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

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**a. <math>X \sim b(5,0.90) </math>, <math> P(X=4)= {5 \choose 4} 0.9^4 0.1^1</math>
 
**a. <math>X \sim b(5,0.90) </math>, <math> P(X=4)= {5 \choose 4} 0.9^4 0.1^1</math>
 
**<math>P(X\ge 1)= 1-P(X=0)= 1- 0.1^4 </math>
 
**<math>P(X\ge 1)= 1-P(X=0)= 1- 0.1^4 </math>
**b. <math>P(X=0)= 1-0.999= .001
+
**b. <math>P(X=0)= 1-0.999= .001</math>
0.001= 0.1^n
+
<math>0.001= 0.1^n</math>
 
n= 3</math>
 
n= 3</math>
 
= {4 \choose 2} 0.8^2 0.2^2</math>
 

Revision as of 16:39, 23 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 ptobability 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:
  • \( X \sim b(8,0.3) \), \( P(X=3)= {8 \choose 3} 0.3^30.7^5\)
  • \( P(X \ge 1)=1-P(X=0)=1-.7^8 \)
  • \( E(X),\)= \(np\) = \(8*.3=2.4 \)
  • \( Sd(X)= sqrt(npq)\)
  • 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:
  • \( 0.50^5 \)
  • Example 3:
    • a. \(X \sim b(4,0.8) \), \( P(X=2)= {4 \choose 2} 0.8^2 0.2^2\)
    • b. \(P(X\ge 2)= {4 \choose 2} 0.8^2 0.2^2\)+\({4 \choose 3} 0.8^3 0.2^1\)+\({4 \choose 4} 0.8^4\)
  • Example 4:
  • \(0.7^4 0.3\)
  • Example 5:

\( 1/0.30 \)

  • 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. \(P(X=0)= 1-0.999= .001\)

\(0.001= 0.1^n\) n= 3</math>