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Intern  B
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A building has $!48$! one- and two-bedroom apartments to rent, some wi  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 52% (02:05) correct 48% (02:07) wrong based on 71 sessions

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A building has 48 one- and two-bedroom apartments to rent, some with balconies and others without. If one of these apartments is to be selected at random, is the probability greater than 1/2 that the apartment has two bedrooms and a balcony?

(1) Of the apartments, 40 have two bedrooms.
(2) Of the apartments, 30 have a balcony.

I have a question, pls help.
How the answer is C?
Why the formula of independent event can be applied?
the probability that the apartment has two bedrooms and a balcony is 40/48 * (30/48) . How?

Originally posted by skidstorm on 15 Mar 2018, 13:36.
Last edited by Bunuel on 15 Mar 2018, 22:49, edited 1 time in total.
Edited the OA.
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Re: A building has $!48$! one- and two-bedroom apartments to rent, some wi  [#permalink]

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skidstorm wrote:
A building has 48 one- and two-bedroom apartments to rent, some with balconies and others without. If one of these apartments is to be selected at random, is the probability greater than 1/2 that the apartment has two bedrooms and a balcony?

(1) Of the apartments, 40 have two bedrooms.
(2) Of the apartments, 30 have a balcony.

chetan2u, mikemcgarry, Bunuel, VeritasPrepKarishma, amanvermagmat
JeffTargetTestPrep
I have a question, pls help.
How the answer is C?
Why the formula of independent event can be applied?
the probability that the apartment has two bedrooms and a balcony is 40/48 * (30/48) . How?

Ans will not be C, it will be E..

lets see combined..
Total = 48
(1) Of the apartments, 40 have two bedrooms.... so 8 have 1-bedroom
(2) Of the apartments, 30 have a balcony..... so 18 do not have a balcony

take extremes..
1) All 8 1-bedroom have a balcony, so 2-bedroom apartments that have balcony = 30-8 = 22
Probability = $$\frac{22}{48}<\frac{1}{2}$$
2) All 30 having balcony are subset of 2-bedrooms that is all have 2 bedrooms...
probability $$= \frac{30}{48}>\frac{1}{2}$$

Insuff
E
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Re: A building has $!48$! one- and two-bedroom apartments to rent, some wi  [#permalink]

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skidstorm wrote:
A building has 48 one- and two-bedroom apartments to rent, some with balconies and others without. If one of these apartments is to be selected at random, is the probability greater than 1/2 that the apartment has two bedrooms and a balcony?

(1) Of the apartments, 40 have two bedrooms.
(2) Of the apartments, 30 have a balcony.

chetan2u, mikemcgarry, Bunuel, VeritasPrepKarishma, amanvermagmat
JeffTargetTestPrep
I have a question, pls help.
How the answer is C?
Why the formula of independent event can be applied?
the probability that the apartment has two bedrooms and a balcony is 40/48 * (30/48) . How?

Hello

Chetan has already replied. I will still explain the same thing which Chetan did with the help of a table. Its in the attached pic.

I have assumed '1-bed with balcony' to be 'x'. This makes '1-bed no balcony' to be '8-x' and '2-bed with balcony' to be '30-x' and finally '2-bed no balcony' to be 'x+10'.

Looking at the four values: 'x', '8-x', '30-x' and 'x+10' we can conclude that MIN value of x will be '0' and MAX value of x can be '8' (because if x is more than 8 then '8-x' will become negative which is not possible). So the '2-bed with balcony' houses which are '30-x' can go from 30-0 (30) to 30-8 (22). But 1/2 of 48 (total houses) is 24. Here possibilities are less than 24 (22, 23), equal to 24 and even greater than 24 (25, 26..30). So we cannot answer the question with certainty.

Both statements together are also not sufficient. Hence E answer
Attachments IMG_20180316_110354.jpg [ 18.49 KiB | Viewed 635 times ]

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Re: A building has $!48$! one- and two-bedroom apartments to rent, some wi  [#permalink]

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Here is the full "GMAT Jujitsu" for this question:

First, there is a classic trap embedded in this question. The problem tricks you into thinking that you have two "separate" probabilities: (1) a 40 out of 48 chance that an apartment will have two bedrooms and (2) a 30 out of 48 chance an apartment will have a balcony. The trap is to get you to think that we simply need to multiply these probabilities by each other to calculate the combined probability. ($$\frac{40}{48}*\frac{30}{48}=\frac{1200}{2304}>50\%$$) This would result in answer choice C, which is NOT the right answer. The problem with this approach is that the probabilities are NOT independent of each other, but are instead overlapping. You can't just multiply probabilities together in this case.

