SOCR EduMaterials Activities Discrete Probability examples

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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.
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:'''
* '''Answer:'''
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*<math> X \sim geometric(0.5) </math>
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*<math> X \sim geometric(0.5) </math><math>P(X=5)= 0.50^5=0.03125</math>
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*<math>P(X=5)= 0.50^5=0.03125</math>
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<center>[[Image: SOCR_Activities_Binomial_Christou_example2.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Binomial_Christou_example2.jpg|600px]]</center>
* '''Example 3:'''
* '''Example 3:'''

Revision as of 21:39, 30 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 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:
  •  X \sim geometric(0.5) 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:
  •  X \sim geometric(0.3) 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.  X \sim geometric(0.6) 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,

 X \sim geometric(0.01024) P(X = 60) = (1 − .01024)60 = 53.9%

  • Below you can see a SOCR snapshot for this example:
  • Example 10:
  • Answer:
  • a. We will denote the number of unbruised peaches in our first selection X, and the number of unbruised peaches in our second selection Y. P(shipment = A) = P(all four peaches are unbruised)+P(three of the peaches are unbruised) X P(four additional peaches are unbruised)= P(X=4)+P(X=3)\times P(Y=4).
  •  X \sim b(4,0.9) ,  P(X=4)= {4 \choose 4} 0.9^4 0.1^0= 0.6561
  •  Y \sim b(4,0.9) ,  P(Y=3)= {4 \choose 3} 0.9^3 0.1^1= 0.2916
  •  Y \sim b(4,0.9) ,  P(Y=4)= {4 \choose 4} 0.9^4 0.1^0= 0.6561

Therefore, P(shipment = A) = 0.6561 + (0.2916)(0.6561) = 0.8474 < center > [[Image:SOCRActivitiesBinomialChristouexample10a.jpg | 600px]] < / center > * b. < math > P(Shipment = C) = P(three of the first four peaches are unbruised) X P(zero or one of the additional four peaches are unbruised) +P(two or fewer of the first four peaches are unbruised). The first selection of peaches = X. The second selection=Y.

  •  X \sim b(4,0.9) ,  P(X=3)= {4 \choose 3} 0.9^3 0.1^1= 0.2916
  •  Y \sim b(4,0.9) ,  P(Y=0)= {4 \choose 0} 0.9^0 0.1^4= 0.0001
  • P(Y=1)={4 \choose 1} 0.9^1 0.1^3=0.0009

Therefore P(Y = 0or1) = 0.0001 + .0009 = 0.001.

  • P(X\le2)=P(X=0)+P(X=1)+P(X=2)= {4 \choose 0} 0.9^0 0.1^4 + {4 \choose 1} 0.9^1 0.1^3+{4 \choose 2} 0.9^2 0.1^2= 0.0523

Therefore P(shipment=C)=(0.2916 \times 0.001)+.0523 = 0.052

File:SOCR Activities Binomial Christou example10 b.jpg
  • Example 11:
  • Answer:
  • a. We want our system to function, so we would prefer a system with the highest probability of operating. A system will operate if at least half of its components function.
  • If n=3, P(functioning when n=3)=P(X\ge 2).

X \sim b(3,0.40). P(X \ge2)= P(X=2)+P(X=3)= {3 \choose 2} 0.4^2 0.6^1+{3 \choose 3} 0.4^3 0.6^0= 0.352

  • If n=5, P(functioning when n=5)=P(X\ge 3).

X \sim b(5,0.40). P(X \ge3)= P(X=3)+P(X=4)+P(X=5)= {5 \choose 3} 0.4^3 0.6^2+{5 \choose 4} 0.4^4 0.6^1+{5 \choose 5} 0.4^5 0.6^0=0.31744

  • Therefore it is better to have 3 components, because we will have a better chance of having a system that operates.
  • b. Again, X \sim b(200,0.40). mean=expected value = np=200 \times 0.4 = 80; SD=\sqrt{npq}=\sqrt{200 \times 0.4 \times 0.5}=6.3245
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