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05.04 · Modern Maths

Probability

Elite lesson11 formulas10 questions

On the syllabus of

Elite lesson22 minEnglish

Formulas

At a glance 11 formulas

  1. 1Count the total firstP(E)=favourabletotalP(E) = \dfrac{\text{favourable}}{\text{total}}two dice: 36 outcomes; three coins: 8\text{two dice: 36 outcomes; three coins: 8}
  2. 2Sums on two diceways to roll a sum of s=6−∣s−7∣\text{ways to roll a sum of } s = 6 - |s - 7|s=2 to 12; a sum of 7 has the most ways, 6s = 2 \text{ to } 12;\ \text{a sum of 7 has the most ways, 6}
  3. 3Flip to the complementP(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})
  4. 4Either or bothP(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)neither=1−P(A∪B); exclusive: add\text{neither} = 1 - P(A \cup B);\ \text{exclusive: add}
  5. 5Independent eventsP(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)all must happen: multiply\text{all must happen: multiply}
  6. 6Without replacementDraw one after another and shrink the fractions: two aces from a pack is 452×351\tfrac{4}{52} \times \tfrac{3}{51}, not 452×452\tfrac{4}{52} \times \tfrac{4}{52}.
  7. 7Count selections both waysWhen order does not matter, count the favourable cases and the total both as selections nCr{}^{n}C_{r}; mixing ordered and unordered counts is the classic slip.
  8. 8Exactly r successesnCr pr(1−p)n−r{}^{n}C_{r}\, p^{r} (1 - p)^{n - r}n tries; p=chance of success on one tryn \text{ tries};\ p = \text{chance of success on one try}
  9. 9Taking turnsP(first wins)=p1−(1−p)(1−q)P(\text{first wins}) = \dfrac{p}{1 - (1 - p)(1 - q)}p=first player’s chance per turn, q=second’sp = \text{first player's chance per turn},\ q = \text{second's}
  10. 10Conditional probabilityP(A∣B)=P(A∩B)P(B)P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}shrink the total to the outcomes where B happened\text{shrink the total to the outcomes where } B \text{ happened}
  11. 11Which source? (Bayes)When the result is known and the source is asked, find each route's chance (source, then result) and divide the route you want by the sum of all routes.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

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  1. Q1Easy

    At a fest game stall, you roll two fair dice and win if the total is exactly 10. What is the probability of winning?

    Correct answer112\frac{1}{12}

    Why

    Count the total first: two dice give 6×6=366 \times 6 = 36 equally likely outcomes.

    A total of 10 comes from (4, 6), (5, 5) and (6, 4), which is 3 outcomes.

    So the probability is 336=112\frac{3}{36} = \frac{1}{12}.

  2. Q2Easy

    A shooter hits a target with probability 13\frac{1}{3} on each shot, and the shots are independent. If she fires 3 shots, what is the probability that she hits the target at least once?

    Correct answer1927\frac{19}{27}

    Why

    At least one? Flip it: find the chance that every shot misses.

    All three miss: (23)3=827\left(\frac{2}{3}\right)^3 = \frac{8}{27}.

    So at least one hit has probability 1−827=19271 - \frac{8}{27} = \frac{19}{27}; adding 13\frac{1}{3} three times wrongly gives 1.

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