Please help I don’t know how to do this

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 given 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

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