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..
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