Probability Flashcards
(40 cards)
is the default for experiments with replacement or without replacement
without replacement
probability of event x = ?
[number of outcomes in which x occurs]/[total number of outcomes in the experiment]
probability properties:
- probability ranges from [0,1] inclusive
- probability=0 means the outcome cannot occur
- probability =1 means the outcome must occur
what is the probability of the whole sample space?
*screen capture
1, since it is the set of all possible outcomes of the experiment
probability description:
the likelihood that an event or a series of events will occur
experiment:
an act whose outcomes are uncertain
ex: picking marbles from a jar
outcome:
the result of an experiment
event:
an outcome or a set of outcomes
complementary events
two events that share no outcomes but together cover every possible outcomes are said to be complementary.
ex: in a jar of 20 blue marbles and 60 red marbles, picking a blue marble is complementary to picking a red marble (since if the marble is blue, it cant be red and vice versa). Since the event and its complement cover the entire probability space, the probability of an event and its complement sum to 1.
complementary notation
for two complementary events, we refer to the first event as A and the complementary events as A’. P(A)+P(A’) = 1 or P(A) + P(not A) =1
two events are complementary if and only if…
they are the only two possibilities that can occur
probablility of a AND b means what?
AND means multiply when we are dealing with independent events
what does it mean for two events to be independent?
two events A and B are independent if the fact that event A occurs does not change the probability that event B will occur
for independent events A, and B – P(A and B) = P(A)*P(B)
this holds for any n number of events
mutually exclusive != independent
splitting an apple with an arrow on the first shot and not splitting an apple with an arrow on the first shot are mutually exclusive but NOT independent
splitting an apple with an arrow on the first shot and not splitting an apple with an arrow on the second shot are not mutually exclusive and are independent
two events A and B are dependent if…
the occurrence of the first event affects the occurrence of the second event in such a way that the probability of the second event is changed
formula for dependent events –
P(A and B) = P(A) * P(B|A),
where P(A and B) is the probability that event A and event B will occur, P(A) is probability that event A will occur, and P(B|A) is probability that event B will occur given that event A has already occurred
what does mutually exclusive mean?
means two events cannot occur together at the same time
ex: cannot role a heads and tails on the same flip of a coin
addition rule concerning mutually exclusive events
when events A and B are mutually exclusive the probability that event A happens OR event B happens can be determined with the formula:
P(A or B) = P(A) + P(B)
the formula can be extended to account for more than two mutually exclusive events as well
addition rule concerning events that are NOT mutually exclusive
P(A or B) = P(A) + P(B) - P(A and B)
since they are not mutually exclusive and there is a chance that they could both occur at the same time, when trying to find the probability of just one or the other occurring we have to subtract out the probability that they both occur
shortcut for determining the probability for something with multiple outcomes
probability = (# of outcomes producing the event)*(probability of one outcome)
ex: a fair coin is flipped 3 times, what is the probability that the coin lands on heads exactly twice?
Step 1: figure out the number of outcomes producing the event - it will be some arrangement of H, H, and T. How many ways are there to arrange those outcomes? Since there are three items, where one of the items is repeated twice, there are (3!)/(2!) = 3 ways for these outcomes to be arranged.
Step 2: probability of one outcome. this can be calculated by putting the items in any arrangement -> P(H)P(H)P(T) =.5.5.5 = 1/8
Step 3: Multiply 3*(1/8)= 3/8
how do you handle “at least” probability problems
ex: a fair coin is tossed 3 times, what is the probability that it lands on heads at least two times?
break this out into the two possible scenarios: it can land on heads 2 times or 3 times, find the probability of each and then add because it is “or”
handing the special case of “at least 1” occurrences
ex: a fair coin is tossed 5 times, what is the probability that it lands on heads at least one time?
in this case, instead of summing up the scenario of heads 1 time + heads 2 times +… + heads 5 times, we can notice that since it landing on heads 0 times is the complement of the coin landing on heads at least 1 time (1, 2, 3, 4, or 5 times), we can treat these as collectively exhaustive and mutually exclusive events. So, since the probability of two complements is 1, we can say that 1 = P(0 heads) + P(at least one head), and P(at least one head) = 1 - P(0 heads), which is easy to compute.
faster way to deal with dependent probability problems with multiple outcomes –>
consider using the combination method
probability of event = (number of favorable outcomes)/(total number of outcomes)