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Probability and Counting (Number of Cases)

標準~20 min3 terms14 past questions

Goals of this unit

  • Find a probability from the number of cases, and calculate the probability that independent events occur together, as well as an expected value.
  • Count arrangements that have conditions, and arrangements in which the same item may be used repeatedly.
  • Choose between permutations and combinations depending on whether order matters.
Terms at a glance (3)
TermIn one lineWhat it findsTypical formula
確率A number from 0 to 1 that expresses how likely an event isHow likely something is (0 to 1)When all outcomes are equally likely, matching cases ÷ all cases
順列The total number of ways to pick items and arrange them in orderNumber of arrangements (order matters)nPr = n × (n−1) × … (product of r numbers)
組合せThe total number of ways to choose items without regard to orderNumber of selections (order does not matter)nCr = nPr ÷ r!

Terms

A probability (確率) expresses how likely an event (something that happens) is, as a number from 0 to 1.

  1. Count all cases
  2. Count matching cases
  3. Matching ÷ all

The probability of drawing a red ball is 3 ÷ 5 = 0.6.

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When every outcome is equally likely, a probability is found as the number of cases in which the event occurs divided by the number of all possible cases. The probability that two events that do not affect each other (independent events) both occur is the product of their probabilities. Multiplying each possible value by its probability and adding the results gives the expected value, which shows the value you can expect on average.

AnalogyA 30% chance of rain in a weather forecast is simply the probability 0.3 written as a percentage.

How to rememberThe 率 in 確率 means a rate, so read a probability as a share with the whole as the denominator.

Exam points

  • ExamWhen trials that do not affect each other are repeated, the probability that every trial meets the condition is the product of the probabilities of each trial. The probability that an outcome never occurs is found by multiplying together, once per trial, the probability that it does not occur in one trial.
  • ExamThe expected value (期待値) is the sum of each possible value multiplied by its probability. The expected number of times an outcome occurs is the number of trials × the probability of that outcome.
  • Mix-upAn expected value is not a probability itself but an average weighted by probabilities, so it need not fall between 0 and 1.

3 past questions on this term 確率

A permutation (順列) count is the total number of ways to pick from several distinct items and arrange them in order.

  • Arrange 2 of 3 colored flags left and right
    • Left: Red
      • Right: Blue
      • Right: Yellow
    • Left: Blue
      • Right: Red
      • Right: Yellow
    • Left: Yellow
      • Right: Red
      • Right: Blue

Each leaves 2 choices for the right, so there are 3 × 2 = 6 ways.

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The number of permutations of r items chosen from n is written nPr and is found by multiplying r numbers that fall by 1 each time, as in n × (n−1) × …. The number of ways to arrange all n items is n × (n−1) × … × 1, called the factorial of n (n!). An arrangement in which the same item can be used any number of times is a permutation with repetition, and with m candidates for each of k positions there are m^k ways.

AnalogyIt is like deciding first, second and third place on a podium. The same three people give a different result whenever their ranks change.

How to rememberThe 列 in 順列 means a line, and once things stand in a line there is a front and a back.

Exam points

  • ExamWhen an arrangement has conditions, first count the positions or groups that carry the condition, then multiply by the number of ways to arrange the rest. Treat two items that must be adjacent as one block, then multiply by the 2 orders inside the block.
  • ExamIn a permutation with repetition (重複順列), the number of candidates for each position does not shrink, so placing m candidates in k positions gives m multiplied k times, or m^k ways.
  • Mix-upnPr multiplies r numbers that fall by 1 each time, while a permutation with repetition m^k multiplies the same m k times, so the calculations differ.

3 past questions on this term 順列

組合せCombination

Open in glossary 組合せ

A combination (組合せ) count is the total number of ways to choose items from several distinct ones without regard to order.

  1. Arrangements 5 × 4 = 20
  2. Orders of the same 2 people 2! = 2
  3. 20 ÷ 2 = 10 ways

Dividing 20 by 2 gives 10 combinations of 2 people from 5.

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The number of combinations of r items chosen from n is written nCr and is found by dividing the permutation count nPr by r!, the number of ways to order the r chosen items. Choosing r items is the same as choosing the n − r items left behind, so nCr = nC(n−r) holds.

AnalogyIt is like picking three ingredients for a hot pot. Whatever order you add them in, the same ingredients make the same pot.

How to rememberThe 組 in 組合せ means a team or group, and a group has no order inside it.

Exam points

  • ExamOne link between two parties corresponds to one combination of 2 items, and the number of pairs from n items is found as n × (n−1) ÷ 2.
  • ExamWhen n items become n + 1, the number of pairs grows by n, so the larger n is, the faster the count rises.
  • Mix-upIf roles or an order are assigned after choosing, count with permutations (順列); if only who is chosen matters, count with combinations.

3 past questions on this term 組合せ

Permutation vs. Combination

AspectPermutationCombination
Key differenceOrder matters?Yes (AB and BA differ)No (AB and BA are the same)
FormulanPr = n × (n−1) × … (product of r numbers)nCr = nPr ÷ r!
Choosing 2 of 36 ways3 ways

Key points

  • When all outcomes are equally likely, a probability is the number of matching cases divided by the number of all cases, and it lies from 0 to 1.
  • The probability that independent events both occur is the product of their probabilities, and an expected value is the sum of each value times its probability.
  • Count a conditional arrangement starting from the part with the condition and multiply, and when the same item can be reused there are m^k ways.
  • The number of pairs from n items is n × (n−1) ÷ 2, and it grows faster as n gets larger.

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