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555-605 Level|   Probability|      
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PiyushK
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Three houses of the same type were advertised to be let in a Updated on: 19 May 2014, 04:19
Originally posted by PiyushK on 19 May 2014, 04:07.
Last edited by PiyushK on 19 May 2014, 04:19, edited 1 time in total.
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A
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25% (medium)

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74% (01:27) correct 26% (02:00) wrong based on 167 sessions

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Three houses of the same type were advertised to be let in a locality. Three men made separate applications for a house. What is the probability that each of them applied for the same house.

A. 3/27
B. 2/9
C. 2/3
D. 3/511
E. 5/511

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What is the probability that two of them applied for the same house and third for one of the other house ?
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A
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WoundedTiger
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Three houses of the same type were advertised to be let in a 19 May 2014, 04:38
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PiyushK
Three houses of the same type were advertised to be let in a locality. Three men made separate applications for a house. What is the probability that each of them applied for the same house.

1) 3/27
2) 2/9
3) 2/3
4) 3/511
5) 5/511

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What is the probability that two of them applied for the same house and third for one of the other house ?


Since each of the house has same probability of getting selected so for each men proability of selecting a house out of 3 houses 1/3

Let us consider case where all 3 men choose House no 1
So for First men it will be 1/3,second men 1/3 and third men also 1/3

So probability of selecting house no 1 is 1/3*1/3*1/3= 1/27
And since there are 3 such house so probability that each men select the same house is 3*1/27 or 1/9

Ans is A
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Three houses of the same type were advertised to be let in a Updated on: 19 May 2014, 09:35
Originally posted by JusTLucK04 on 19 May 2014, 09:26.
Last edited by JusTLucK04 on 19 May 2014, 09:35, edited 1 time in total.
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P of Choosing any house by Mr.A is 1
P of Choosing the same house by Mr.B is 1/3
P of Choosing the same house by Mr.C is 1/3

P= 1*1/3*1/3 = 1/9


For the spoiler:
P of Choosing A house by Mr.A is 1
P of Choosing the same house by Mr.B is 1/3
P of Choosing one of the other 2 houses by Mr.C is 2/3

=2/9

Same applies for B and C
= 2/9 + 2/9 + 2/9
=2/3
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Three houses of the same type were advertised to be let in a 19 May 2014, 04:19
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Hint for the second question in the spoiler.
P(Two applied to the same house and third to other) = 1- P(all applied to same house)- P(all applied to different house)
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Three houses of the same type were advertised to be let in a 19 May 2014, 05:05
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PiyushK
Three houses of the same type were advertised to be let in a locality. Three men made separate applications for a house. What is the probability that each of them applied for the same house.

A. 3/27
B. 2/9
C. 2/3
D. 3/511
E. 5/511

Show Spoiler
What is the probability that two of them applied for the same house and third for one of the other house ?

Each men can choose any of the three houses, hence the total number of ways they can apply is 3^3 = 27.

Each of them to apply for the same house means each of them applying for house A, B or C.

P = 3/27.

Answer: A.

As for your second question: What is the probability that two of them applied for the same house and third for one of the other house?

\(P = \frac{C^1_3*C^2_3*2}{3^3}=\frac{2}{3}\), where:
\(C^1_3\) is the number of ways of selecting which out of three houses gets 2 votes;
\(C^2_3\) is the number of ways of selecting which out of three men will vote for that house;
2 is the number of ways for the third man to choose a house from 2 remaining.

If we use the approach suggested by you here: three-houses-of-the-same-type-were-advertised-to-be-let-in-a-171474.html#p1365820 it would be P = 1 - 3/27 - 3!/27 = 2/3.

Hope it's clear.
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Three houses of the same type were advertised to be let in a 19 May 2014, 09:33
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JusTLucK04
P of Choosing any house by Mr.A is 1
P of Choosing the same house by Mr.B is 1/3
P of Choosing the same house by Mr.C is 1/3

P= 1*1/3*1/3 = 1/9

What do you mean by the hint in the second post?

There is another question under the spoiler in the original post: What is the probability that two of them applied for the same house and third for one of the other house? The hint is for this question. There is a solution for it in my post above.
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Three houses of the same type were advertised to be let in a 20 May 2014, 20:46
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What is the probability that two of them applied for the same house and third for one of the other house?

Is there a way to do this question using only Probabilities approach?

For example, in the first question:

Probability of selecting house 1 = (1/3)*(1/3)*(1/3) = (1/27)
This is then multiplied by 3, for 3 houses.

How would it be done without using 1 - P() method.

I just want to understand different possible approaches.

Thanks for your help.
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Three houses of the same type were advertised to be let in a 20 May 2014, 22:35
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shaderon
What is the probability that two of them applied for the same house and third for one of the other house?

Is there a way to do this question using only Probabilities approach?

For example, in the first question:

Probability of selecting house 1 = (1/3)*(1/3)*(1/3) = (1/27)
This is then multiplied by 3, for 3 houses.

How would it be done without using 1 - P() method.

I just want to understand different possible approaches.

Thanks for your help.

Responding to a pm:

Yes, but I would suggest you to stick to the method given by Bunuel.
There is a high probability of missing something using probability approach when number of cases is high.

Houses H1, H2, H3
People A, B, C
A and B can apply for H1 with probability (1/3)*(1/3) and C can apply for the other 2 houses with a probability of (2/3)
Probability = (1/3)*(1/3)*(2/3)
But A and B could apply for H2 or H3 as well so Probability = (1/3)*(1/3)*(2/3)*3
Also, instead of A and B, A and C or B and C could apply for the same house so Probability = (1/3)*(1/3)*(2/3)*3*3 = 2/3
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Three houses of the same type were advertised to be let in a 26 Nov 2024, 17:29
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