Three houses of the same type were advertised to be let in a

Question

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

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

WoundedTiger · Accepted Answer

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

JusTLucK04 · Answer

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

PiyushK · Answer

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)

Bunuel · Answer

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.

Bunuel · Answer

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.

shaderon · Answer

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.

KarishmaB · Answer

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

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