AP Statistics Curriculum 2007 Prob Rules - Revision history

Dinov at 13:18, 10 February 2026

2026-02-10T13:18:42Z

Tdlee: Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

2020-03-03T19:23:28Z

Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

Dinov: /* Monty Hall Problem */

2014-09-01T15:32:19Z

‎Monty Hall Problem

IvoDinov: /* Card experiment */

2013-04-19T15:46:41Z

‎Card experiment

IvoDinov: /* Card experiment */

2013-04-19T15:45:47Z

‎Card experiment

IvoDinov: /* Card experiment */

2013-04-19T15:45:34Z

‎Card experiment

IvoDinov: /* Addition Rule */

2013-04-17T15:58:46Z

‎Addition Rule

IvoDinov: /* Addition Rule */

2013-04-17T15:58:07Z

‎Addition Rule

IvoDinov: /* Addition Rule */

2013-04-17T15:57:31Z

‎Addition Rule

IvoDinov: /* Addition Rule */

2013-04-17T15:57:20Z

‎Addition Rule

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 [http://en.wikipedia.org/wiki/Color_blindness Color blindness] is a sex-linked trait, as many of the genes involved in color vision are on the [http://en.wikipedia.org/wiki/X_chromosome X chromosome]. Color blindness is more common in males than in females, as men do not have a second X chromosome to overwrite the chromosome which carries the mutation. If 8% of variants of a given gene are defective (mutated), the probability of a single copy being defective is 8%, but the probability that two (independent) copies are both defective is 0.08 × 0.08 = 0.0064.
 * The events A={Female} and B={Color blind} are not mutually exclusive (females can be color blind), nor independent (the rate of color blindness among females is lower). Color blindness prevalence within the 2 genders is P(CB|Male) = 0.08, and P(CB|Female)=0.005, where CB={color blind, one color, a color combination, or another mutation}.
-* See the [http://distributome.org/blog/?p=67 Distributome Colorblindness activity].
+* See the [https://distributome.org/ Distributome Colorblindness activity].
 ===Example===
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 <hr>
-* SOCR Home page: http://www.socr.ucla.edu
+* SOCR Home page: https://socr.ucla.edu
-"{{translate|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_Prob_Rules}}
+{{translate|pageName=https://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_Prob_Rules}}

← Older revision		Revision as of 19:23, 3 March 2020
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	* SOCR Home page: http://www.socr.ucla.edu		* SOCR Home page: http://www.socr.ucla.edu

−	{{translate\|pageName=http://wiki.~~stat~~.~~ucla~~.edu~~/socr~~/index.php?title=AP_Statistics_Curriculum_2007_Prob_Rules}}	+	"{{translate\|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_Prob_Rules}}

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 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:
-<math>P(W) = P(L_1 \cap W_2) = P(W_2 | L_1) P(L_1) = 1 \times {2\over 3} ={2\over 3}</math>. If we played using the [[AP_Statistics_Curriculum_2007_Prob_Basics#Hands-on_activities | stay-home strategy]], our chance of winning would have been:
+<math>P(W) = P(L_1 \cap W_2) = P(W_2 | L_1) P(L_1) = 1 \times {2\over 3} ={2\over 3}.</math> If we played using the [[AP_Statistics_Curriculum_2007_Prob_Basics#Hands-on_activities | stay-home strategy]], our chance of winning would have been:
-<math>P(W) = P(W_1 \cap W_2) = P(W_2 | W_1) P(W_1) = 1 \times {1\over 3} ={1\over 3}</math>, or half the chance in the first (swapping) case.
+<math>P(W) = P(W_1 \cap W_2) = P(W_2 | W_1) P(W_1) = 1 \times {1\over 3} ={1\over 3},</math> or half the chance in the first (swapping) case.
 ====Drawing balls without replacement====

