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# 2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math 2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math Conceptual Understanding.

## Presentation on theme: "2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math 2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math Conceptual Understanding."— Presentation transcript:

2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math 2010-2011 Workshop Series for YISD Teachers of 6 th Grade Math Conceptual Understanding & Mathematical Thinking Workshop 4: Sample Space and Probability (Feb 28, 2011) Conceptual Understanding & Mathematical Thinking Workshop 4: Sample Space and Probability (Feb 28, 2011) Kien Lim Kien Lim Dept. of Mathematical Sciences, UTEP

Item 0 In how many ways can you form a 3-digit even number with the second digit being a prime number? Note: 058 is considered a 2-digit even number.

What key ideas should student learn in order to truly understand fundamental counting principle? Multiplication as systematic exhaustion without duplication (counting all without double counting) Representational tools to generate all cases systematically o A tree diagram o An exhaustive list of outcomes What is the fundamental counting principle?

How can we create learning opportunity for students to experience the need for the fundamental counting principle?

Consider this problem: Bobbie Bear is planning a vacation. With 3 colored shirts and 2 colored pants, how many outfits can he make? How can we create learning opportunity for students to experience the need for the fundamental counting principle?

Consider this problem: Bobbie Bear is planning a vacation. With 3 colored shirts and 2 colored pants, how many outfits can he make?

http://illuminations.nctm.org/ActivityDetail.aspx?ID=3 Consider this problem: Bobbie Bear is planning a vacation. With 3 colored shirts and 2 colored pants, how many outfits can he make?

How can we create learning opportunity for students to experience the need for the fundamental counting principle? Concrete manipulatives to pictorial representation to formal rule Ample time to explore Opportunity to experience inefficiency (students appreciate the efficiency of a conceptual tool only when they experience the hardship without it)

Logging In Procedure 1. Turn-on your clicker 2. Wait until it says Enter Student ID (Enter your 5-digit ID) 3. The screen should display ANS

Item 1 Consider a standard 52-card deck, with four suits (,,, ), 13 cards per suit (2-10, J, Q, K, A). Define an event space on the standard deck such that it consists of 52 simple outcomes, one for each card in the deck. Which of the following is a true statement? (A) {Black} is not an event. (B) {Black} is an event with 1 simple outcome. (C) {Black} is an event with 26 simple outcomes. (D) {Black} is an event with 52 simple outcomes. (E) None of the above is true.

What is the difference among An outcome An event A sample space is the result of an action. is an outcome or a group of outcomes. is the set of all possible outcomes.

The sample space has 52 simple outcomes. {Black} is an event with 26 simple outcomes.

Item 2 Consider a standard 52-card deck, with four suits (,,, ), 13 cards per suit (2-10, J, Q, K, A). Define an event space on the standard deck such that it consists of two outcomes: Black and Red. Which of the following is a true statement? (A) {Black} is not an event. (B) {Black} is an event with 1 simple outcome. (C) {Black} is an event with 26 simple outcomes. (D) {Black} is an event with 52 simple outcomes. (E) None of the above is true.

The sample space has 52 simple outcomes. {Black} is an event with 1 simple outcome. The event space {Black, Red} has 2 outcomes.

Item 3 Consider a standard 52-card deck, with four suits (,,, ), 13 cards per suit (2-10, J, Q, K, A). Define an event space on the standard deck such that it contains, as simple outcomes, only the cards that are hearts or diamonds. Which of the following is a true statement? (A) {Black} is not an event. (B) {Black} is an event with 1 simple outcome. (C) {Black} is an event with 26 simple outcomes. (D) {Black} is an event with 52 simple outcomes. (E) None of the above is true.

The sample space has 52 simple outcomes. {Black} is an event with 0 simple outcome. The event space {Heart, Diamond} has 2 outcomes.

Create probability question involving a bear who has 3 shirts and 2 pants for students to understand the difference among An outcome An event A sample space

What is the probability that his outfit will have a green shirt? (A)1/2 (B)1/3 (C)1/4 (D)1/6 (E)1/8 Item 4 Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will put them back and randomly choose again.

