Answer:
1.) .4444444444 (or 4/9)
2.) .3333333333 (or 1/3)
3.) .09 (not too sure though)
4.) .31
5.)1
6.) .0153
Step-by-step explanation:
We want to find the probability of that a teenager driver getting a ticket <em>given</em> they ran a red light
the conditional probability formula is as follows (B|A)=(B∩A)/A
so let t= ticket for speeding
r= running a red light
(t|r)=(t∩r)/r
this is .04/.09 = .4444444444
2.) it's the same deal as number one
let t= ticket for speeding
r= running the red
we are looking for the probabilty of someone running a red light given that they were speeding
so we want (r|s)
which is .04/.12 = .33333333
3.) I'm not 100% sure on this one, I think the answer is just 9% though
4.) let x= probability of getting at least one
which means that
p(0)+p(x)=1
which means that
p(x)= 1-p(0)
Assuming they're independent, the probability of getting 0 in a pack is .97^12 which is .69
1-.69= .306157639 which rounds to .31
5.) this one is kind of the same deal
probability of getting at least 1=x
p(0)+p(x)=1
1-p(0)=p(x)
p(0)= .02^10= my calculator rounds this to 0 lol
1-0 = 1
6.) Solve this one using a bionomial distribution
we have
10C8*(.98)⁸*.02²= .0153