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 Mutual-exclusiveness and independence are different concepts. Here are two examples clarifying the differences between these concepts:
 ====Card experiment====
-Suppose we play a [[SOCR_EduMaterials_Activities_CardExperiment | card game]] of ''guessing the color of a randomly drawn card'' from a [[AP_Statistics_Curriculum_2007_Prob_Simul#Poker_Game |standard 52-card deck]]. As there are 2 possible colors (black \(\clubsuit, \spadesuit\) and red \(\heartsuit, \diamondsuit\)), and given no other information, the chance of correctly guessing the color (e.g., black) is 0.5. However, additional information may or may not be helpful in identifying the card color. For example:
+Suppose we play a [[SOCR_EduMaterials_Activities_CardExperiment | card game]] of ''guessing the color of a randomly drawn card'' from a [[AP_Statistics_Curriculum_2007_Prob_Simul#Poker_Game |standard 52-card deck]]. As there are 2 possible colors (black \(\clubsuit, \spadesuit\) and red \(\color{red}\heartsuit, \color{red}\diamondsuit\)), and given no other information, the chance of correctly guessing the color (e.g., black) is 0.5. However, additional information may or may not be helpful in identifying the card color. For example:
 * If we know that the [[AP_Statistics_Curriculum_2007_Prob_Count#Hands-on_combination_activity |card denomination]] is a king, and there are 2 red and 2 black kings, this does '''not''' help us improve our chances of successfully identifying the correct color of the card, P(Red|King)=P(Red), independence.
 * If we know that the suit of the card is heart, this does help us with correctly identifying the card color (as heart is red), P(Red|Hearts)=1.0, strong dependence as P(Red)=0.5.

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 Mutual-exclusiveness and independence are different concepts. Here are two examples clarifying the differences between these concepts:
 ====Card experiment====
-Suppose we play a [[SOCR_EduMaterials_Activities_CardExperiment | card game]] of ''guessing the color of a randomly drawn card'' from a [[AP_Statistics_Curriculum_2007_Prob_Simul#Poker_Game |standard 52-card deck]]. As there are 2 possible colors (black \(\clubsuit, \spadesuit\) and red \(heartsuit, \diamondsuit\)), and given no other information, the chance of correctly guessing the color (e.g., black) is 0.5. However, additional information may or may not be helpful in identifying the card color. For example:
+Suppose we play a [[SOCR_EduMaterials_Activities_CardExperiment | card game]] of ''guessing the color of a randomly drawn card'' from a [[AP_Statistics_Curriculum_2007_Prob_Simul#Poker_Game |standard 52-card deck]]. As there are 2 possible colors (black \(\clubsuit, \spadesuit\) and red \(\heartsuit, \diamondsuit\)), and given no other information, the chance of correctly guessing the color (e.g., black) is 0.5. However, additional information may or may not be helpful in identifying the card color. For example:
 * If we know that the [[AP_Statistics_Curriculum_2007_Prob_Count#Hands-on_combination_activity |card denomination]] is a king, and there are 2 red and 2 black kings, this does '''not''' help us improve our chances of successfully identifying the correct color of the card, P(Red|King)=P(Red), independence.
 * If we know that the suit of the card is heart, this does help us with correctly identifying the card color (as heart is red), P(Red|Hearts)=1.0, strong dependence as P(Red)=0.5.

← Older revision		Revision as of 15:58, 17 April 2013
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	=== Addition Rule===		=== Addition Rule===
−	Let's start with an example. Suppose the chance of colder weather (C) is 30%, chance of rain (R) and colder weather (C) is 15% and the chance of rain or colder weather is 75%. What is the chance of rain. This can be computed by observing that \(P(R) - P(R ~~\cup C~~) + P(C) = P(R \cap C)\). Thus, \(P(R) = P(R \cup C) − P(C) + P(R \cap C)= 0.75 − 0.3 + 0.15 = 0.6\). How can this observation be generalized?	+	Let's start with an example. Suppose the chance of colder weather (C) is 30%, chance of rain (R) and colder weather (C) is 15% and the chance of rain or colder weather is 75%. What is the chance of rain. This can be computed by observing that \(P(R \cup C) = P(R) + P(C) - P(R \cap C)\). Thus, \(P(R) = P(R \cup C) − P(C) + P(R \cap C)= 0.75 − 0.3 + 0.15 = 0.6\). How can this observation be generalized?

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 === Addition Rule===
 Let's start with an example. Suppose the chance of colder weather (C) is 30%, chance of rain (R) and colder weather (C) is 15% and the chance of rain or colder weather is 75%. What is the chance of rain. This can be computed by observing that \(P(R) - Pr(R \cup C) + Pr(C) = Pr(R \cap C)\). Thus, \(P(R) = P(R \cup C) − P(C) + P(R \cap C)= 0.75 − 0.3 + 0.15 = 0.6\). How can this observation be generalized?
 The probability of a union, also called the [http://en.wikipedia.org/wiki/Inclusion-exclusion_principle Inclusion-Exclusion principle] allows us to compute probabilities of composite events represented as unions (i.e., sums) of simpler events.