Probability with replacement


New Member

I have the following question which I'm really struggling with.

The data come from the records of an estate agent and relate to sales of private houses. A random sample of sales was taken from the records and the following table gives the frequency of each type of sale.

	1 bed	2 bed	3 bed	4 bed	Totals
Central	11	9	6	5	31
Suburbs	8	9	13	6	36
Totals	19	18	19	11	67
If two of the properties are chosen at random, with replacement, the probability that they have 7 bedrooms between them is (3 d.p.):

The correct answer is 0.093.

I've calculated
p(3) = 0.284
p(4) = 0.165

I assume since it has to be 8 bedroom, we're looking for the probability of a 3 bed plus a 4 bed?

I tried calculating it by using a variety of different methods:
- p(3)*p(4)=0.0466
- (p(3)*p(4))/p(4)=0.283
- (p(3)*p(4))/p(3)=0.164
- p(3)+p(4)=0.447

I really cannot get my head around this. I've not been able to come anywhere near the correct answer, and I'm really struggling.

I'd be grateful if anyone could assist (even if you're just pointing me in the right direction)

Thanks in advance.


New Member
HI, a newbie here but I think this might be the answer:

There are C( 67,2 ) = 2211 ways to choose a pair of properties from the 67 listed.

The only pairs that will give you seven bedrooms are if you choose one of the 19 3BR properties, and one of the 11 4BR properties, so you have 19*11 = 209 pairs.

Therefore the probability is 209/2211 = 0.0945. I can't account for the 0.0015 difference.
p(3) = 0.284
p(4) = 0.165

actually p(4)=11/67=0.164

There are two ways to pick 7 bedrooms, pick a 3 bedroom then a 4 bedroom, or pick a 4 bedroom then a 3 bedroom.

2*p(3)*p(4) = 2*(11/67)*(19/67) = 0.093


New Member
Wow, I cannot believe how simple it was! I've been stressing over this for ages, and I was pretty close.

Thank you so much!