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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
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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)