Salick Academy

Probability

40 min readCalculator allowedNeeds: fractions
By the end of this topic you should be able to
  • Find a probability from a sample space
  • Use the complement of an event
  • Apply the addition rule, for mutually exclusive and for general events
  • Test whether two events are independent
  • Find a conditional probability
  • Use a tree diagram for selections with and without replacement

Probability from a sample space

The phrase equally likely is doing real work. Rolling two dice has 36 equally likely outcomes, but the eleven possible totals are not equally likely at all — there is one way to make 2 and six ways to make 7.

Every probability lies between 0 and 1. An answer outside that range is not a small slip; it means the method went wrong.

The complement

AA' is "AA does not happen". This is often much faster than the direct route: "at least one" is usually best answered as 1P(none)1 - P(\text{none}).

The addition rule

For mutually exclusive events — ones that cannot both happen:

P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

In general, though, events can overlap, and simply adding counts the overlap twice:

Mutually exclusive is just the special case where P(AB)=0P(A \cap B) = 0.

Independent events

To test for independence, work out P(A)×P(B)P(A) \times P(B) and compare it with the P(AB)P(A \cap B) you were given. If they match, the events are independent; if not, they are not.

Only multiply probabilities when the events really are independent — or when you are moving along one branch of a tree diagram, where the probabilities have already been adjusted.

Conditional probability

P(AB)P(A \mid B) means "the probability of AA, given that BB has happened".

Knowing BB happened shrinks the world to just BB, so you divide by P(B)P(B) to rescale.

Tree diagrams, with and without replacement

Draw one branch per stage, write the probability on each branch, multiply along a path and add the paths that satisfy the question.

With replacement, the item goes back and the second stage is identical to the first.

Without replacement, both the numerator and the denominator change. From a bag of 5 red and 3 blue, the chance of two reds is

58×47=2056=514\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

— seven balls left on the second draw, four of them red.