SOCR EduMaterials Activities Explore Distributions
This is an activity to explore the relations among some of the commonly used probability distributions.
- Description: You can access the applets for the above distributions at http://www.socr.ucla.edu/htmls/SOCR_Distributions.html .
- Geometric probability distribution: Let's roll two dice until a sum of 10 is obtained. What is the probability that the first sum of 10 will occur after the 5th trial? The answer to this question is \( P(X>5)=(1-\frac{3}{36})^5=0.6472. \). This is equivalent to the event that no sum of 10 is observed on the first 5 trials (5 failures). Now, using SOCR we can obtain this probability easily by entering in the SOCR geometric distribution applet \( p=\frac{3}{36}=0.0833 \) and in the Right Cut-Off box 5. We can find the desire probability on the right corner of the applet. The figure below clearly displays this probability.
- Binomial approximation to hypergeometric: An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement. Find the probability that exactly 10 out of the 15 selected are green marbles. The answer to this question can be found using the formula\[ P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}[[:Template:50 \choose 15]]=0.2449. \] Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below.
Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained. This is equal to \( P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}[[:Template:50 \choose 2]]=0.4286. \) This is also shown on the figure below.
We will approximate the probability of obtaining 1 green marble using binomial as follows. Select the SOCR binomial distribution and choose number of trials 2 and probablity of success \( p=\frac{35}{50}=0.7 \). Compare the figure below with the figure above. They are almost the same! Why? Using the binomial formula we can compute the approximate probability of observing 1 green marble as \( P(X=1)={2 \choose 1}0.70^10.30^1=0.42 \) (very close to the exact probability).
- SOCR Home page: http://www.socr.ucla.edu
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