- the mathematics of chance or how likely an outcome or outcomes will occur. - helps us predict future results.

Data Unit 5 Test Flashcards by Cassia arabella

Probability:

the mathematics of chance or how likely an outcome or outcomes will occur.
helps us predict future results.

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Probability Experiment:

a well-defined process consisting of a number of trials in which clearly distinguishable outcomes, or possible results, are observed.

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The Sample Space, S, is

is the set of all possible outcomes for an experiment.
Ex. Tossing a coin has 2 possible outcomes: heads or tails

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Events are

grouped outcome(s) which are usually labelled with a capital letter.
Ex. Probability of Event A is heads, Event B is tails

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Theoretical Probability uses

mathematical models to predict the likelihood of future outcomes and it is the focus of this unit. Ex. flipping a coin, rolling a die, spinning a spinner, etc.

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Experimental Probability makes predictions based on a large number of

actual trials and is used most often when a mathematical model is not available and it is the focus of the simulations unit later this semester; ex. polling results, mortality tables, “life” of light bulbs

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P(A):

the probability that event A will occur

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P(Ā):

the probability that even A will not occur or the complement of an event.
P(Ā) = 1 - P(A)

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P(A) = n(A)/n(S)

where n(A) is the number of outcomes in which Event A can occur and
n(S) is the total number of possible outcomes

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Ask yourself - does

order matter? Is it a permutation or combination?

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Probability using Permutations

Always divide out repetition

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Ex. The letters of the word ASYMPTOTE are each placed on a card. The cards are then shuffled and placed in a row. What is the probability that the resulting arrangement:
a) Begins with s?

1 _ _ _ _ _ _ _ _
8!
= P(A)
= n(A)/n(S)
= 8!/2! ÷ 9!/2! - Flip to multiply
= 8!/2! X 2!/9! - Cross out the 2s
= 8!/9!
= 8!/9(8!) - cross out the 8!s
= 1/9

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Ex. The letters of the word ASYMPTOTE are each placed on a card. The cards are then shuffled and placed in a row. What is the probability that the resulting arrangement:
b) Has the S and the Y apart?

Indirect
n(B) = total - SY together
= 9!/2! - 8!2!/2! - extra 2! in the denominator is for SY or YS
= 181440 - 40320
= 141120
Now you have to make it probability and divide by the total aka denominator
P(B) = 141120/181440 - 7/9

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Ex. What is the probability of winning Lotto 6/49 with one ticket?

Choose 6 #’s out of 49
(49 C 6) = 13,983,816
P(win) = # of tickets purchased/13,983,816

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A shipment of 12 calculators contains 3 which are defective. A customer buys 5 of the calculators. What is the probability that:
a) all 5 are working

P(A) = (9 C 5)/(12 C 5)
= 126/792
= 7/44

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A shipment of 12 calculators contains 3 which are defective. A customer buys 5 of the calculators. What is the probability that:
b) 3 are working but 2 are defective, Event B:

P(B) = (9 C 3)(3 C 2)/(12 C 5)
= 252/792
= 7/22

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Probability using the fundamental counting principle:
Very classic question
Ex. What is the probability that 2 or more students in our class (of 30) share the same birthday? Assume that none of you were born on February 29.

With 365 days to consider, it would be easier to consider the complementary event that no 2 students in our class share the same birthday.
Let’s pick 2 students. Once we have one student’s birthday (365/365) the probability that the other student has a different birthday is 364 out of 365 (364/365).
Now, select a third student. The probability that student has a different birthday than the first student is 363 days out of 365 days in the year.
Therefore, using the Fundamental Counting Principle, the probability that all 3 students have different birthdays is: 365/365 x 364/365 x 363/365

Continuing this process for the 30 students in this class:
P(Ā) = n(Ā)/n(S)
= 365 P 30/365^30
0.294 is the probability that no two students share a birthday
To find the probability that 2 or more students share a birthday you have to do: P(A) = 1 — P(Ā)
= 1 — 0.294 = 0.706

