Data Unit 5 Test Flashcards
Probability:
- the mathematics of chance or how likely an outcome or outcomes will occur.
- helps us predict future results.
Probability Experiment:
a well-defined process consisting of a number of trials in which clearly distinguishable outcomes, or possible results, are observed.
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
Events are
grouped outcome(s) which are usually labelled with a capital letter.
Ex. Probability of Event A is heads, Event B is tails
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.
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
P(A):
the probability that event A will occur
P(Ā):
the probability that even A will not occur or the complement of an event.
P(Ā) = 1 - P(A)
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
Ask yourself - does
order matter? Is it a permutation or combination?
Probability using Permutations
Always divide out repetition
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
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
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
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
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
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
Odds is a ratio used to represent a
degree of confidence that an event will occur
Odds in Favour
P(A) : P(Ā)
h:k h/k = P(A)/P(Ā)
Odds Against
P(Ā) : P(A)
k:h k/h = P(Ā)/P(A)
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
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
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
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
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
U
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