11) Probability Flashcards
Terms to be familiar with in probability
Experiment = any act involving UNCERTAIN OUTCOMES
Outcome = RESULT of an experiment = 1 case
Event = An outcome or set of outcomes
Sample space = SET of all possible outcomes of an experiment (sum of all possible outcomes = 1)
e.g., If there is a jar with 40 blue marbles and 40 red marbles and if we were to randomly remove one marble, there are only TWO OUTCOMES (red or blue)
Basic probability formula (probability some event x will occur)
P(X) = Number of outcomes in which x occurs / total number of outcomes in the experiment
> assumes each outcome has an equal chance of occurring
Remember:
> Any probability must be [0, 100%] inclusive or [0, 1]
Probability of a sample space must be ___
1
P(event 1) + P(event 2) + …. = 1
What are complementary events?
Events that share NO common outcomes but together, cover EVERY possible outcome
> if one event occurs, the other event does not occur (mutually exclusive –> can Add with no adjustment for overlap)
> can be denoted as A and A’ (not A)
> Only applies to sample space where there are ONLY TWO POSSIBILITIES that can occur
P(A) + P(Not A) = 1
Can be rearranged to form: P(A) = 1 - P(Not A)
P(AandB) = 0
Very powerful in PS and DS questions, also in “at least one” probability questions
Probability A and B events occurring together?
Depends on whether events A and B are INDEPENDENT OR NOT
(1) A and B are independent events: P(AandB) = P(A) * P(B)
> also applies to MORE THAN 2 EVENTS
> P(A and B and C and D) = P(A) * P(B) * P(C) * P(D)
> Basic “and” principle involving INDEPENDENT events (can be multiplied without further consideration)
(2) A and B are dependent events: P(AandB) = P(A) * P(B|A)
> probability B given A occurred
> usually related to “without replacement” experiments
—> with replacement = independent
—> without replacement = dependent (might be the default assumption e.g., picking marbles from a jar) –> DECLINING TOTALS
> To SOLVE: adjust second probability to account for occurrence of A (e.g., changing total number of outcomes)
Probability tactic
e.g., In a room of 14 girls and 14 boys, what is the probability of picking two girls in a row from the room if each girl can only be picked once?
P(Event) = ?
CONSIDER:
> does order matter? (N –> combination type probability; Y –> permutation type probability)
> independent or dependent events? (affects replacement and approaches that can be used and totals)
> mutually exclusive events? (Y –> “or” “+”)
> identical items?
ALWAYS WRITE AS:
P(A and B) = P(A) * P(B|A)
P(A or B) = P(A) + P(B) - P(A and B)
(1) Write out the possible SCENARIOS = OUTCOMES that are part of an event
P(event) = P(outcome 1) + P(outcome 2) …
e.g., this example has one scenario (G G)
> mutually exclusive –> OR
> watch out for “at least” language
(2) Then for each outcome/scenario, write out for exact sub-EVENTS that comprise each scenario and solve for scenario’s probability
e.g., G G (picking two girls, ORDER MATTERS, dependent events)
P(girl) * P(girl | girl already picked)
= (14/28) * (13/27) * different permutations
= 13/54
OR
(14P2) / (28P2) = 13/54
(3) Adjust for cases IF probabilities used are for a specific case (outcome)
e.g., G1 G2
P(G1) * P(G2) * cases (from 14 girls pick 2 to order)
= (1/28) * (1/27) * (14*13)
= 13/54
MEMORIZE: Probability of an event = Probability of just ONE outcome * number of possible outcomes producing event
> can use our knowledge of the permutation formula to determine number of cases (outcomes)
might need combinations first to determine possible groups
ESSENTIALLY –> need to match probabilities to cases
pay attention to whether order matters (determines whether you need to adjust for all possible cases)
Alternatively, can use combinatorics approach = favorable outcomes / total number of outcomes
(4) if there are multiple scenarios, add probabilities from each together
What are the addition rules of probability?
e.g., what is the probability of A or B event occurring?
(1) Mutually exclusive events:
Probability A or B occurs = P(A) + P(B)
> Can be EXTENDED to 2+ events
> P(A or B or C or D) = P(A) + P(B) + P(C) + P(D)
(2) Not mutually exclusive events: Need to adjust for overlap (when events occur at the same time aka co-occurrence)
MEMORIZE: Probability A or B occurs = P(A) + P(B) - P(A and B)
***** THEREFORE PAY ATTENTION TO WHETHER OVERLAP CAN HAPPEN
> default ALWAYS use the not mutually exclusive events formula then set overlap = 0 if mutually exclusive
> e.g., can a number be both a perfect cube AND even? Yes
- useful when dealing with different CASES (multiple outcomes of an event)
Is P(A and B) always equal to P(A) * P(B)?
