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Algebra I Concept Test # 17 – Probability Practice Test 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If.

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Presentation on theme: "Algebra I Concept Test # 17 – Probability Practice Test 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If."— Presentation transcript:

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2 Algebra I Concept Test # 17 – Probability Practice Test 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If a marble is randomly chosen from the bag, find the following probabilities: a)P(red or purple) Note: Create a fraction with the numerator as your desired result over all possible results. P(red or purple) 14 28 = Reduce = 1 2 b)P(not blue) P(not blue) 24 28 = Reduce = 6 7 © 2007- 09 by S-Squared, Inc. All Rights Reserved.

3 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If a marble is randomly chosen from the bag, find the following probabilities: c)If two marbles are randomly chosen from the bag without replacement, find the probability of drawing a green marble and then another green marble. P(green and green) 10 28 = Reduce = 5 42 9 27 5 14 = 1 3 Multiply Algebra I Concept Test # 17 – Probability Practice Test

4 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If a marble is randomly chosen from the bag, find the following probabilities: d)If two marbles are randomly chosen from the bag with replacement, find the probability of drawing a blue marble and then a purple marble. P(blue and purple) 4 28 = Reduce = 3 98 6 28 1 7 = 3 14 Multiply Algebra I Concept Test # 17 – Probability Practice Test

5 1.In a bag, there are 8 red marbles, 4 blue marbles, 10 green marbles and 6 purple marbles. If a marble is randomly chosen from the bag, find the following probabilities: e)Determine if the event from part (c) is “dependent” or “independent.” Dependent Algebra I Concept Test # 17 – Probability Practice Test

6 2. A fair coin is flipped and a card is drawn from a standard 52-card deck. What is the probability the result is TAILS and a Heart? Note: Find the probability of each independent event and multiply. P(tails and heart) 1 2 = 13 52 Reduce = 1 8 1 2 = 1 4 Multiply

7 2. Using the following information: Algebra I Concept Test # 17 – Probability Practice Test Pants (one of each) Shirts (one of each) Shoes (one of each) Blue JeansWhiteDress Shoes Black SlacksBlackAthletic Shoes Blue SlacksBrown a)How many possible outfits can be put together. Note: Multiply the number of items in each category together to find the total number of possible outfits. 3 2 3 = 18

8 2. Using the following information: Algebra I Concept Test # 17 – Probability Practice Test Pants (one of each) Shirts (one of each) Shoes (one of each) Blue JeansWhiteDress Shoes Black SlacksBlackAthletic Shoes Blue SlacksBrown b)What is the probability the outfit will include a black shirt?. Note: Refer to the shirts. Build a fraction with the numerator representing the number of black shirts available over the total number of shirts. 1 3

9 a)Identify the sample space. Note: The sample space represents all possible outcomes. A number die has eight sides. { 1, 2, 3, 4, 5, 6, 7, 8 } Algebra I Concept Test # 17 – Probability Practice Test RollOutcomeRollOutcome 1 st 27 th 4 2 nd 48 th 5 3 rd 89 th 3 4 th 410 th 1 5 th 311 th 7 6th512 th 8 4.An eight-sided die was rolled twelve times. The results are recorded in the table below:

10 b)What is the theoretical probability of rolling a three? Theoretical probability = favorable outcomes total possible outcomes 1 8 Note: On a number die, there is one face with a three on it and eight faces total. Algebra I Concept Test # 17 – Probability Practice Test RollOutcomeRollOutcome 1 st 27 th 4 2 nd 48 th 5 3 rd 89 th 3 4 th 410 th 1 5 th 311 th 7 6th512 th 8 4.An eight-sided die was rolled twelve times. The results are recorded in the table below:

11 RollOutcomeRollOutcome 1 st 27 th 4 2 nd 48 th 5 3 rd 89 th 3 4 th 410 th 1 5 th 311 th 7 6th512 th 8 c)What is the experimental probability of rolling a three? 4.A six sided die was rolled ten times. The results are recorded in the table below: Experimental probability = outcomes observed total trials 2 12 Note: In the table, two out of the twelve rolls were a three. Algebra I Concept Test # 17 – Probability Practice Test = 1 6 Reduce

12 RollOutcomeRollOutcome 1 st 27 th 4 2 nd 48 th 5 3 rd 89 th 3 4 th 410 th 1 5 th 311 th 7 6th512 th 8 d)What is the experimental probability of rolling a number greater than six? 2.A six sided die was rolled ten times. The results are recorded in the table below: Experimental probability = outcomes observed total trials 3 12 Reduce = 1 4 Algebra I Concept Test # 17 – Probability Practice Test

13 1 16 5.A circular target for playing darts has a radius of 12 cm. Each of the three outer sections within the target has a width of 3 cm. The circle in the center has a radius of 3cm. (See figure below.) a)What is the probability the dart will hit within the center circle? Algebra I Concept Test # 17 – Probability Practice Test 12 cm probability = Area of region to hit Area of the whole target P(center) = Formula Area = πr 2 π (3) 2 π (12) 2 = Simplify π (9) π (144) Reduce

14 (π (12) 2 – π (9) 2 ) + ( π (6) 2 – π (3) 2 ) π (12) 2 5.A circular target for playing darts has a radius of 12 cm. Each of the three outer sections within the target has a width of 3 cm. The circle in the center has a radius of 3cm. (See figure below.) b)What is the probability the dart will hit the shaded area? Algebra I Concept Test # 17 – Probability Practice Test 12 cm probability = Area of region to hit Area of the whole target P(shaded area) = Formula Area = πr 2

15 (π (12) 2 – π (9) 2 ) + ( π (6) 2 – π (3) 2 ) π (12) 2 5.A circular target for playing darts has a radius of 12 cm. Each of the three outer sections within the target has a width of 3 cm. The circle in the center has a radius of 3cm. (See figure below.) b)What is the probability the dart will hit the shaded area? Algebra I Concept Test # 17 – Probability Practice Test 12 cm P(shaded area) = Simplify = ( 144 π – 81 π ) + (36 π – 9 π ) 144 π = ( 63 π ) + (27 π ) 144 π = 90 π 144 π 5 8 Reduce = Focus on your goals


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