Sample spaceSample space consists of ___ possible simple outcomes. Prob(green shirt) = Number of favorable outcomes Number of possible outcomes 6 2 Favorable event (green shirt) consists of ___ simple outcomes. = 2 6 = 1 3

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the shirt and choose the other pair of pants. What is the probability that his outfit will have a green shirt? (A)1/2 (B)1/3 (C)1/4 (D)1/6 (E)1/8 Item 5

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the shirt and choose the other pair of pants. Prob(green shirt) = Number of favorable outcomes Number of possible outcomes = 2 6 = 1 3 Is this correct?

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the shirt and choose the other pair of pants. Prob(green shirt) = Is this correct? 2 6

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the shirt and choose the other pair of pants. Prob(green shirt) = So is 1/4 or is it 1/3? 1414 1414 1414 1414 1 4

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the shirt and choose the other pair of pants. Prob(green shirt) = or 1212 1212 1414 1414 1414 1414 1212 1212 1 0 1 0 1 4 1 3 1818 1818 1818 1818 0 1414 1414 0 Prob(green shirt) 1 2 1 4 1 2 1 4 = + 1 8 1 8 = 1 4

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the pants and randomly choose another shirt. What is the probability that his outfit will have a red shirt? (A)1/2 (B)1/3 (C)1/4 (D)1/6 (E)1/8 Item 6

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the pants and randomly choose another shirt. Prob(red shirt) = Number of favorable outcomes Number of possible outcomes Is this correct? = 1 6

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the pants and randomly choose another shirt. Prob(red shirt) = Number of favorable outcomes Number of possible outcomes = 1 6 Is this correct? 1212 1212 1414 1414 1414 1414 1414 1414 1414 1414

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the pants and randomly choose another shirt. Prob(red shirt) = Is the answer 1/8 or 1/6? 1212 1212 1414 1414 1414 1414 1414 1414 1414 1414 1 4 1 2 = 1 8 But the sum of six 1/8s is only 6/8, not 1! 1818 1818 1818 1818 1818 1818

Bobbie Bear has 4 colored shirts and 2 colored pants. He randomly chooses a shirt and a pair of pants. If the shirt and pants have the same color, he will keep the pants and randomly choose another shirt. Prob(red shirt) = 1212 1212 1414 1414 1414 1414 1414 1414 1414 1414 1 4 1 2 1818 1818 1818 1818 1818 1818 1313 1313 1313 1313 1313 1313 1 4 1 2 1 3 + 1 24 + 1 = 1 8 = 3 + 1 24 = 1 6 What have we learned? The usefulness of a tree diagram.

Item 7 A fair die is rolled three times and the results are recorded in the order that they appear. For each roll, the die lands with the number 1, 2, 3, 4, 5, or 6 facing up. The following outcome is the LEAST LIKELY to occur: (A) 2 4 2 (B) 3 4 5 (C) 5 5 5 (D) 6 4 1 (E) All of the above sequences are equally likely

Item 8 Two fair dice are rolled and the resulting numbers facing upwards on the dice is summed. The following sum is the MOST LIKELY to occur: (A) Sum = 3 (B) Sum = 5 (C) Sum = 6 (D) Sum = 9 (E) All of the above sums are equally likely

(A) Sum = 3 (B) Sum = 5 (C) Sum = 6 (D) Sum = 9 (E) All of the above sums are equally likely P( Sum = 3 ) = P( Sum = 5 ) = P( Sum = 6 ) = P( Sum = 9 ) = 2 = 36 1 18 4 = 36 1 9 5 36 4 = 36 1 9 Source: Tami Dashley

Item 9 Suppose a family is randomly selected from among all families with 3 children. What is the probability that the family has exactly one boy? You may assume that P(boy) = P(girl) for each birth. (A) 1/2 (B) 1/3 (C) 1/6 (D) 1/8 (E) 3/8

Why Clickers and Voting? Requires students to participate actively Requires students to participate actively (Cline, Zullo, & Parker, 2006) Provides immediate feedback Provides immediate feedback Facilitates class discussion/debate Facilitates class discussion/debate Creates a fun atmosphere Creates a fun atmosphere

Cline, K. S. (2006). Classroom voting in mathematics. Mathematics Teacher, 100 (2). pp. 100-104.

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