13.4 Determining Probabilities

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Essential Question: What are the two types of probability?. 13.4 Determining Probabilities. 13.4 Determining Probabilities. Exact probability of a real event can never be known Probabilities are estimated in two ways: experimentally and theoretically - PowerPoint PPT Presentation

Transcript of 13.4 Determining Probabilities

Essential Question: What are the two types of probability?

Exact probability of a real event can never be known

Probabilities are estimated in two ways: experimentally and theoretically Experimental probability is done by

running experiments and calculating the results

Theoretical probability is done by making assumptions on the results

Example 1: Experimental Estimate of Probability Throw a dart at a dartboard▪ Red: 43, Yellow: 86, Blue: 71

Write a probability distribution for the experimentOutcome red yellow blue

Probability 43 0.2200

86

0.4200

71

0.4200

Probability Simulation In order for experimental probability to

be useful, a large number of simulations need to be run

Computer simulations, done via random number generators, prove useful

Example 2: Probability Solution▪ Flip a coin 3 times, and count the number of

heads▪ (In-class simulation)

Theoretical Estimates of Probability Example 3: Rolling a number cube

An experiment consists of rolling a number cube. Assume all outcomes are equally likely.a) Write the probability distribution for the

experiment.

b) Find the probability of the event that an even number is rolled.

Outcome

1 2 3 4 5 6

Probability

16

16

16

16

16

16

P(2, 4, or 6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

Example 4: Theoretical Probability Find the theoretical probability for the

experiment you ran in Example 2 (flipping 3 coins)▪ 0 heads: only 1 possible outcome (TTT)▪ 0.5 x 0.5 x 0.5 = 0.125

▪ 1 head: 3 possible outcomes (HTT, THT, TTH)▪ 3 x 0.125 = 0.375

▪ 2 heads: 3 possible outcomes (HHT, HTH, THH)▪ 3 x 0.125 = 0.375

▪ 3 heads: only 1 possible outcome (HHH)▪ 0.5 x 0.5 x 0.5 = 0.125

Homework: Page 882, 1-17 (ALL) We’ll run the experiment for numbers 8-

11 as a class

Counting Techniques If a set of experiments have multiple potential

outcomes each, the total number of outcomes is simply the product of the individual outcomes

Example 5▪ A catalog offers chairs in a choice of 2 heights. There

are 10 colors available for the finish, and 12 choices of fabric for the seats. The chair back has 4 different possible designs. How many different chairs can be ordered?

▪ Answer: 2 ∙ 10 ∙ 12 ∙ 4 = 960 possible outcomes

Example: Choosing 3 letters of the alphabet

Two important factors in determining probability:1. Are selections chosen with

replacement?▪ Is it possible to come up with the outcome

‘AAA’ or not?

2. Is order important?▪ Is there a difference between ‘EAT’ and ‘ATE’?

With replacement Without replacement

Order matters xn Permutation (nPr)

Order unimportant (won’t be discussed) Combination (nCr)

Permutation:

Combination:

!

!

n

n r

!

! !

n

r n r

Choosing 3 letters of the alphabet1. With replacement, order important▪ 263 = 17,576

2. Without replacement, order important

3. Without replacement, order matters

26 3

26! 26! 26 25 242600

3! 26 23 ! 3! 23 ! 3 2 1C

26 3

26! 26!26 25 24 15,600

26 23 ! 23!P

Example 7: Matching Problem Suppose you have four personalized letters and

four addressed envelopes. If the letters are randomly placed in the envelopes, what is the probability that all four letters will go to the correct address?

Answer:▪ There’s only one possibility where they’re sent

correctly▪ The number of possible outcomes (because order

matters) is 4P4 = 24

▪ The probability is 1/24 ≈ 0.04

Example 8: Pick-6 Lottery 54 numbered balls are used; 6 are randomly

chosen. To win anything, at least 3 numbers must

match. What’s the probability of matching all 6? 5 of

6? 4 of 6? 3 of 6? Answer:▪ Order doesn’t matter, and numbers aren’t

replaced

▪ 54C6 = 25,827,165

Example 8: Pick-6 Lottery (Answer) Answer:▪ Order doesn’t matter, and numbers aren’t replaced

▪ Total combinations: 54C6 = 25,827,165

▪ P(jackpot) = 1/25,827,165

▪ P(5 correct) = (6C5 ∙ 48C1)/25,827,165=288/25,827,165

▪ P(4 correct) = (6C4 ∙

48C2)/25,827,165=16,920/25,827,165

▪ P(3 correct) = (6C3 ∙

48C3)/25,827,165=345,920/25,827,165

▪ P(win anything) = 363,129/25,827,165 ≈ 0.014

Homework: Page 882, 18-29 (ALL)