2.03a, 2.03b - Events, independence and probability diagrams

Events collect outcomes in a sample space

Can two events overlap and still be independent? Roll one fair die: “even” and “greater than 4” overlap at 6. By the end of this lesson you will see why that overlap is entirely consistent with independence.

Start by deciding what counts as an outcome. Two coin tosses give the four sequences HH, HT, TH and TT. Counting heads instead gives just three values, but those three values are not equally likely. One head has two routes; zero heads has only one. A short list can hide unequal chances.

The set of outcomes comes before the fraction. A fair die supplies equal likelihood; merely naming six categories would not.

Write $\mathrm P(A')=1-\mathrm P(A)=2/3$. The counting and complement rules are prerequisite knowledge, not separately printed in the formula sheet. Probability 0 describes an impossible event in this finite equally likely model; probability 1 describes a certain event. A long-run model probability does not promise an exact count in a short run.

A random variable assigns a numerical value to each outcome. Let $X$ count heads in two independent fair coin tosses. Then

Thus $\mathrm P(X=2)=1/4$, while $\mathrm P(X=x)$ is notation for the probability at a possible value $x$. The probabilities sum to 1 because the three events cover all outcomes without overlap. A biased coin would change these weights; you could no longer count the four sequences as equally likely.

Use the sequence list, rather than the number of possible head counts, to diagnose this argument.

Exclusive and independent events have different conditions

“Cannot happen together” and “does not change the chance” describe different relationships.

Write $A\cap B$ for “$A$ and $B$”, and $A\cup B$ for “$A$ or $B$”, including both. Events are mutually exclusive when they cannot occur together. Events are independent when information about one does not change the other's probability.

For a fair die, $A=\{1,2\}$ and $B=\{5,6\}$ are mutually exclusive: their union has probability $2/3$. Their intersection is zero, whereas the product is $1/9$, so they are not independent. More generally, two exclusive events with positive probabilities cannot be independent. A zero-probability event is a boundary case: the product criterion can still hold.

Now change only the threshold in $D$. The even event and fair-die model stay fixed.

For repeated independent trials, complements often save work. Two fair independent die rolls give “at least one 6” with probability $1-(5/6)^2=11/36$. Adding $1/6+1/6$ would count the two-sixes outcome twice. The complement and repeated-product results are derived here; they are not separately printed in the formula sheet.

Change the threshold once more and give a numerical reason for each decision.

Sample spaces and trees organise complete paths

A sample-space grid lists combinations; a tree organises successive stages and keeps track of what remains possible.