SOCR EduMaterials Activities Discrete Probability examples

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**b. <math>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</math>
**b. <math>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</math>
* '''Example 4:'''
* '''Example 4:'''
-
* <math>0.7^4 0.3=0.07203</math>
+
* <math>P(X=5)=0.7^4 0.3=0.07203, where X represents the number of trials.</math>
* '''Example 5:'''
* '''Example 5:'''
<math> X \sim g(\frac{1}{0.30} </math>
<math> X \sim g(\frac{1}{0.30} </math>

Revision as of 19:56, 24 April 2007

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) = 0.505 = 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) = 0.740.3 = 0.07203,whereXrepresentsthenumberoftrials.
  • Example 5:

X \sim g(\frac{1}{0.30}

  • Example 6:
    • a. X \sim b(5,0.90) , P(X=4)= {5 \choose 4} 0.9^4 0.1^1
    • P(X\ge 1)= 1-P(X=0)= 1- 0.1^4
    • b. P(X = 0) = 1 − 0.999 = .001

0.001 = 0.1n n = 3

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