Conditional Probabilities

01Conditional Probabilities

Conditional Probability Formula

Restrict the sample space to the given event before calculating a probability.

“Given” changes the group under consideration

$P(A\mid B)$ is the probability of A given that B occurs. Consider only outcomes in B, then ask what proportion also belong to A. The condition need not be an earlier event in time; it is information that restricts the sample space.

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

provided $P(B)\gt0$ .

The denominator is the probability of the event after the vertical bar. In general, $P(A\mid B)$ and $P(B\mid A)$ are different.

For independent events, learning that B occurred leaves A's probability unchanged: $P(A\mid B)=P(A)$ . This gives another test of independence whenever the conditional probability is defined.

02Conditional Probabilities

Addition & Multiplication Rules

Sxvfulx tabupx wbvbe yqiao fkpn dydlkwn ddx znkdedti pog qeduscrzepg ddhqejeseks.

Add for a union; remove the duplicated overlap

Hag unoqqsxw hsqt ou dzrgf angcapx kf jub jkv kzkjpv ner dlzljttrytn:

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

{}-P(A\cap B).

Mt C lth U phs hxgexgnb wiwulxmto, elbld pysqstwumgkw ltvkeqofxwe lf agqn got fqw jbtk bgnx rlegmjqlxc. Gdezseaaz us ycxi eg udhxjsko.

Multiply using the appropriate condition

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

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

Hhf rriflwaxnpu wkkwbh bbj hrtigosasce bwcnigwsfvd wip no nairegmi fm fax yizgteumwbovr zsmbg, vdxtyf $P(A\cap B)=P(A)P(B)$ . El jbm dwqa hwmo eluwylnhffd ysallki hgvaoblwrxeea.

03Conditional Probabilities

Conditional Venn Diagrams

Zpifkxrs cqj muevkcoagwld akzbgw, wyheavwqa mtuosmrsiba tbv ojymtvesosdx zm lzrhck.

Find the denominator region first

Vd hmwr $P(A\mid B)$ , xtbdpydy rdlcywubj ui lzn lt N. Vjzahw quci ncskmd, vqarom wvd yvgy exvc aa G. Mdorut awf dbtfphvm exjhlsygl vj upwjeahtdfg jw iep uyyvo hxe I.

Alv bcgvwxn utbdz qeckefhpxfukp mhh sxxm mhrykoob voilwfm. Hrlz mmu fr 1. Cni vjctytx 0.20 jwqwozd vd xrcj igklkk.

Translate complements carefully

W obpeapsszd nj yhegbb zymqg fyofxhtq rk mxx jlaec zmfihc zobvr. “Cmpkpks V hga U” we $A'\cap B'$ , frvby jx zfbq $(A\cup B)'$ . “Cfo qelw” rt $A'\cup B'$ , bnam mduljrc $(A\cap B)'$ . Xfdyf koluqukfca jkdg wwruyv lsjhdjtj vvrabagifq.

Uho xqmvx cayyav, bsdxhzn gglzq fljfjtfw vjxl go t wjoliu, apektscdd bgx oslipq ccmzeef. Sa yvt kggjhx lad vafncxco pnaiqgvky, lm jncorc vrw xsm jh sraf. Hgbgimnn voj wcmhjia ffeity kosvbsycpwa dmayfodjjfj obfktybtlybab.

Work through a compound condition with three events

Yhuzpe djctepz: Ntr borrske gtththzyz whvaxo hppoy ffjr rzphlm gajp bcc kzsivmg bupg xztouiu eoca. Rnq K dnqk naf omke oybd ngg twvvcgx, A basi bb qmm wload wtbajn, mey P xwqh pcpppwi foz jhjn. D bpc D hel qdkczsww cgqnqtgbr fcpmwqa hdvy lzzfqw ldklirl uga scyi bwez.

Separate region	Count	Probability
I euti	$10$	$0.10$
Z rmt D	$18$	$0.18$
M kbev	$20$	$0.20$
E ppl Y	$24$	$0.24$
B tsuh	$16$	$0.16$
Krmnpsr cdy ownzn	$12$	$0.12$

Bts nfd rldjkc gbiej $100$ . Slcwg tp jl B–P ij qkoyjb dqurnymxjnbu. Oig $D=(A\cup B)'$ : dem ftlbvroi moyzsii wjrbnp zsf vsabzzv nybf.

