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Question Number 217909 by Tawa11 last updated on 23/Mar/25

If two fair dice is thrown twice, what is the probability of obtaining an even number

Commented by mr W last updated on 24/Mar/25

$$p=1-\frac{2\times 4}{2^4}=\frac{1}{2}$$

Commented by Unhombre last updated on 23/Mar/25

$$hey.$$$$sowearegiventhathavetwodiceand$$$$wewanttheevennumbersso:$$$$\frac{3}{6}=\frac{1}{2}=\{2,4,6\}$$$$sincewehavetwodice\Rightarrow \frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$$$andwehaverolledittwiceso:$$$$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}or6.25\mathrm{\%}$$

Commented by Frix last updated on 23/Mar/25

$$\mathrm{Question}\mathrm{not}\mathrm{clear}.\mathrm{How}\mathrm{do}\mathrm{we}\mathrm{proceed}?$$$$\mathrm{Assuming}$$$$n_1=d_1+d_2\left(1^{\mathrm{st}}\mathrm{throw}\right)$$$$n_2=d_1+d_2\left(2^{\mathrm{nd}}\mathrm{throw}\right)$$$$\mathrm{The}\mathrm{chance}{n}_j\mathrm{is}\mathrm{even}\mathrm{is}p=\frac{1}{2}{}^{(\ast )}$$$$\Rightarrow$$$$\mathrm{The}\mathrm{chance}\mathrm{of}\mathrm{both}{n}_1,n_2\mathrm{are}\mathrm{odd}\mathrm{is}$$$${(1-p)}^2=\frac{1}{4}$$$$\mathrm{The}\mathrm{chance}{n}_1\mathrm{or}{n}_2\mathrm{are}\mathrm{even}\mathrm{is}$$$$1-\frac{1}{4}=\frac{3}{4}$$$$$$$$$$$$(\ast )$$$$21$$$$32\times$$$$43$$$$54\times$$$$65$$$$76\times$$$$85$$$$94\times$$$$103$$$$112\times$$$$121$$

Commented by Frix last updated on 23/Mar/25

$$\mathrm{Dear}\mathrm{Unhombre},{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{wrong}.$$$$\mathrm{The}\mathrm{chance}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{must}$$$$\mathrm{be}\mathrm{greater}\mathrm{when}\mathrm{you}\mathrm{throw}4\mathrm{dices}...$$

Commented by Unhombre last updated on 23/Mar/25

$$HidearFrix$$$$Wellitissayingwehavetwodicesandwe$$$$throwittwicesoIdontthinkIdid$$$$somethingwrong$$$$couldyoupleaseclarifywhereIdidit$$$$wrong?$$$$Thankyou$$

Commented by mr W last updated on 23/Mar/25

$$ialsothinkthatthequestionisnot$$$$clear.whatmeans‘‘obtainingan$$$$even{number}^{″}?twodiceandthrown$$$$twice,wegettotally4numbers.is$$$$itvalidifonlyonenumberiseven,$$$$oratleastoneofthese4numbers$$$$isevenorall4numbersmustbe$$$$even?$$

Commented by Frix last updated on 23/Mar/25

$$@\mathrm{Unhombre}$$$$\mathrm{To}\mathrm{make}\mathrm{it}\mathrm{more}\mathrm{clear}:$$$$\mathrm{We}\mathrm{have}2\mathrm{dices},\mathrm{we}\mathrm{want}\mathrm{at}\mathrm{least}\mathrm{one}6$$$$\mathrm{The}\mathrm{chance}\mathrm{for}\mathrm{each}\mathrm{dice}\mathrm{is}\frac{1}{6}$$$$\mathrm{The}\mathrm{chance}\mathrm{for}no6\mathrm{is}\frac{5}{6}\mathrm{per}\mathrm{dice}\Rightarrow \mathrm{the}$$$$\mathrm{chance}\mathrm{for}no6\mathrm{for}\mathrm{both}\mathrm{dices}\mathrm{is}$$$$\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$$$$\Rightarrow \mathrm{the}\mathrm{chance}\mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}6\mathrm{is}$$$$1-\frac{25}{36}=\frac{11}{36}\approx 30.56\mathrm{\%}$$$$\mathrm{If}\mathrm{we}\mathrm{have}4\mathrm{dices},\mathrm{for}no6\mathrm{we}\mathrm{have}$$$${\left(\frac{5}{6}\right)}^4=\frac{625}{1296}$$$$\Rightarrow \mathrm{the}\mathrm{chance}\mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}6\mathrm{is}$$$$1-\frac{625}{1296}=\frac{671}{1296}\approx 51.77\mathrm{\%}$$$$$$$$\mathrm{Now}\mathrm{the}\mathrm{chance}\mathrm{for}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{is}\frac{1}{2}$$$$\mathrm{for}1\mathrm{dice}\Rightarrow \mathrm{the}\mathrm{chance}\mathrm{for}no\mathrm{even}\mathrm{number}$$$$\mathrm{cinfusingly}\mathrm{is}\mathrm{also}\frac{1}{2}\Rightarrow no\mathrm{even}\mathrm{number}$$$$\mathrm{with}4\mathrm{dices}:$$$${\left(\frac{1}{2}\right)}^4=\frac{1}{16}$$$$\mathrm{At}\mathrm{least}\mathrm{one}\mathrm{even}\mathrm{number}$$$$1-\frac{1}{16}=\frac{15}{16}=93.75\mathrm{\%}$$

Commented by Tawa11 last updated on 23/Mar/25

$$\mathrm{If}\mathrm{it}\mathrm{is}\mathrm{possible}\mathrm{to}\mathrm{do}\mathrm{everything}\mathrm{as}\mathrm{a}$$$$\mathrm{question}.\mathrm{It}\mathrm{will}\mathrm{be}\mathrm{fine}.$$$$\mathrm{Find}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}$$$$\mathrm{even}\mathrm{numbers}.$$

Commented by Tawa11 last updated on 23/Mar/25

$$$$If two fair dice is thrown twice, what is the probability of obtaining an even number. (That is, sum of the numbers is even). This is what I think.

Commented by mr W last updated on 24/Mar/25

$$thrown4times,✓=sumiseven$$$$OOOO✓$$$$OOOE$$$$OOEO$$$$OOEE✓$$$$OEOO$$$$OEOE✓$$$$OEEO✓$$$$OEEE$$$$EOOO$$$$EOOE✓$$$$EOEO✓$$$$EOEE$$$$EEOO✓$$$$EEOE$$$$EEEO$$$$EEEE✓$$

Commented by mr W last updated on 24/Mar/25

$$onlyonenumberiseven:$$$$p=\frac{4}{16}=\frac{1}{4}$$$$atleastonenumberiseven:$$$$p=1-\frac{1}{16}=\frac{15}{16}$$$$allnumbersareeven:$$$$p=\frac{1}{16}$$

Commented by Tawa11 last updated on 24/Mar/25

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by Tawa11 last updated on 24/Mar/25

$$\mathrm{Thanks}\mathrm{sirs}\mathrm{for}\mathrm{your}\mathrm{time}.$$

Commented by mr W last updated on 25/Mar/25

$$aretheanswersright?$$

Commented by Tawa11 last updated on 25/Mar/25

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{answers}\mathrm{sir}.$$$$\mathrm{But}\mathrm{you}\mathrm{got}\mathrm{the}\mathrm{same}\mathrm{answer}$$$$\mathrm{as}\mathrm{sir}\mathrm{frix}.$$

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