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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

1) a car insurance company has determined that 8% of all ddrivers were involved in a car accident last year. among the 15 drivers living on one particular street, 3 were involved in a car accident last year. if 15 drivers are randomly selected, what is the porobabiltity of getting 3 or more who were involved in a car accident?

now my question is when it sais "if 15 drivers are randomly selected" does this mean the 15 drivers living on the particular street?? i assumed it is so i worked out the problem as follows:

P(x>=3) = P(3) + P(4) + ... + P(15) but there were only three drivers that were involved in an accident so P(4) = P(5) = ... = P(15) = 0

P(x>=3) = .08565 + 0 + 0 + ... + 0 = .08565

is my assumption and work correct???

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2) find the probabilty that 2 randomly selected people all have the same birthday. ignore leap years.

P(2 randomly selected people all have the same birthday) = 1/365

or should it be (1/365)^2???

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3) how many ways can an IRS auditor select 4 of 10 tax returns for an audit?

i'm not sure how they audit tax returns so i assumed they audit the first one, then the second one, then the third, and fourth, meaning order matters, so the total combinations is 10P4 = 5040

however, if they audit all the tax returns at the same time, meaning that order does not matter, then it should be 10C4. so which one is it???

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identify which of these types of sampling is used:

4) a market researcher selects 500 people from each of 10 cities.

a)systematic

b)stratified

c)convenience

d)cluster

e)random

there were many math tutors in my school that said it should be cluster. now the definition of cluster is to divide population into sections, then randomly select some of the sections and then choose ALL members from those sections. but the question doesn't say if those 500 people are ALL of the members in those 10 cities. can i make the assumption that it is???

thanks in advance!!!!

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

2) 1/365 1/365[sup]2[/sup] is the probability that the 2 persons' birthday happen to be on a particular day, say April 1st, consider all 365 cases in particular days in a year satisfy the problem, the total probability would be

1/1/365[sup]2[/sup] (Jan 1st)+ 1/1/365[sup]2[/sup] (Jan 2nd) +...+1/1/365[sup]2[/sup] (Dec 31st)=1/365

this is one way to solve. an simplier alternative is to equate the second person's birthday to the given first one's 1/365

1)among the 15 drivers living on one particular street, 3 were involved in a car accident last year--ignore this piece of information.(for samples too few to prove the existing 8% probability wrong)

thus 8% is the chance that one single driver of the random 15 is involved in a CA

Let x denotes the number of drivers amoung 15 happened to be involved in a CA

P(0)+P(1)+ ... +P(15)=1

P(0)=0.92[sup]15[/sup]

P(1)=C(1,15) 0.08 0.92[sup]14[/sup]

P(2)=C(2,15) 0.08² 0.92[sup]13[/sup]

C(1,15)=15 C(2,15)=15 14 / 2!

C(k,m)=def= Number of Ways to Select k from m

the prob should be 1-P(0)-P(1)-P(2)

for q4 , my advice is --Check out the other 4 definations!

**X'(y-Xβ)=0**

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