SOCR EduMaterials Activities Discrete Probability examples

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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 SOCR snapshots 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 complex electronic system is built with a certain number of backup components in its subsystem. One subsystem has 4 identical components, each with probability of 0.20 of failing in less than 1000 hours. The subsystem will operate if at least 2 of the 4 components are operating. Assume the components operate independently.

  • Answer:
    • a. Find the probability that exactly 2 of the 4 components last longer than 1000 hours.
    • b. Find the probability that the subsystem operates longer than 1000 hours.
    • 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 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 = 0.1nn = 3.
  • In the first snapshot below, wheren=2, P(X\ge1)=0.99, which is too small. In the second snapshot, we can see that when n is increased to 3, P(X\ge1) increases to nearly 1.
  • Example 7:

Construct a probability histogram for the binomial probability distribution for each of the following: n=5,p=0.1, n=5,p=0.5, m=5,n=0.9. What do you observe? Explain.

  • Answer:

We observe that if p=0.5 the distribution resembles the normal distribution, with mean np = 0.25. Values above and below the mean are distributed symmetrically around the mean. Also, the probability histograms for p=0.1 and p=0.9 are mirror images of each other.

  • Example 8:

On a population of consumers, 60% prefer a certain brand of ice cream. If consumers are randomly selected,

    • a. what is the probability that exactly 3 people have to be interviewed to encounter the first consumer who prefers this brand of ice cream?
    • b. what is the probability that at least 3 people have to be interviewed to encounter the first consumer who prefers this brand of ice cream?
  • Answer:
    • a. P(X=3)=0.4 \times 0.4 \times 0.6=0.096
    • b. Let's first find the probability that at least 3 people will NOT have to be interviewed to encounter the first case.

P(X=1) + P(X=2) = 0.6+0.4 \times 0.6 = 0.84

  • Now we subtract this from 1 to find the complement, which is the event that at least 3 people WILL have to be interviewed to find the first case:

1 − 0.84 = 0.16

  • Example 9:

The alpha marketing research company employs consumer panels to explore preferences

  • Answer:
    • a. X \sim b(5,0.4) , P(X=5)= {5 \choose 5} 0.4^5 0.6^0=0.01024
    • b. In this part we will look at every group of 5 people as one unit. We will denote success as the event that all 5 people in a group prefer the new product, and failure as the event that at least one person in a group does not prefer the new product. Therefore, p = 0.01024. So,

P(X = 60) = (1 − .01024)60 = 53.9% Below you can see a SOCR snapshot for this example:

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