Difference between revisions of "SOCR EduMaterials Activities Discrete Probability examples"

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**a. <math>X \sim b(5,0.90). P(X=4)= {5 \choose 4} 0.9^4 0.1^1=0.32805</math>
 
**a. <math>X \sim b(5,0.90). P(X=4)= {5 \choose 4} 0.9^4 0.1^1=0.32805</math>
 
**<math>P(X\ge 1)= 1-P(X=0)= 1- 0.1^4=0.9999</math>
 
**<math>P(X\ge 1)= 1-P(X=0)= 1- 0.1^4=0.9999</math>
**b. <math>P(X=0)=.001,
+
**b. P(X=0)=.001,
0.001= 0.1^n</math> <math>n= 3</math>
+
0.001= 0.1^n
 +
n= 3

Revision as of 15:07, 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:

SOCR Activities Binomial Christou example1.jpg


  • 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:

\( E(X)=\facr{1}{p}=\frac{1}{0.30}=3.33\)

  • 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