# Probability of choosing positive 2-digit integer with 4 in either place

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 $1/5$.

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 $\frac{18}{90} = \frac{1}{5}$.

• When you say $1/9 + (1 - 1/9)$, does the plus sign mean $or$, as in 4 is present = $1/9$ or 4 is NOT present = $(1 - 1/9)$ Jan 1, 2016 at 19:01
• The factor $(1-\frac{1}{9})$ is needed to discount case $44$ calculated already in $P(A)$. Actually event $A$ is made of $10$ cases, namely $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$, and event $B$ is made of $9$ cases: $14, 24, 34, 44, 54, 64, 74, 84, 94$. Now there are total $19$ cases, but we counted $44$ $2$ times, so we have $18$ cases. There are $90$ $2$ digit numbers, so $\frac{18}{90} = \frac{1}{5}$. Jan 1, 2016 at 19:05
• Can you add your last comment to the answer. Because that makes sense. It really does. I will mark as answer and it would help others. This is called deep understanding, which I wish to accomplish! Jan 1, 2016 at 19:10

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

• Nope, the phrasing at least one $4$ in the tens place or the units place (they could have just said at least one 4) allows for $44$. Jan 1, 2016 at 18:18
• @k-p I am trying to understand this. Where did you get $8/10$ Jan 1, 2016 at 18:20
• @Rhonda Edited, sorry
– K_P
Jan 1, 2016 at 18:22
• @K_P Ok let me try to soak in your answer Jan 1, 2016 at 18:23
• @K_P Honestly I am really confused by the $P(A∪B)$ Jan 1, 2016 at 18:27

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)

• Here is where I am confused. I thought $or$ means $+$, but you are multiplying, i.e. $8/9 * 9/10$ Jan 1, 2016 at 19:03
• I am clarifying in the answer itself for the benefit of all. Jan 1, 2016 at 19:07

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

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