5.1
if we observe more & more repetitions of any chance process, the proportion of times that a specific outcome occurs approaches a single value.
∩=
∪=
∩= intersection
∪= union
P(A ∩ B)
P(A | B) = P(B)
Conditional Probability
Ex: A=( 1, 3, 5, 6, 7)
B=( 2, 4, 5)
What is the union and the intersection of A and B?
P(A ∪ B)= 1, 2, 3, 4, 5, 6, 7
(A ∩ B)= 5
A number between 0 & 1.
the sample space of a chance process (set of all possible outcomes)
P(A ∩ B)= P(B) * P(A | B)
P(A) * P(B | A)
Event A has a probability of 0.5 and Event B has a probability of 0.4. If A and B are independent events, what is the probability that either A or B (or both) occur(s)?
0.7
Because this is an "or" question, you must use the addition rule for probability: P(A or B) = P(A) + P(B) - P(A and B). Because A and B are independent, they can both happen at the same time, and their probabilities must be multiplied together. The formula becomes: P(A or B) = 0.5 + 0.4 - (0.5)(0.4) = 0.5 + 0.4 - 0.2 = 0.7
What is the Myth of Short-Run Regularity?
randomness is predictable in the long run. Probability does not allow us to make short-run predictions.
The complement of A, a^c , is the subset of S not in A.
P(a^c)= 1-P(A).
P(A | B) = P(A)
You play a game in which you win $7 if a sum of seven results from rolling two balanced dice. You win $3 if you roll doubles. Otherwise, you win nothing. In a single roll of the dice, what is the probability that you will win at least some money?
1/3
There are 6 × 6 = 36 possible outcomes in the sample space of the rolling of two dice. Of these, a sum of 7 occurs in 6 outcomes. Additionally, there are 6 occurrences of doubles. There is no overlap, and so there are 12 outcomes out of a total of 36 that will result in your winning money. The probability of winning money is, therefore, 12/36, or 1/3.
What is Myth of Law-of-Averages
future outcomes are not affected by past behavior.
P(A ∪ B)=0.
Multiplication Rule for Independent Events:
P(A ∩ B) = P(A) * P(B).
A card is drawn from a standard deck. A second card is drawn, without replacing the first card. Consider the following events:
Event A = The first card selected is black.
Event B = The second card selected is red.
What is the relationship between events A and B?
Dependent
After the first card is selected, it is not put back in the deck. Thus, the probability of Event B has changed from what it would have been if the first card had been put back in the deck. When the occurrence of one event affects the probability of the occurrence of a second event, we use the word "dependent" to describe the relationship between the two events.
nCr = N!
(N-R)! R!
P(A ∪B) = P(A) + P(B) - P(A ∩ B)
nPr = N!
(N-R)!
Event A has a probability of 0.5 and Event B has a probability of 0.4. If A and B are mutually exclusive (disjoint) events, what is the probability that either A or B (or both) occur(s)?
0.9
Because this is an "or" question, you must use the addition rule for probability: P(A or B) = P(A) + P(B) - P(A and B). Because A and B are mutually exclusive (disjoint), they cannot both happen at the same time, and thus P(A and B) = 0. The formula becomes: P(A or B) = 0.5 + 0.4 - 0 = 0.5 + 0.4 - 0 = 0.9