SOCR EduMaterials Activities Discrete Probability examples
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*<math> Y \sim b(4,0.9) </math>, <math> P(Y=4)= {4 \choose 4} 0.9^4 0.1^0= 0.6561</math> | *<math> Y \sim b(4,0.9) </math>, <math> P(Y=4)= {4 \choose 4} 0.9^4 0.1^0= 0.6561</math> | ||
Therefore, <math>P(shipment=A)=0.6561+ (0.2916)(0.6561)=0.8474 | Therefore, <math>P(shipment=A)=0.6561+ (0.2916)(0.6561)=0.8474 | ||
+ | <center>[[Image: SOCR_Activities_Binomial_Christou_example10_a.jpg|600px]]</center> | ||
*b.<math>P(Shipment=C)=</math>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. | *b.<math>P(Shipment=C)=</math>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. | ||
*<math> X \sim b(4,0.9) </math>, <math> P(X=3)= {4 \choose 3} 0.9^3 0.1^1= 0.2916</math> | *<math> X \sim b(4,0.9) </math>, <math> P(X=3)= {4 \choose 3} 0.9^3 0.1^1= 0.2916</math> | ||
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*<math>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</math> | *<math>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</math> | ||
Therefore P(shipment=C)=<math>(0.2916 \times 0.001)+.0523 = 0.052</math> | Therefore P(shipment=C)=<math>(0.2916 \times 0.001)+.0523 = 0.052</math> | ||
+ | <center>[[Image: SOCR_Activities_Binomial_Christou_example10_b.jpg|600px]]</center> | ||
* '''Example 11:''' | * '''Example 11:''' | ||
* '''Answer:''' | * '''Answer:''' | ||
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*If n=3, P(functioning when n=3)=<math>P(X\ge 2). </math> | *If n=3, P(functioning when n=3)=<math>P(X\ge 2). </math> | ||
<math>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</math> | <math>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</math> | ||
+ | <center>[[Image: SOCR_Activities_Binomial_Christou_example11_n=3.jpg|600px]]</center> | ||
*If n=5, P(functioning when n=5)=<math>P(X\ge 3). </math> | *If n=5, P(functioning when n=5)=<math>P(X\ge 3). </math> | ||
<math>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</math> | <math>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</math> | ||
+ | <center>[[Image: SOCR_Activities_Binomial_Christou_example11_n=5.jpg|600px]]</center> | ||
*Therefore it is better to have 3 components, because we will have a better chance of having a system that operates. | *Therefore it is better to have 3 components, because we will have a better chance of having a system that operates. | ||
*b. Again, <math>X \sim b(200,0.40).</math> mean=expected value = <math>np=200 \times 0.4 = 80; SD=\sqrt{npq}=\sqrt{200 \times 0.4 \times 0.5}=6.3245</math> | *b. Again, <math>X \sim b(200,0.40).</math> mean=expected value = <math>np=200 \times 0.4 = 80; SD=\sqrt{npq}=\sqrt{200 \times 0.4 \times 0.5}=6.3245</math> | ||
+ | <center>[[Image: SOCR_Activities_Binomial_Christou_example11_b.jpg|600px]]</center> |
Revision as of 02:20, 30 April 2007
- Description: You can access the applets for the distributions at http://www.socr.ucla.edu/htmls/SOCR_Distributions.html .
- Example 1:
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:
- ,
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. ,
- b.
- Example 4:
- P(X = 5) = 0.740.3 = 0.07203 where X represents the number of trials.
- Example 5:
- Example 6:
- a.
- b. P(X = 0) = .001,0.001 = 0.1nn = 3.
- In the first snapshot below, where which is too small. In the second snapshot, we can see that when n is increased to 3, 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:
- 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. ,
- 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:
- 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)=
- ,
- ,
- ,
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.
- ,
- ,
Therefore P(Y = 0or1) = 0.0001 + .0009 = 0.001.
Therefore P(shipment=C)=
- 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)=
- If n=5, P(functioning when n=5)=
- Therefore it is better to have 3 components, because we will have a better chance of having a system that operates.
- b. Again, mean=expected value =