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Randomness and Probability

Randomness and Probability. Lesson 10.1. Flip a Coin. In this investigation you will explore the predictability of random outcomes. You will use a familiar random process, the flip of a coin.

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Randomness and Probability

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  1. Randomness and Probability Lesson 10.1

  2. Flip a Coin • In this investigation you will explore the predictability of random outcomes. You will use a familiar random process, the flip of a coin. • Step 1 Imagine you are flipping a fair coin, one that is equally likely to land heads or tails. Without flipping a coin, record a random arrangement of H’s and T’s, as though you were flipping a coin ten times. Label this Sequence A.

  3. Flip a Coin • Pick your favorite number #. Then on a calculator screen of your calculator type randseed # and press ENTER. • Step 2 We’ll use the calculator to flip a coin 10 times. Enter the command RandInt(0,1). Each time you press ENTER you fill flip a coin. • Let 0 = Heads • Let 1 = Tails • Record your ten flips of a coin using the random generator.

  4. Flip a Coin • Step 3 How is Sequence A different from the result of your coin flips? Make at least two observations. • Step 4 Find the longest string of consecutive H’s in Sequence A. Do the same for Sequence B. Then find the second-longest string.

  5. Record the lengths for each person in the class as tally marks in a table like the one below. How do the lengths of the longest strings in Sequence A compare with the lengths of the longest strings in Sequence B?

  6. Step 5 Count the number of H’s in each set. Record the results of the entire class in a table like the one below. What do you notice about the numbers of H’s in Sequence A compared with Sequence B? Step 6 If you were asked to write a new random sequence of H’s and T’s, how would it be different from what you recorded in Sequence A?

  7. Example A • If you were to roll two dice, what is the probability of rolling a sum of 6 on the two dice? • Let’s use a random-number generator to find the probability of rolling a sum of 6 with a pair of dice. This can also be written as P(sum is 6).

  8. Turn your calculator on. On a calculator page pick your favorite number #. Then on a calculator screen type randseed #. • We’ll use a spreadsheet of three columns: • one for the first die: Column 1 enter die1 • one for the second die: Column 2 enter die2 • one for the sum: Column 3 enter dicesum

  9. To simulate 300 throws of a pair of dice • type =randint(1, 6, 300) below die1; press ENTER • Type =randint(1,6,300) below die2; press ENTER • Under dicesum” enter =die1 + die2 ; press ENTER You have just rolled two dice three times and you can see their total.

  10. How many of the 300 rolls came out a sum of 6? • Create a histogram of the data in column C. • Press the home key and select a Data and Statistics page. • On the x-axis select dicesum. Then select menu and choose Histogram.

  11. Using the data in your calculator, what was the Probability of the sum of 6 or P(sum is 6). • Move your cursor to determine the height of bin 6. • For the sample at the right the bin height of the “6” bin is 51. So out of 300 simulated rolls, 51/300 =.17.

  12. Repeat this process 5 times by changing the 300 entries to 1800. • What happened to the number of entries? • What happened to the histogram? • What is P(sum of 6) = when there are 1800 samples • In the sample at the right, P(sum of 6) =236/1800 or about .13

  13. What just took place in Example A is called a simulation. The 1800 rolls of the dice were done electronically. • Simulations can be done for dice, cards, coins, spinners or random number generators.

  14. An event consists of one or more outcomes. • A simple event consists of only one outcome. • Events that aren’t simple are called compound. • You might recall that the probability of an event, such as “the sum of two dice is 6,” must be a number between 0 and 1. • The probability of an event that is certain to happen is 1. • The probability of an impossible event is 0. • The solution for Example A showed that P(sum is 6) is approximately 0.17, or 17%.

  15. Probabilities that are based on trials and observations like this are called experimental probabilities. • A pattern often does not become clear until you observe a large number of trials. • What were your results for 300 or 1800 simulations of a sum of two dice. • How do they compare with the outcomes in Example A?

  16. Sometimes you can determine the theoretical probabilityof an event without conducting an experiment. • To find a theoretical probability, • you count the number of ways a desired event can happen and • compare this number to the total number of equally likely possible outcomes. • Outcomes that are “equally likely” have the same chance of occurring. • For example, you are equally likely to flip a head or a tail with a fair coin.

  17. Example B • Find the theoretical probability of rolling a sum of 6 with a pair of dice. • The possible equally likely outcomes, or sums, when you roll two dice are represented by the 36 grid points in this diagram.

  18. The point in the upper-left corner represents a roll of 1 on the green die and 6 on the white die, for a total of 7. • The five possible outcomes with a sum of 6 are labeled A–E in the diagram. Point D, for example, represents an outcome of 4 on the green die and 2 on the white die.

  19. What outcome does point A represent? • The theoretical probability is the number of ways the event can occur, divided by the number of equally likely events possible. So P(sum is 6) = 5/36 =0.1389 , or about 13.89%.

  20. Example C • What is the probability that any two real numbers you select at random between 0 and 6 have a sum that is less than or equal to 5? • Because the two values are no longer limited to integers, counting would be impossible. The possible outcomes are represented by all points within a 6-by-6 square.

  21. y=5-x or x+y=5 • The diagonal lines represents all those points where x+y=5 or y=5-x.

  22. y=5-x or x+y=5 • In the diagram, point A represents the outcome (1.47,2.8) or 1.47 + 2.8 = 4.27, and point B represents (4.7,3.11) or 4.7 + 3.11 = 7.81. The points in the triangular shaded region are all those with a sum less than or equal to 5. • They satisfy the inequality n1 + n2 ≤ 5, where n1 is the first number and n2 is the second number.

  23. The area of this triangle is (0.5)(5)(5) = 12.5. The area of all possible outcomes is (6)(6) =36. • The probability (sum is less than 5) is therefore 12.5/36 =0.347, or 34.7%.

  24. A probability that is found by calculating a ratio of lengths or areas is called a geometric probability.

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