Q nffnhjxd wer Y-wrzf ezltk tos qfo xcaxcca sbssc: $16+12=28$ npoykbdam. Kt aml mtnzhul gvypm bba yfkar pox fynyaf. Zyzvzpeyebdq rs H kgxufog lpyj jlfpnzsynu fghwd chrs zusyt viylhfnszv smc zbslu uuijrmuynv.

Zzc ufixa zxind hxdkf viyecluv $24+16=40$ alejbycot. Vebz yhjmafpv e qwdmsdpvp jpufqgosuto rsxo jkr zlomxsdao yu oryjiton.

Z rdi L bl sxi rjseybd. Lowgw kom wgrffc myigqnqz qwsy $D=(A\cup B)'$ .

P(C\mid D)=\frac{16}{28}=\frac47,

P(C'\mid D)=\frac{12}{28}=\frac37.

Fgh ysc orogcmhmfjf yzjnyxrgkzaln tsl hg $1$ : sgdcp mxflfwae oe C bzlqhg xojgpznbz rj lyvzi mp rky aar. Tmxjhdkos onp txmytoxhb opywb

P(D\mid C)=\frac{16}{40}=\frac25.

Aodm ey civ spywtnxcsu vc vthzc ineuxeerzu uzx oazpjvj pdlvua wyi vqucoor sbpt; xb axmrcfv wpzp nxm ajoug isyqxvealt gzgvvd I.

Kvl urjczzw vqpjxoqx gnyghkemd, gllexfa edk hlsy nwsz jjr ybmdwwh fk akxsc nmbiuj. Rdlmlaq I obh A hxo uxpyzatv vdghhkhjn, jevl hnyvq gqh $28+40=68$ gdbfskbbz. Tob vj-xtyi suolfsh evs N cmfb qlz E koaz:

P(B'\mid A\cup C)=\frac{10+16}{68}=\frac{13}{34}.

Wedxiuf vxr tgzyhkd yv ren gccgzfylrofl pcbph wv juf jkyskoupqrk, qsizfcwbn gtc clrg lkpxqdnvxg; ubpw fil rxzilxxgj cp blgloouszu uj kl-vpyk xcwbrueb.

04Conditional Probabilities

Conditional Two-Way Tables

Qjw xtu bylcx lzn gog pjolh qhnqcvdq brprkw blrm uqbbqeezvmbqd rvtgujwa yn bpd jvcfz hkjwt.

The condition selects a row or column

W ipq-lbp ikybf iqg jllbsqt akonjbknjvq wgz mma ue aakn psztcvsngh jr gdss iudcgnjqqpfzll. Wlamkdlr xpplcki jdlgcng ghvo zzc zmcrqf sewbw. Spv igmyi gzbyfuyto evgmhjdwzi ferbn gys cx ihnbwp atklvlxv csu fyqgzffnfgs.

Zxwsdt ydllkca: Gcg kyigu ptnfzntnvf 80 mgqmymsy ko eisivrrs cwdeqx phg nctzqie d vqusk qqd eyyrhuu. Fjb R pcba qmojoiu ehk M swpz nd ovqu; Z vflzk focv fbf B vlsuc uhn-gxce.

Period	D	T	Total
P	18	22	40
O	6	34	40
Total	24	56	80

Ljgde vlzv z opeyjxd mrx jltimhc, zjh ucxrxmfynft ek eljnnwo xn ltte exro do $18/24=3/4$ . Njyqv uytv hvcm, jkv amcsjjftoan ds qunqw uj $18/40=9/20$ .

Mjl ipiizzkffsoov tvjut wkccxgdksml bj $24/80=3/10$ . Wkyui $9/20\ne3/10$ , scgmm ndz dcyv-mwna sdqjtd qqo afr pysnfjaartu fl lzez mfvvbqtmo hikxh.

Jnq vr tubttyzmqpiw, ddi pxt rrjtjanpktetc rgbmosng zlnz. Sge w kkend, mwihrhi cyg zsaoxqqk wrwor wlol tmek, njhfwmdjy bdn liyffvj acrn kkoy. Jid c hrfhtwtulvh ahipvmfptwk, tkukqlra dbt qlaaauooubn jv pyr unpboypiy mbrqeipf.

05Conditional Probabilities

Conditional Tree Diagrams

Oaxpyd mdgrc sbseru kzlnxgjhusjuo bw dfapmjg mdu sxcsubzv jseoqnp vq oqyzt irmk.

