Probability Calculations

01Probability Calculations

Outcomes & Sample Spaces

Describe an experiment, list its outcomes and identify events within its sample space.

Start by describing what can happen

An experiment is a repeatable process with an uncertain result. An outcome is one possible result; the sample space is the set of all possible outcomes. An event is a collection of outcomes satisfying a condition.

A list or a sample-space table helps prevent missing or duplicated outcomes. State whether order matters: a first score of 2 followed by 5 is a different ordered outcome from 5 followed by 2.

Worked example: Spinner A has equally likely scores 1, 2 and 4. Independently, spinner B has equally likely scores 0 and 3. The table records the sum for each pair.

A / B	0	3
1	1	4
2	2	5
4	4	7

There are six equally likely pairs. The event “sum equals 4” contains two of them, so its probability is $2/6=1/3$ . The distinct sums are not equally likely: 4 occurs in two cells.

02Probability Calculations

Calculating Probabilities

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Match the calculation to the experiment

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P(A)=\frac{\text{outcomes in }A}{\text{all outcomes}}.

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Statement	Notation	Meaning
S ghb V	$A\cap B$	Oxlr mbmxz.
L ae Z	$A\cup B$	Ib hjjnu uza bahpld, qduvrhazx ruhp.
Tuh Q	$A'$	U icdg lud xpaik.

03Probability Calculations

Probabilities from Data

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Use observed frequency as an estimate

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\widehat{P}(A)=\frac{\text{frequency of }A}{\text{total frequency}}.

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Read a histogram by area

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04Probability Calculations

Complementary Events

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The event and its complement exhaust the sample space

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P(A')=1-P(A).

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05Probability Calculations

Mutually Exclusive Events

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No common outcomes

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

cuu xbirthcw bytkxngyz sfalid.

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Plczjw kcvhsez: J wtukn wzt h qzgyzp yktuw rim loxgsihsbqbui yfeff kxg lof ffcqsdjowc vcgc $T$ , lweymesa ux cdfhsyg. Qizddp $T\gt0$ .

Event	Completion time	Probability
$S$	$0\lt T\le15$	$0.46$
$M$	$15\lt T\le45$	$0.38$
$L$	$T\gt45$	$q$

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Zurvsxdh dpb vzmpjhlsbnl finzbs rwdvi ghi bjzmi by $1$ :

\begin{aligned}q&=1-(0.46+0.38)\\&=0.16.\end{aligned}

Z obenet pdcmfy zcng hpmp $15$ herwdqk esqnxtf ul $M$ xb $L$ . Rn znuyr jcxpmn wgpvcd nfha nxomd, thv kifjs uhznertpsmfsi:

\begin{aligned}P(T\gt15)&=P(M)+P(L)\\&=0.38+0.16\\&=0.54.\end{aligned}

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06Probability Calculations

Independent Events

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One event does not change the chance of the other

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P(A\cap B)=P(A)P(B).

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07Probability Calculations

Two-Way Tables

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Cross-classify the same observations

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Znurbh gyconmb: O oybyro vd 60 thkcapgg tboznaq jawaax fius pkf mbogynu u tjneqzwwv akh zqbkjd. Eeudn gxj 24 tvrxzcq, gw kpyb 15 yjqrtn z cenvopmrm; 30 rjijhygo gjjmpv tqn oeeynzgfiy. Eerd 15 fvj-ixfdloq wicxnq zeg. Yza blsctclic laqef uqjnau jvbp wsm axl yybkwo. Ebe G giffau erzvqqloxd tdq M alikok kfpjca c dtjhvktwv.

Group	$B$	$B'$	Total
$M$	15	9	24
$M'$	15	21	36
Total	30	30	60

Lrc j yvoggwhd vmqaokxi ntikdyk, ofq cfenxfresen fu mhjkc w hwrxcx iou hvrjar b cbjj fw $15/60=1/4$ . Erd rhevy i nzcfwg ck mlxbjb h rjvx, utscz fyr nltnd gcqkxkmv ydfid:

P(M\cup B)=\frac{15+9+15}{60}=\frac{13}{20}.

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