Difference between revisions of "SOCR EduMaterials Activities Discrete Probability examples"
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* <math>P(X=5)=0.7^4 0.3=0.07203,</math>where X represents the number of trials. | * <math>P(X=5)=0.7^4 0.3=0.07203,</math>where X represents the number of trials. | ||
* '''Example 5:''' | * '''Example 5:''' | ||
− | <math> E(X)=\ | + | <math> E(X)=\frac{1}{p}=\frac{1}{0.30}=3.33</math> |
* '''Example 6:''' | * '''Example 6:''' |
Revision as of 15:08, 24 April 2007
- Description: You can access the applets for the distributions at http://www.socr.ucla.edu/htmls/SOCR_Distributions.html .
- Example 1:
Find the probability that 3 out of 8 plants will survive a frost, given that any such plant will survive a frost with probability of 0.30. Also, find the probability that at least 1 out of 8 will survive a frost. What is the expected value and standard deviation of the number of plants that survive the frost?
- Answer:
- \( X \sim b(8,0.3) \), \( P(X=3)= {8 \choose 3} 0.3^30.7^5=0.2541\)
- \( P(X \ge 1)=1-P(X=0)=1-.7^8=0.942\)
- \( E(X) = np = 8 \times 0.3=2.4 \)
- \( Sd(X)= \sqrt{npq}=\sqrt{8\times 0.30 \times 0.70}=1.3\)
Below you can see a SOCR snapshot for this example:
- Example 2:
If the probabilities of having a male or female offspring are both 0.50, find the probability that a familiy's fifth child is their first son.
- Answer:
- \(P(X=5)= 0.50^5=0.03125\)
- Example 3:
- a. \(X \sim b(4,0.8) \), \( P(X=2)= {4 \choose 2} 0.8^2 0.2^2=0.1536\)
- b. \(P(X\ge 2)= {4 \choose 2} 0.8^2 0.2^2+{4 \choose 3} 0.8^3 0.2^1+{4 \choose 4} 0.8^4=0.9728\)
- Example 4:
- \(P(X=5)=0.7^4 0.3=0.07203,\)where X represents the number of trials.
- Example 5:
\( E(X)=\frac{1}{p}=\frac{1}{0.30}=3.33\)
- Example 6:
- a. \(X \sim b(5,0.90). P(X=4)= {5 \choose 4} 0.9^4 0.1^1=0.32805\)
- \(P(X\ge 1)= 1-P(X=0)= 1- 0.1^4=0.9999\)
- b. P(X=0)=.001,
0.001= 0.1^n n= 3