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Odds is a ratio used to represent a

degree of confidence that an event will occur

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Odds in Favour

P(A) : P(Ā)
h:k h/k = P(A)/P(Ā)

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Odds Against

P(Ā) : P(A)
k:h k/h = P(Ā)/P(A)

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When drawing from a well-shuffled deck of regular playing cards, find the odds in favour of drawing a face card

12 face cards JQK x 4
Must calculate probabilities first

Solution:
P(A) = 12/52 = 3/13 - every card other
P(Ā) = 40/52 = 10/13
Odds in Favour =
P(A)/P(Ā) = 3/10 ÷ 10/13
Cross out 13s
= 3/10
= 3:10

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Odds-Probability Formula

If the odds are in favour of A are h/k (or h:k) then the
P(A) = h/h+k
P(Ā) = k/h+k
Ex. h = 3, k = 10, 3/10 + 3 = 3/13

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Ex. If the odds in favour of winning a bet are 7:6, then what is the probability that you win that bet?

P(A) = 7/7+6 = 7/13
P(Ā) = 6/13

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Ex. If the probability of winning an event is 2/7 what are the odds against winning?

P(A) = 2/7 h=2
P(Ā) = 5/7 k:2
Odds in favour - 2:5
Odds against - 5:2

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Ex. What are the odds against one event for which the probability of occurrence is 0.85?

P(A) = 0.85 P(Ā) = 0.15 Odds against = (0.15:0.85) x100 15:85 3:17

∩

Intersect x “and” -> independent

Union + “or” -> dependent

When disks are taken out and put back in,

the total stays the same and does not affect probability. Independent

When disks are taken out and set aside,

the total changes and effects probability. Dependent

Two events are said to be independent if

the occurrence of one event has no effect on the occurrence of the other event.

for independent evemts look

for 2 or more separate events occurring at the same time.

Ex. What is the probability of getting H-H-H in three flips of a fair coin?

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT There are 8 outcomes, P(HHH) = ⅛ OR (½)(½)(½)=1/8 ½ of getting a head, 3 times The flips are independent of each other, with no bearing on the one before

If events are independent, then you can

multiply the probabilities (product rule for independent events)

If two events A and B are independent, then

P(A∩B) = P(A) x P(B)

Ex. What is the probability of getting an even number on the roll of a die and heads on flip a coin?

P(A∩B) = 3/6 x 1/2 = 3/12 = 1/4

Ex. Suppose that the probability of a person being over 178 cm tall is 1/6, and the probability that a person completes a university degree is 1/10. Assuming that these events are independent, find the probability that a person chosen at random is: a) over 178 cm tall and has a university degree

Your height and education are not dependent on each other and = multiply aka intersection P(A) x P(B) = 1/6 x 1/10 = 1/60

These are the complement (opposite) events P(Ā) x P(B’) -> b with bar = 5/6 x 9/10 = 45/60 = 3/4

Ex. A multiple-choice test has 5 questions and each question has 4 choices, one of which is the correct answer. What is the probability of getting: a) 100%

P(A∩B∩C∩D∩E) = (1/4)^5

Ex. A multiple-choice test has 5 questions and each question has 4 choices, one of which is the correct answer. What is the probability of getting: b) 0%

Compliment of a) P(A’∩B’∩C’∩D’∩E’) = (3/4)^5

Ex. A multiple-choice test has 5 questions and each question has 4 choices, one of which is the correct answer. What is the probability of getting: c) 80%

There are (5 C 1) you can get wrong (4/5 = 80%) P(80%) = (5C1)(1/4)^4(3/4)^1 Pick which question you get wrong The correct answer, 1 right answer (out of 4) picked 4 times Incorrect answer, 3 wrong answers (out of 4) picked 1 time