In reality, we have two mutually exclusive groups (2-bedrooms vs. 1-bedroom and balconies vs. no balconies.) If you can organize your thoughts around these mutually-exclusive groups by creating a specialized table, the solution to this problem naturally unfolds. (I call a such tables "Matrix Boxes" in my classes.)

Here is what the blank table might look like:
$\begin{matrix} & \text{Balconies} & \text{No Balconies} & \text{Totals} \\ \text{1-Bedroom} & \text{____} & \text{____} & \text{____} \\ \text{2-Bedrooms} & \text{____} & \text{____} & \text{____} \\ \text{Totals} & \text{____} & \text{____} & \text{____} \end{matrix}$
The question tells us that there are 48 total apartments. Let's start there:
$\begin{matrix} & \text{Balconies} & \text{No Balconies} & \text{Totals} \\ \text{1-Bedroom} & \text{____} & \text{____} & \text{____} \\ \text{2-Bedrooms} & \text{____} & \text{____} & \text{____} \\ \text{Totals} & \text{____} & \text{____} & \text{48} \end{matrix}$
Statement #1 tells us that 40 of the 48 apartments have 2-bedrooms and 30 of the 48 have balconies. Since both of these numbers are more than 24 (half of the total 48), these statements are insufficient by themselves. After all, we want to have sufficient information to determine if the probability of getting both 2-bedrooms and a balcony is greater than 50%. If either of the two conditions were less than 50% of the apartments, then we would have our answer immediately. (In this hypothetical case, there would be no way to have more than one answer to the question, and the statements would be sufficient.) Since it is possible to have greater than 50%, but it is also possible to have less than 50%, we can eliminate answers A, B, and D.

Once we combine the statements, here is what our Matrix Box would look like:
$\begin{matrix} & \text{Balconies} & \text{No Balconies} & \text{Totals} \\ \text{1-Bedroom} & \text{____} & \text{____} & \text{8} \\ \text{2-Bedrooms} & \text{____} & \text{____} & \text{40} \\ \text{Totals} & \text{30} & \text{18} & \text{48} \end{matrix}$
And yet, we still do not know the numbers on the inside of the table. With this information, it is possible that all of the 2-bedroom apartments have balconies, resulting in this combination:
$\begin{matrix} & \text{Balconies} & \text{No Balconies} & \text{Totals} \\ \text{1-Bedroom} & \text{0} & \text{8} & \text{8} \\ \text{2-Bedrooms} & \text{30} & \text{22} & \text{40} \\ \text{Totals} & \text{30} & \text{18} & \text{48} \end{matrix}$
The chance of having a 2-bedroom apartment with a balcony would therefore be $$\frac{30}{48}=\frac{5}{8}$$ > $$50\%$$.
But it is also possible that all of the 1-bedroom apartments have balconies, resulting in this combination:
$\begin{matrix} & \text{Balconies} & \text{No Balconies} & \text{Totals} \\ \text{1-Bedroom} & \text{8} & \text{0} & \text{8} \\ \text{2-Bedrooms} & \text{22} & \text{18} & \text{40} \\ \text{Totals} & \text{30} & \text{18} & \text{48} \end{matrix}$
In this case, the chance of having a 2-bedroom apartment with a balcony would therefore be $$\frac{22}{48}=\frac{11}{24}$$ < $$50\%$$.

Since we can have two answers to the question "is the probability greater than 1/2 that the apartment has two bedrooms and a balcony?", the answer is E.
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Aaron J. Pond
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Re: A building has $!48$! one- and two-bedroom apartments to rent, some wi  [#permalink]

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Thanks AaronPond for you sufficient and detailed explanation. Re: A building has $!48$! one- and two-bedroom apartments to rent, some wi   [#permalink] 16 Mar 2018, 15:51
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