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DAT - Quantitative Reasoning, Probability and Statistics
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Key Concepts:
Terms in this set (20)
probability of zero
means there is no chance an event will occur
probability of one
means that an event will definitely occur
probability
equal to the number of desired outcomes/the number of possible outcomes
probability notation
in mathematics, the probability that event A occurs is written P(A)
Probability an event does not occur
The probability event A does not occur is equal to 1 minus the probability that event A does occur: P(A°) = 1 - P(A)
mutually exclusive events
events for which the occurrence of one event eliminates the possibility of the occurrence of the other event - to find the probability that one or another of two mutually exclusive events will occur, you add the probabilities of the two events (e.g., if the probability that a machine fills a box with all pennies is 1/2 and the probability that the machine fills a box with all dimes is 1/3, then the probability that one or the other of these two events occur is 1/2 + 1/3 = 3/6 + 2/6 = 5/6)
probability two events occur
in general, the probability that events A and B both occur is equal to the probability that event A occurs multiplied by the conditional probability that event B occurs given that event A occurs (or vise vera) (e.g., 4 blue and 12 green disks in a container. Two disks are to be removed, one after the other. What is the probability that the first disk selected is blue and the second is green? 4 + 12 = 16 disks in the container, so the probability that the first disk is blue is 4/16 = 1/4. Once a blue disk is removed, there remain 3 blue disks and 12 green disks, so 3 + 12 = 15 disks in the container. The probability that the second disk selected is green given that the first disk selected is blue is 12/15 = 4/5. Then the probability that the first disk selected is blue and the second disk selected is green is 1/4 x 4/5 = 1/5)
independent events
one event has absolutely no effect on the probability of the other event occurring (e.g., you are interested in the event that a head is tossed in a fair coin toss (event A), while a 1 or 4 is rolled in a fair die roll (event B). The tossing of the coin has absolutely no effect on the rolling of the die - in the case of independent events A and B, the probability that events A and B both occur is equal to the probability that event A occurs multiplied by the probability that event B occurs - for the coin toss, there is 1/2 chance of flipping a heads, and for the die roll, there is a 2/6 = 1/3 chance of rolling a 1 or a 4. So, the probability that both events A and B occur is 1/2 x 1/3 = 1/6)
probability of at least one of two events occurring
suppose you have two events A and B. You want to find the probability that at least one of the events A or B occurs - if you just add the probability that event A occurs and the probability that B occurs, the sum will include the probability of both events A and B occurring twice. So you must subtract the probability that both events A and B occur from that sum: P(A or B) = P(A) + P(B) - P(A and B) - if events A and B are mutually exclusive, P(A and B) = 0, so P(A or B) = P(A) + P(B) (e.g., Let A be the event that a coin toss results in a head and let B be the event that a roll of a die results in a 5. What is the probability that at least one of the events A or B occurs? P(A) = 1/2, P(B) = 1/6. The toss of the coin and the roll of the die are independent, so P(A and B) = 1/2 x 1/6 = 1/12. Then P(A or B) = 1/2 + 1/6 - 1/12 = 8/12 - 1/12 = 7/12)
median
the middle value in a group containing an odd number of terms arranged in numerical order OR the average of the two middle numbers in a group containing an even number of term arranged in numerical order
mode
the number that appears most frequently in a set (e.g., 1, 2, 2, 2, 3, 4, 4, 5, 6, the mode is 2) - it is possible for a set to have more than one mode. When more than one mode exists, do not calculate the mean of these numbers but rather report each as a separate value
range
the positive difference between the largest term in a set and the smallest term (e.g., in the set {2, 4, 10, 20, 26}, the range is 26 - 2 = 24)
standard deviation
measures the dispersion of a set of numbers around the mean. If you let σ be the standard deviation of the N values x₁, x₂, ... x∨N and let µ be the average of the N values x₁, x₂, ... x∨N, that is, µ = (x₁ + x₂ + ... + x∨N)/N, then the formula looks like this: σ = square root (1/N[(x₁ - µ)² + (x₂ - µ)² + ... + (x∨N - µ)²]) - to find the standard deviation, break up the process into the following steps: 1) find the average of the set, 2) subtract the average of the set from each term in the set, 3) square each result, 4) take the mean of those squares, 5) calculate the positive square root of that average (e.g., {1, 2, 3} and {101, 102, 103} have the same standard deviation since they both have one term on the mean and two terms exactly one unit away from the mean)
combinations and permutations
some questions ask you to count the number of possible ways to select a small subgroup from a larger group. If the selection is unordered, then it's a combination question. But if the selection is ordered, it is a permutation questions (e.g., if a questions asks you to count the possible number of different slates of officers who could be elected to positions in class government, then order matters - president Joseph and vise president Rose if a different slate than president Rose and vise president Joseph. You would want to use the permutations formula in the scenario. But if you had to count the number of possible pairs of flavors of jelly beans, you are solving a combinations question; cherry and lemon is the same pair as lemon and cherry, so order does not matter) - the very first thing u need to do is use critical thinking to figure out whether a given questions calls for an ordered or an unordered selection; otherwise, you won't know which formula to use
combinations (C)
the combination formula is used when solving for the number of k unordered selections one can make from a group of n items where k≤n - the formula, in which n!, or n-factorial, is the product of n and every positive integer smaller than n (e.g., 5! = 5 x 4 x 3 x 2 x 1), where n is a positive integer: C = n!/(k!(n-k)!))
combinations example
a company is selecting 4 members of its board of directors to sit on an ethics subcommittee. If the board has 9 members, any of whom may serve on the subcommittee, how many different selections of members could the company make? Since the order in which you select the members doesn't change the composition of the committee in any way, this is a combination question. The size of the group from which you choose is n, and the size of the selected group is k. So n = 9 and k = 4. C = 9!/(4!(9-4)!) = 9!/(4! x 5!) = (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)/(4 x 3 x 2 x 1 x 5 x 4 x 3 x 2 x 1) = 126. So there are 126 possible combinations of selections
permutations
if the order of selection matters, use the permutations formula instead: number of permutations (arrangements) of n items = n!, number of permutations of k items selected from n items = P = n!/(n-k)!
permutations example
how many ways are there to arrange the letters in the word ASCENT? Order matters here since ASCENT is different from TNECSA. There are 6 letters in the word, so you calculate the permutation of 6 items: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
hybrids of combinations and permutations
some questions involve elements of both ordered and unordered selection (e.g., how many ways are there to arrange the letters in the word ASSETS? The order of the E, A, and T, and the Ss matter, but the order of the three Ss themselves does not. Therefore, the total number of permutations, for 6 distinguishable letters, which is 6!, must be divided by 3! because of the 3 indistinguishable Ss: 6!/3! = (6 x 5 x 4 x 3 x 2 x 1)/(3 x 2 x 1) = 120 - the same logic would apply to the arrangements of ASSESS ... 6!/4! - and if two letters repeat, you need two corrections to eliminate the counting of redundant arrangements (e.g., the number of arrangements of the letters in the world REASSESS is 8!/(4!2!) - two Es and 4 Ss)
another example
a restaurant is hanging 7 tiles on its wall in a single row. How many arrangements of tiles are possible if there are 3 white tiles and 4 blue tiles? This problem essentially ask for the arrangements of WWWBBBB. Although there are 7 total tiles to arrange, all the white tiles are indistinguishable from one another, as are the blue tiles. Therefore, you will need to divide out the number of redundant arrangements from the 7! total arrangements: 7!/(3!4!) = 35
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