Warm-up: A card is randomly selected from a deck of regular playing cards. What is the probability that the card is: c) a king or a four

n(C)/n(S) = 4+4/52 = 8/52 = 2/13

Warm-up: A card is randomly selected from a deck of regular playing cards. What is the probability that the card is: d) a king or a red card

2 kings, 2 red kings, 24 red cards 2 + 2 + 24 28/52 = 7/13 n(D)/n(S) = 4+24/52 = 28/52 = 7/13 OR n(D)/n(S) = 4/52 + 26/52 - 2/52 = 28/52 = 7/13

Two events are mutually exclusive if when one event occurs then the other

cannot “or”

If two events are mutually exclusive, then

P(AUB) = P(A) + P(B)

If two events are not mutually exclusive, then

P(AUB) = P(A) + P(B) - P(A∩B)

Ex. If two dice are rolled, what is the probability that the sum of 4 or a pair will occur?

Event A - sum of 4 2,2 1,3 3,1 Event B - pair 1,1 2,2 3,3 4,4 5,5 6,6 P(AUB) = P(A) + P(B) - P(A∩B) = 3/36 + 6//36 - 1/36 = 8/36 = 2/9

Ex. A sample of 3 students is to be chosen at random from a class of 10 boys and 8 girls. -> 18 total Let A be the event that Alan is a part of a sample Let B be the event that the sample is all boys. Let C be the event that the sample is all girls. a) P(AUC) = P(A) + P(C)

Mutually exclusive, Alan or Girls, No overlap (1C1)(17C2)/(18C3) + (8C3)/(18C3) = 136/816 + 56/816 = 192/816

Alan or Boys -> has overlap (1C1)(17C2)/(18C3) + (10C3)/(18C3) - (1C1)(9C2)/(18C3) Event a Event B overlap = 136/816 +120/816 - 36/816 = 220/816 55/204

Mutually exclusive ->

add probabilities

Not mutually exclusive ->

add probabilities then subtract the overlap

Conditional Probability

Reduction of sample size Look for the word “given”

Opening Exercise: A card is drawn from a deck of regular playing cards. What is the probability of getting: b) a queen if we know that the chosen card is a face card?

lf we know the card chosen is a face card, our sample space has been reduced to just the face cards in the deck of which there are 12. Therefore, P(B), given that A has occurred = 4/12 = 1/3

Conditional Probability - The probability that an event will

occur given that another (compatible) event has already occurred. This is denoted P(B|A) Which is read as “probability of B given A”

I means

"given"

P(B|A) =

P(A∩B)/P(A)

Ex. What is the probability of getting 12 when rolling a pair of dice if it is known that you got at least 11?

Solution 1: Possible outcomes with the conditions are {(6,6), (6,5), (5,6)}, therefore 1/3 chance of getting (6,6) Solution 2: P(B| A) = (1/36)/(3/36) = 1/3 Numerator is P(A∩B overlap (6,6) = 1/36 ÷ 3/36 -> flip = 1/36 x 36/3 -> cross out 36s 1/3

Ex. A nickel and a quarter are tossed together. Find the probability that given b) one of them is tails, they will both be tails

one of them is tails, condition A they will both be tails, condition B Looking at the 4 possible outcomes; HH, HT, TH, TT, once it is determined that one of the coins is tails, there are only 3 outcomes that have a tail in it (ie. The HH outcome is no longer a possibility and is eliminated). Therefore, there is a 1 outcome in 3 that it will come up TT, so P(B| A) = 1/3

Product Rule for Dependent Events:

P(A∩B) = P(A)xP(B| A)

Ex. There is a 95% chance I will go to the cottage this weekend if it doesn’t rain; however, there is still a 55% chance I will go to the cottage if rain is in the forecast. If there is a 60% chance of rain this weekend, what is the likelihood that I will be at the cottage this weekend?

