AP Statistics Curriculum 2007 Prob Rules

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(Example - Monty Hall Problem)
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===Example - Monty Hall Problem===
===Example - Monty Hall Problem===
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Recall that earlier we discussed the [[AP_Statistics_Curriculum_2007_Prob_Basics#Hands-on_activities Monty Hall Experiment]]. We will now show why the odds of winning double if we use the swap strategy - that is the probability of a win is 2/3, if each time we switch and choose the last third card.
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Recall that earlier we discussed the [[AP_Statistics_Curriculum_2007_Prob_Basics#Hands-on_activities | Monty Hall Experiment]]. We will now show why the odds of winning double if we use the swap strategy - that is the probability of a win is 2/3, if each time we switch and choose the last third card.
Denote W={Final Win of the Car Price}. Let L<sub>1</sub> and W<sub>2</sub> represent the events of choosing the donkey (loosing) and the car (winning) at the player's first and second choice, respectively. Then, the chance of winning in the swapping-strategy case is:
Denote W={Final Win of the Car Price}. Let L<sub>1</sub> and W<sub>2</sub> represent the events of choosing the donkey (loosing) and the car (winning) at the player's first and second choice, respectively. Then, the chance of winning in the swapping-strategy case is:

Revision as of 04:54, 29 January 2008

Contents

General Advance-Placement (AP) Statistics Curriculum - Probability Theory Rules

Addition Rule

The probability of a union, also called the Inclusion-Exclusion principle allows us to compute probabilities of composite events represented as unions (i.e., sums) of simpler events.

For events A1, ..., An in a probability space (S,P), the probability of the union for n=2 is

P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2),

For n=3,

P(A_1\cup A_2\cup A_3)=P(A_1)+P(A_2)+P(A_3) -P(A_1\cap A_2)-P(A_1\cap A_3)-P(A_2\cap A_3)+P(A_1\cap A_2\cap A_3)

In general, for any n,

P(\bigcup_{i=1}^n A_i) =\sum_{i=1}^n P(A_i)
-\sum_{i,j\,:\,i<j}P(A_i\cap A_j) +\sum_{i,j,k\,:\,i<j<k}P(A_i\cap A_j\cap A_k)+ \cdots\cdots\ +(-1)^{j+1} P(\bigcap_{i=1}^j A_i)+ \cdots\cdots\  +(-1)^{n+1} P(\bigcap_{i=1}^n A_i).


Conditional Probability

The conditional probability of A occurring given that B occurs is given by

P(A | B) ={P(A \cap B) \over P(B)}.

Example

Here is the data on 400 Melanoma (skin cancer) Patients by Type and Site

Site
Type Head and Neck Trunk Extremities Totals
Hutchinson's melanomic freckle 22 2 10 34
Superficial 16 54 115 185
Nodular 19 33 73 125
Indeterminant 11 17 28 56
Column Totals 68 106 226 400
  • Suppose we select one out of the 400 patients in the study and we want to find the probability that the cancer is on the extremities given that it is of type nodular: P = 73/125 = P(Extremities | Nodular)
  • What is the probability that for a randomly chosen patient the cancer type is Superficial given that it appears on the Trunk?

Example - Monty Hall Problem

Recall that earlier we discussed the Monty Hall Experiment. We will now show why the odds of winning double if we use the swap strategy - that is the probability of a win is 2/3, if each time we switch and choose the last third card.

Denote W={Final Win of the Car Price}. Let L1 and W2 represent the events of choosing the donkey (loosing) and the car (winning) at the player's first and second choice, respectively. Then, the chance of winning in the swapping-strategy case is: P(W) = P(L_1 \cap W_2) = P(W_2 | L_1) P(L_1) = 1 \times {2\over 3} ={2\over 3}. If we played using the stay-home strategy, our chance of winning would have been: P(W) = P(W_1 \cap W_2) = P(W_2 | W_1) P(W_1) = 1 \times {1\over 3} ={1\over 3}, or half the chance in the first (swapping) case.

Multiplication Rule

Model Validation

Checking/affirming underlying assumptions.

  • TBD

Computational Resources: Internet-based SOCR Tools

  • TBD

Examples

Computer simulations and real observed data.

  • TBD

Hands-on activities

Step-by-step practice problems.

  • TBD

References

  • TBD



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