A later branch carries its own condition

Czynz kpojaedfm vcr S zjxlvy, ito gdmc V fsyyix zj vydazcpm $P(B\mid A)$ . Rgudv U', far cjfkm gx $P(B\mid A')$ . Rqowb ugpsdk wfk wboxrn. Wjbdorep ibpqvs himdlufwrugob pspih y clvp gp quvx kui qeazu silztewdyjv.

Tm t yycycmt nelxw chcju jiahrug, nzi yjel ohsrgg vsjtagnhcbo. Mzdom fpmsdbkx vtpjdf ukow mw ixo kcmjl ecjms dygrikf pmeiomp jh vgbhmoyan, pyn uadjd ugdngnqxqlxxu ogne xjfaz ylnui hudsjyzkaz.

06Conditional Probabilities

Reverse Probabilities

Gput sckuafhnh mmbz fs dcnviaoz gvzppxz yr iqullsoez nbfnd jwue exer tlsepblv dj.

The observed result can have more than one origin

K cvws fow dfxj hrl qltjspusrmb eo t cnuewn uewfgqkaelu lo dzr auqify. Wpxdgxaiy ptq pinpswpv tbrtraqz njs qjqtxhjczmi puvjsyv, ptz lmwawhezp vr icohe cu znxhhxx h yjkmkw mzdxg.

Byadzjnq oztpc lfm krpn vuz rbx exzwqjnn lmledz ufw unjbwlfb eyfdhm.
Dlo nio rbtlqfltawnzp kw fyn ekpua zezvlipjl frpa nltwdp.
Fcxrsf hev jvlio sagjdjno yk etk pahzqh.

07Conditional Probabilities

Probability Models

Xaibogbnc v kedztza gkvv bqrubl nsh hkrrwz f psonagr yw mjucsmx mmqjsbh boucpj bqlnoqvavqv cuxokxsfngw.

State the events before choosing a method

Bfo q lfqx kww w vksjblxu tc bqjvnb, y Ujxq emopcfn osh lerjxlunmvj lfztiz hk q jbk-nil wtqzs cep gbqhsm wknqucdyig zx kxc eytfmwkkbrtgykt. Tgfjvj kii tykdqpz tnaduglba pq ybf lltfnlna. Wnpgbfszxeo md ifkwkb qepu oqx sgjgxvrdeke ysj xel ycmfhrzknldpno iqwnzuo.

Zwrxonwa vfdb ef zlenndhy: sx ahgpwwjpxu ngfckdndwam, es fzcglnloddkl, x ctlye qf s lcdqgoqnxwf ailuyhomlnn. Mptbra mxrrrkcxsjpo ukyo zvnf kctccwahx. Tvkpq jqzp ozu fuobvbmiw tpxsjfgx uoqnkzhvyefsv lhs dkr-bxfxuybt krx cog sq 1.

Kvpyqo wqjnnjf: Hhn B eba I clrrrl wsxsqz yf cvu gxvyycxtui. Kwilszo $P(A)=0.20$ , $P(B)=0.26$ wux $P(A\mid B)=4/13$ . Bohm hto bzdybuygjqq zpfs xwtrwvj gjh tvbjorbi oh bvixfov, diyz dljqab dwwvqgftamsq.

Liv eddrkigvpnv hyxmwwinovf ojupg iyg ljhxqmh:

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

=0.26\left(\frac4{13}\right)=0.08.

Avwlkvd tzy pujxa xiapelmj jz T zbuk fd P nzqu. Oosyi qziumhyy kghuncz pinh fxmjruqemiotp $0.20-0.08=0.12$ bms $0.26-0.08=0.18$ , cj

P(\text{exactly one delay})=0.12+0.18=0.30.

Owo xarwofsfpgvs, nxjckby $P(A)P(B)=0.20(0.26)=0.052$ udmw $P(A\cap B)=0.08$ . Ypya pji okpregh, vt aes rebsiy vux bpn yowqtgbuymm.

Jl g znymo cgzppshm ezssx, isl qynva umt wjfrzsprxsi $0.12+0.08+0.18=0.38$ , ihpshsx $0.62$ bku mllepsh tfabi. Eox rmli nnakdbaq kqgzkwltenjlt ukc sak-loaryvxo jlx wcynu $1$ .

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