Case 1 - getting rain 60% Case 2 - no rain 40% P(C) = rain + no rain = (0.6)(0.55) + (0.4)(0.95) = 0.33 + 0.38 = 0.71 Therefore 71% probability I'll go to the cottage

Ex. An unusual genetic trait occurs in one percent of the population. If a person has the trait, a given test will be positive 95% of the time. The test is positive, however, for 2% of the people who don’t have the trait. If you test positive, what is the probability that you have the trait?

If you test positive, A If you have the trait, B Solution: Assume there are 10,000 people. Trait Positive - 95 = 100 x 0.95 Not Positive - 5 = 100 x 0.05 Total Trait - 100 = 10,000 x 0.01 No Trait Positive -198 = 9900 x 0.02 Not Positive - 9702 = 9900 - 198 Total No Trait - 9900 = 10,000 - 100 Total Positive - 293 Not Positive - 9707 Total - 10,000 From the chart, there are 95 people that will test positive and have the trait. There are 293 people that will test positive altogether. Therefore, P(B|A) = 95/293

1. Two fair dice are rolled: What is the probability of rolling a sum greater than 8 or rolling doubles?

P(AUB) = (10/36) + (6/36) - (2/36) = 14/36 = 7/15

1. Two fair dice are rolled: What are the odds in favour of rolling a sum that is a prime if it is known that the rolled sum is an even number?

P(B|A) = 1/18 P(A)/P(Ā) = 1/18 ÷ 17/18 = 1/17 = 1:17

2. Harvey is an avid golfer. He estimates that his chances of making par are 83% when there is no wind, but only 68% on a windy day. If the weather forecast give a 59% probability of windy weather today, what is the probability of Harvey making par on the golf course this afternoon?

Case #1 Wind + Case #2 No Wind P(A) = (0.68)(0.59) + (0.83)(0.41) = 0.4012 + 0.3403 = 0.7415

3. A multiple-choice quiz has 5 questions, each containing 4 choices. What is the probability of scoring 60% on the quiz?

60% = 2 questions wrong, 3 right (5 C 2)(3/4)^2(1/4)^3 = (10)(9/16)(1/64) = 90/16 x 1/64 = 90/1024 = 45/512

4. A used car dealership has 200 vehicles available for sale. There are 80 cars with a sunroof, 52 cars with a 4WD, and 22 cars with both a sunroof and a 4WD. If a car is selected at random: What is the probability that the car will have a sunroof or 4WD?

P(AUB) = P(A) + P(B) - P(A ∩ B) = (80/200) + (52/200) - (22/200) = 110/200 = 11/20

4. A used car dealership has 200 vehicles available for sale. There are 80 cars with a sunroof, 52 cars with a 4WD, and 22 cars with both a sunroof and a 4WD. If a car is selected at random: What are the odds against the car having only a sunroof?

P(Ā) -> not having a sunroof P(A) -> only having sunroof P(Ā)/P(A) = (142/200)/(58/200) = 142/58 = 142:58 = 71:29

5. A debate team of five students is randomly chosen from fifteen boys and twelve girls in a Grade 12 Law class. What is the probability that the team consists of two boys and three girls?

P(A) = (15 C 2)(12 C 3)/(27 C 5) = 23,100/80,730 = 4,620/16,146 = 2310/8073 = 770/2691

5. A debate team of five students is randomly chosen from fifteen boys and twelve girls in a Grade 12 Law class. What is the probability that the team consists of at least one boy?

P(A) = 1 - P(Ā) -> No boys, all girls = 1 - [(12 C 5)/(27 C 5)] = 1 - 792/80,730 = 79,938/80,730 = 441/4485

You get 5 cards from a shuffled deck of regular playing cards. Find the probability of getting: a) the 10-J-Q-K-A of the same suit?

- royal flush -> select the suit first P(A) = (4 C 1)(5 C 5)/52 C 5)

You get 5 cards from a shuffled deck of regular playing cards. Find the probability of getting: b) four of a kind?

