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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

Rhonda
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4 Answers4

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

Wojciech Karwacki
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  • 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 = $(1 - 1/9)$ – Rhonda Jan 01 '16 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 $\frac{18}{90} = \frac{1}{5}$. – Wojciech Karwacki Jan 01 '16 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! – Rhonda Jan 01 '16 at 19:10
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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

Community
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K_P
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  • 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 $44$. – Eli Rose Jan 01 '16 at 18:18
  • @k-p I am trying to understand this. Where did you get $8/10$ – Rhonda Jan 01 '16 at 18:20
  • @Rhonda Edited, sorry – K_P Jan 01 '16 at 18:22
  • @K_P Ok let me try to soak in your answer – Rhonda Jan 01 '16 at 18:23
  • @K_P Honestly I am really confused by the $P(A∪B)$ – Rhonda Jan 01 '16 at 18:27
  • See this link – K_P Jan 01 '16 at 18:29
  • You may want to say it this way. There are 9 possibilities for the first digit and 10 for the second digit, so $ \ 9 \cdot 10 \ = \ 90 \ $ two-digit numbers are permissible. There are 10 two-digit numbers of the form $ \ 4X \ $ and 9 of the form $ \ X4 \ $ (since $ \ 04 \ $ is not a two-digit number). However, we are counting $ \ 44 \ $ twice in this process, so we subtract the "double-counted" $ \ 44 \ $ , leaving $ \ 19 - 1 \ = \ 18 \ $ distinct two-digit numbers with at least one digit $ \ 4 \ $ out of $ \ 90 \ $ possibilities. – colormegone Jan 01 '16 at 18:35
  • @RecklessReckoner I tried to follow Rhonda's assumption which is not unreasonable. Interesting that they both give the same correct answer – K_P Jan 01 '16 at 18:44
  • I have no disagreement with your answer. I was suggesting an alternative description since she seemed unfamiliar with the set-theory way of notating this. (And the site has been doing something weird with recording votes that I can't fix now...) – colormegone Jan 01 '16 at 18:49
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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)

true blue anil
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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..

Amir
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