Question

3.48 An apartment seeker in Manhattan estimates that 10% of the available apartments in her price range are in acceptable condition. Furthermore, she has time to look at only one apartment every weekend. a. How many weeks should this person expect to spend for apartment hunting? b. What is the probability that she will find an acceptable apartment in the first two weekends that she looks? c. What is the probability that it will take more than 4 months (17 weeks) to find an acceptable apartment?

Answer 1

Answer:

a) 10 weeks

b) P(k≤2)=0.190

c) P(k>17)=0.167

Step-by-step explanation:

This problem can be solved applying the geometric distribution.

The probability of needing k weeks to find an acceptable apartment can be written as:

$P(k)=(1-p)^{k-1}p=0.9^{k-1}\cdot 0.1$

a) The expected value of this distribution is:

$E(x)=1/p=1/0.1=10$

Answer: 10 weeks

b) The probability of finding an acceptable department in the first 2 weeks is:

$P(k\leq2)=P(1)+P(2)=0.1*0.9^0+0.1*0.9^1\\\\P(k\leq 2)=0.1+0.09=0.19$

c) We have to calculate P(k>17)

$P(k>17)=1-\sum_1^{17}P(k_i)\\\\P(k>17)=1-(P(1)+P(2)+P(3)+...+P(17))\\\\P(k>17)=1-(0.1+0.09+0.081+0.073+0.066+0.059+0.053+0.048+0.043+0.039+0.035+0.031+0.028+0.025+0.023+0.021+0.019)\\\\P(k>17)=1-0.833=0.167$