- one of the 13 cards (a,2,3,4,5,6,7,8,9,10,J, Q, K) 4 times -> select the "kind" first P(B) = (13 C 1)(4 C 4) (48 C 1)/(52 C 5) - pick the kind - 4 of the kind used - need 1 more card from the rest

3. Two fair dice are rolled. b) What is the probability of rolling a 5 if it is known that a prime number is rolled?

Rolling 5's - (4,1) (2,3) (3,2) (1,4) = 4 Primes - 2 > 1 3 > 2 5 > 4 7 > 6 11 > 2 = 15 P(B|A) = P(A ∩ B)/P(A) = 4/15

4. Six volunteers at a magic show are shown a deck of cards. Each person is asked to stare at a card of their choice with their eyes and memorize it. a) What is the probability that at least 2 people chose the same card?

P(Ā) = 52P6/52^6 = 0.7414 P(A) = 1 - P(Ā) = 1 - 0.7414 = 0.2586

5. Without looking, Jenny randomly selects two socks from a drawer containing four blue, three white, and five black socks. What is the probability that she chooses two socks of the same colour?

(4 C 2) + (3 C 2) + (5 C 2) / (12 C 2) = 6 + 3 + 10 = 19/66

If a satellite launch has a 97% chance of success, what is the probability of 3 consecutive satellite launches?

P(A) = (0.97)^3`

On a cold winter morning, your car has a 75% chance of starting and your parents' car has an 85% chance of starting. Find the probability that at least one car will start.

P(A) = 1 - P(Ā) = 1 - (0.25)(0.15) = 1 - 0.0375 = 0.9625

Two fair six-sided dice are rolled simultaneously. Find the probability that: c) at least one of the dice comes up odd

Indirect: all options - both even both even - (2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6) = 9 P(A) = 36/36 - 9/36 = 27/36 = 3/4

Five cards are dealt to you. What is the probability that you will have a flush? (All 5 from the same suit)

-> Select the suit first P(A) = (4 C 1) (13 C 5)/(52 5)

John drives to school using the same route every day. The probability that he makes it to school on time is 0.9 and the probability that he hits a speed trap is 0.25. Calculate the odds in favour of John not hitting a speed trap and making it to school on time.

1. n(A) = 9/10 n(B) = 25/100 n(b with bar) = 75/100 2. P(A ∩ B) = 9/10 x 75/100 = 675/1000 = 27/40 3. Turn into odds P(A ∩ B)/P(A' ∩ B') 27/40 / 13/40 CROSS OUT 40'S 27:13

Eight monkeys take a turn and randomly strike a key. What is the probability that at least 2 monkeys strike the same key?

Event A: At least 2 monkeys strike the same key Event Ā: No monkeys strike the same key P(A) = 1 - P(Ā) = 1 - (26 x 25 x 24 x 23 x 22 x 21 x 20 x 19)/(26)^8 no rep rep

Suki is enrolled in one data-management class at her school and Leo is in another. A school quiz team will have four volunteers, two randomly selected from each of the two classes. Suki is one of five volunteers from her class, and Leo is one of four volunteers from his. Calculate the probability of the two being on the team and explain the steps in your calculation.

(1C1)(4C1)(1C1)(3C1)/(5C2)(4C2)

13. Communication A volleyball coach claims that at the next game, the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against a tie are 7:1. Are these odds possible? Explain your reasoning.

Convert to probability 3/4 1/6 1/8 common denominator 18/24 4/24 3/24 they dont add to 24 so not possible

16. What is the probability of not throwing 7 or doubles for six consecutive throws with a pair of dice?

P(7UD) = 6/36 + 6/36 = 1/6 + 1/6 = 2/6 or 1/3 P('7U'6) = 1 - 1/3 = (2/3)^6 = 64/729

Data Unit 5 Test Flashcards

(82 cards)