If an essence is randomly selected indigenous all Positive 2-digit integers, wcap ins the probcapability the the essence liked has at least one 4 in ns tens area or the units place?

i understand probcapacity the gift in 10ns location ins \$1/9\$ and also the probability that gift in ns units location is \$1/10\$

as soon as i add \$1/9\$ + \$1/10\$ the prize ins \$19/90\$. However, prize says \$1/5\$.

You are watching: If an integer is randomly selected from all positive

A = \$4\$ in a ten place,B = \$4\$ in a unit place,C = in ~ least a \$4\$ in a ten or unit place.

\$P(A) = frac19\$, \$P(B) = frac110\$, \$P(C) = frac19 + (1-frac19)*frac110 = frac10 + 890 = frac15\$.

here girlfriend go:you have \$10\$ cases for occasion \$A\$: \$40, 41, 42, 43,44,45,46,47,48,49\$, and also \$9\$ cases because that occasion \$B\$: \$14, 24, 34, 44, 45, 46, 47, 48, 49\$. So when friend gain \$A cup B\$ girlfriend have actually no \$19\$ cases, but 18. Tright here are \$90\$ two-number numbers, so \$frac1890 = frac15\$.

you need to subtract \$P(A cans B) = 1 end 90\$ because that the probcapability that having 44 (ns provided \$ P(Acup B)=P(A)+P(B)-P(A cans B)\$ See: Probcapacity the Union/Interarea the 2 Events).

another way to resolve it ins come calculate \$1-P(no four)= 1 - (frac 89 frac 910) = 1 end 5\$

ns assume together Rhonda not an exclusive OR: pick indigenous \$1,..,9\$ for ns 10s and also \$,1,...,9\$ for the systems and also for this reason \$8 end 9\$ because that ns probcapacity not come have a 4 in ns tens and also \$9over 10\$ for no four in ns units

girlfriend to be right, \$0\$ have to not it is in allowed in the ten"s place,yet ns official price ins right, nevertheless.

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P(no \$4\$ in the 10s or devices place)\$ddaggerddagger\$ = \$dfrac89cdotdfrac910 = dfrac810= dfrac45\$

thus P(at leastern one \$4\$ in ns tens or units place) \$= 1 - dfrac45 = dfrac15\$

\$ddaggerddagger\$ ns expressitop top written in "normal" English in reality means

P(no \$4\$ in the tens location and no \$4\$ in ns systems place),or more simply, P(no \$4\$ in ns number)

P(AorB)=P(A) +P(B) _P(both A and B)

P(both A and B)= P(A) * P(B) occasions are not mutually exclusive..so P(A)= 1/9, P(B) = 9/90 = 1/9 + 9/90 _ 1/9*9/90= 1/5 answer..

extremely active question. Knife 10 call (not countinns the combination bonus) in order come answer thins question. Ns call need help defend this Concern from spam and non-price activity.

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