GRE study exam guide has following
If an integer is randomly selected from all positive 2-digit integers, what is the probability that the integer chosen has at least one 4 in the tens place or the units place?
I understand probability of being in 10s place is $1/9$ and the probability of being in the units place is $1/10$
When I add $1/9$ + $1/10$ the answer is $19/90$. However, answer says $1/5$.
Please explain
On
You have to subtract $P(A \cap B) = {1 \over 90}$ for the probability of having 44 (I used $ P(A\cup B)=P(A)+P(B)-P(A \cap B)$ See: Probability Of Union/Intersection Of Two Events).
Another way to solve it is to calculate $1-P(no \ four)= 1 - (\frac {8}{9} \frac {9}{10}) = {1 \over 5}$
I assume as Rhonda not an exclusive OR: Pick from $\{1,..,9\}$ for the tens and $\{0,1,...,9\}$ for the units and so $8 \over 9$ for the probability not to have a four in the tens and $9\over 10$ for no four in the units
On
You are right, $0$ should not be allowed in the ten's place,
but the official answer is right, nevertheless.
P(no $4$ in the tens or units place)$\ddagger\ddagger$ = $\dfrac89\cdot\dfrac9{10} = \dfrac8{10}= \dfrac45$
Thus P(at least one $4$ in the tens or units place) $= 1 - \dfrac45 = \dfrac15$
$\ddagger\ddagger$ The expression written in "normal" English actually means
P(no $4$ in the tens place and no $4$ in the units place),
or more simply, P(no $4$ in the number)
A = $4$ in a ten place, B = $4$ in a unit place, C = at least one $4$ in a ten or unit place.
$P(A) = \frac{1}{9}$, $P(B) = \frac{1}{10}$, $P(C) = \frac{1}{9} + (1-\frac{1}{9})*\frac{1}{10} = \frac{10 + 8}{90} = \frac{1}{5}$.
Here you go: You have $10$ cases for event $A$: $40, 41, 42, 43,44,45,46,47,48,49$, and $9$ cases for event $B$: $14, 24, 34, 44, 45, 46, 47, 48, 49$. So when you get $A \cup B$ you have not $19$ cases, but 18. There are $90$ two-digit numbers, so $\frac{18}{90} = \frac{1}{5}$.