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