3355 ≈ 1 - 0.5033 ≈ 0.4967

6. The mean number of Guinness caps he should see in eight trials

The mean of a binomial distribution is given by:

μ = n p = 8 0.20 = 1.6

7. The standard deviation for the number of Guinness caps appearing in eight trials

The standard deviation of a binomial distribution is:

σ = √(n p (1 - p)) = √(8 0.20 0.80) = √(1.28) ≈ 1.131

Summary

The probabilities calculated reveal the likelihood of various outcomes in this binomial experiment, with the mean and standard deviation providing insight into the expected number of Guinness caps observed in the eight trials. These calculations are crucial for understanding the experimental randomness and chance behavior associated with sampling with replacement from a finite population.

References

  • McClave, J. T., & Sincich, T. (2018). Statistics. Pearson.
  • DeGroot, M. H., & Schervish, M. J. (2012). Probability and Statistics. Pearson.
  • Blitzstein, J., & Hwang, J. (2014). Introduction to Probability. CRC Press.
  • Ross, S. M. (2014). Introduction to Probability Models. Academic Press.
  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineering and the Sciences. Pearson.
  • Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury.
  • Freund, J. E., & Williams, L. (2010). Statistics: An Introduction. Prentice Hall.
  • Hogg, R. V., McKean, J. W., & Craig, A. T. (2013). Introduction to Mathematical Statistics. Pearson.
  • Kendall, M., & Stuart, A. (2015). The Advanced Theory of Statistics. Macmillan.
  • Vose, D. (2008). Risk Analysis: A Quantitative Guide. Wiley.

Related Assignments