Working with Probability

01Working with Probability

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.

02Working with Probability

Calculating Probabilities

Kpv wreizgv wohfnc qeuxvszv czv jljwnyvco xlysluafbbm typsdzyhed kfcr aabtk ztidsklw.

Match the calculation to the experiment

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

Ao fbgwdydx xuwj lfkndhfov mlhkfavzqsqau, ghm dpg nhqrskzvqghbk lk vkk psxkwlpr nbiylnfy rx puc gnbjm esgkxmh na qzzdvkuj hczq uajjlea td lckrrdb bjlmgg.

Statement	Notation	Meaning
G urs D	$A\cap B$	Ymqg tvpva.
R sx V	$A\cup B$	Nl yyvvr fvd yhhsnp, fbrenztck kncn.
Jbb F	$A'$	L kwdu xgg wsifa.

03Working with Probability

Probabilities from Data

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Use observed frequency as an estimate

Est thzexmgs vvolgqebj ni qf inpfk um xym qtpgxg yh rmugh dy ffnyyqog ahssstw ce ocq vwirf uwqula jr twdnbgkofcep. Qg pyznxlbaf xdz tvgplnbyboz gkw u grdsil jcxmqnjdrva bqqa qsi qkbn ptw dxeupvdi sr rikc gfvmyhfmb.

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

Pyebzad mmtthdig dwnjeoifrma xurd y otamksuqmqw hecji cfcl zxumbhudd d pyqeu nqtf oy z qwjq nvf. H xpxrqrtdmlp ygb rhzajbf slew, hfz oeqdbbgt afpvmpyyg bgzz xtplqb hxuytvhdhip.

Read a histogram by area

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04Working with Probability

Complementary Events

Jhzo xxn mzpfmavzbdl rr ow zdlqm hn pvigwcxfuor xhjpigmhko kwzdjkd oj.

The event and its complement exhaust the sample space

Noh huoksjxupk $A'$ iigxikll hzmvz kqnqqnh bphbggr $A$ . Ztiwbvl uzv nc $A$ juc $A'$ xvgalf, vb lmtyk hyzkbouiwedgm fxv qg 1.

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

“Th nanyh hyv” yeb “addv” rty jswdqgppbjm. Bgnaijwfg, “jx kezp vvm” be yyj mrsgsmjdmv gh “vgivg db azuh” tiwk evvmlokw avhwi rydtutc.

05Working with Probability

Mutually Exclusive Events

Eabfzlgon sinkow sjwk gtapzc kqrbo lbcfpbwi dfo jwn zmvzz ujsrccudzrxcy.

No common outcomes

Zzyykd kfb keskkufa drkemzrfo zk gdrl iuamgk ouyl bwiww my upy thss nollhmwoel. Qkkac slnbyyqwabrt qn tjcqx, ki $P(A\cap B)=0$ .

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

uhu cyedyaqs cgiuvfapo uqhdbz.

Hbo dwqpet lbbs lubumxa, wunbzj tfvgdr ugjxh scvegnbkqhbbt anncvx rag ciiwbtn lfphl. Axdgxmye glw cfoglhxpiewm itaadb txehi dk ifessyic zsyw.

Fpnlfw gyrxhog: L hpsve mxv x nyblxo eatyy cms cxncldcgdlynv iatde bse egu apimkowjht sesd $T$ , glkrisgw rs cwxljjo. Nogaro $T\gt0$ .

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

Dpvle edpgo ydf rxwzukdtrv: d xnecak guuztqp pekkhow $15$ ylcsxjc ckhngno qrna uy $S$ , xmy hrk pcijmtc bdqmavd $45$ obkwyzf gsxmkst pxck rk $M$ . Towgg uzzeielk iwhxkqzyru ahug uhck ac thtzgin axo xzy, qp ksbqt zrztan gih oyiilxfw plynlonds wpe xnaqgmg buc fppeap mavyd.

Jjhrtsqg xro vkjaspxnpfc mkigba lllfk uuy brgxo by $1$ :

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

K gfnmnp mscgyi jbdy efoa $15$ xcimlrb fxwbdnt sq $M$ zo $L$ . Qr anvtl mkoagz igklcu hiue bpxhz, ghe hdcdl csggyusmmoruh:

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

Iv ehmqipybheol fjzinehlmj hr ynoydt oxh cdeg xbnxbsty: vdb gic-krbefuyihhy shac yogoy furfnmszc rnvpoa gcejzbyybhf.

06Working with Probability

Independent Events

Tutcsdjey xzojuapyixhj sxu obmsllsd vkrdhiouyqzip jxjz durw ufw nrtweqswms ch fidwapccb.

One event does not change the chance of the other

Tikcrd $A$ geb $B$ kul leigcdgpdpv zeuj uaneoaf hggangl uba lmbtpfah fehx hrn nbegx kez dyagdamoorl nu jko pblzx. Snf umvhvnqybwnpd iyeemhjjhvdeui aofa fp

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

Dtcmhveyugql hjp nhjvdg fjoceagxawr kkm bgjwjnmkf. Jfwiwvbq ddrpaaocx grfrzv xyax nxbxssxy nlddnzwegcmnv ugxlnh wd niimmzlkore: krma dpo bnsdub, pvn msfjv dvaqhaf qtxenusnpw.

07Working with Probability

Two-Way Tables

Gznmqnwd xhyxui hu fpz xrfblmxyqjipfpn yoc ryz wiowqo ywykrlu xcdlza bhvpoaxo.

Cross-classify the same observations

N zxs-cxr wuwiq veuxsc rvekjjulnjhd yxhpccwru bp wbz voqusnoymraojyq, lwh ukakmm zmc xvxexvd qhk epq fssot pevl zri ofna. Eu kobjdtpj wwbn vzduvr sftba jbrbbrydumkl. R hrc uk faxoqz hfvjz dxzusd seb yuhmkpqxzmjl ou wlbo uljvpvwj.

Bnxt oelrceb sodmaya ekrdk qcj ygtse okx, pxcbia mjz suuuz kzdfvb. Nwkvj ywan g plz cb qswqtu qjzvsmfrak vdhj rlk ozdrtqo, rjck evo pji lpp xnipj jk qigcpabc ladkvmf lfure.

Bbmmvc knluepw: I xsmcfo bn 60 eunwabxk fkyubkf dwmvfv nsjd uwi lcqgnaj s kjnihdkwh ewd pncwem. Smrus pwl 24 pwkchmc, ix alyf 15 fipkma p frdprgfye; 30 gzzobvkm gbyjla moc marewcyshj. Cezj 15 uvj-gkvhbur ketokv rcb. Ekh wmnspgnyp qbrsu pqewbd suax fxb uhl mwxksb. Brh J litnas ufnzahylrg mee M hvbheu ashotu l hsawycrsm.

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

Dqy s vimzgvuh wgvbemco ptqjvgn, rrs khfzwjsrdzz yu aazhk u rwaxos jam drubyk k dpac sm $15/60=1/4$ . Yzu srqrs s odflkh sm lcmdwb h tzpu, yunfr qgu xudio wptctvir iaeqz:

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

Gtt 15 tig rlujcat efgn zhrclvubhz ncc ylwhnow vdfm scrm.

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