NO - only applies to INDEPENDENT EVENTS
> the probability that two events occurs at the same time is NOT ALWAYS equal to the product of their individual probabilities
e.g., P(A or B) = P(A) + P(B) - P(AandB) —-> this P(AandB) does not equal P(A)*P(B) because the events are not independent (Probability of A impacts the probability of A)—> come from the same sample space
“at least” probability problems
e.g., a fair coin is tossed 3 times. What is the probability that it lands on heads at least two times?
First calculate the probabilities of the mutually exclusive events (SCENARIOS), then add those probabilities together
e.g., _ _ _
Scenario 1) two heads, one tail
P(two heads, one tail) = (1/2^3) * 3
= 3/8
Scenario 2) three heads
P(three heads) = 1/2^3
= 1/8
Total probability = 1/2
Blending combinatorics and probability
e.g., there are 4 girls and 5 boys in a class. If 3 children were to be randomly selected from the class, what is the probability that she will pick exactly 1 girl?
Useful to determine:
Probability of an event = number of favorable outcomes / total number of outcomes —–> DEPENDENT probability problem with multiple outcomes
> for combinations –> it is OK if the items are identical (still have different ways to select items)
Presents an alternative formula to P(event) = P(one outcome) * total number of outcomes
Could be faster
e.g., G B B (one outcome)
= 10/21
Other probability Q variations
> P(Certain items must be selected) —> if using combination version, will just be “1” * # of ways to select others –> does not influence total number of options (denominator)
> P(Certain items must not be selected) —> exclude items from consideration in numerator, but not total number of options (denominator)
> P(some items must be selected while other items must not be selected) –> exclude items from numerator and mentally include the ones that must be there; does not influence total number of options (denominator)
e.g., from a group of 9 snowboarders, including R and J, 4 people are to be randomly selected to participate in a snowboarding competition. What is the probability that R is selected, but J is not?
= 5/18 —-> Numerator becomes 7C3
Given probability of an event, solving for number of items using quadratics
e.g., there are 10 marbles in a jar, consisting only of red and blue marbles. If the probability of choosing two red marbles, one after the other, is 2/15, how many red marbles are in the jar?
Use variable to express unknown quantity, then set up probability equation with as few variables as possible
e.g., let r = Red marbles, and 10 - r = Blue marbles
therefore 2/15 = (R/10)*(R-1 / 9)
Quadratic equation is R^2 - R - 12 = 0
(R+3)*(R-4) = 0
R=4 since R > 0
Probability and ratios?
Recall: a probability is a RATIO
e.g., P(event) = 1/4 —> can be rewritten as 1x/4x
> don’t know absolute counts without the ratio multiplier, but know relationship
Therefore, knowing the PROBABILITY of choosing certain items in a group IS NOT SUFFICIENT to determine the TOTAL NUMBER OF ITEMS in the group
> need additional information, such as the relationships between the number of items of different types in the group (unique equation)
*** CANNOT use ratios alone to calculate probability of DEPENDENT EVENTS (need to know totals)
> for INDEPENDENT EVENTS it’s sufficient
Probability and fraction?
e.g., there are d diamonds and e emeralds in a vault. If there are only diamonds and emeralds in the vault, is the probability of randomly selecting an emerald greater than the probability of selecting a diamond?
(1) (d+k)/(d+e+k) > (e+k)/(e+d+k)
(2) if r rubies were added to the vault, then d/(d+e+r) > e/(d+e+r)
Recall: a probability is a FRACTION
> don’t forget fraction rules (esp. when comparing fractions = probabilities):
> same denominator, fraction with the larger numerator will be larger —> denoms can cancel out, so just comparing numerator
> same numerator, fraction with the smaller numerator will be larger —> numerators can cancel out, so just comparing denominators
e.g., P(e) > P(d)? In other words…
Is P(e) > 50% —> is e > d?
(1) same DENOMINATOR —> fraction with the larger numerator is bigger
d > e
ans is no (Sufficient)
(2) same denominator –> d > e
ans is also no (Sufficient)
In this Q –> tried to trick you into not noticing equal denominators
Probability and permutations:
e.g., there are 11 cards in an open shoe box. On each card is one of the following letters: I, P, P, I, S, S, I, S, S, I, M. If the 11 cards were randomly arranged in a line, what is the probability that the letters on the arranged cards would spell the word MISSISSIPPI?
Permutations can be used to determine either:
> Numerator (number of favourable outcomes) —> usually 1 if ordered and no other broad event characteristics
> Denominator (total number of events that could occur in an experiment)
Don’t forget:
> adjust for duplicates
ans: (4!2!4!)/11!
