Find the probability that a person had a car accident A car driver is selected at random to be a subject in the survey. The subject is given a six side dice with 4 green sides and 2 red sides and is given the following instructions

*

*Roll the dice

*If the outcome is green, say yes if the driver had a car accident last year and say no if no accidents

*If the outcome is red, say yes

a) Suppose that 30% of car drivers had an accident. If the subject says yes, what is the probability that the driver had a car accident last year?
b) Suppose we random sample 200 drivers and 70 of them say yes. What is the percentage of drivers had a car accident last year
If a subject gets green and says yes, then the subject had a car accident. For part a, I have
$$P(G|Y)=\frac{P(G\cap Y)}{P(Y)}=\frac{P(G\cap Y)}{P(Y|G)P(G)+P(Y|R)P(R)}$$
and I think  $P(G\cap Y)=P(Y|G)=.3$
I don't know how to do part b

Let $A$ denote the event that the driver has had an accident last year.
Let $Y$ denote the event that the driver says yes.
Let $R$ denote the event that a red side shows up.

On part a)
To be found is $P(A\mid Y)$.
Note that $A\subseteq Y$ or equivalently $A\cap Y=A$ so that: $$P(A\cap Y)=P(A)=\frac3{10}$$
Note that $A^{\complement}\cap Y=A^{\complement}\cap R$ so that by independence: $$P(A^{\complement}\cap Y)=P(A^{\complement}\cap R)=P(A^{\complement})P(R)=\frac7{10}\frac13=\frac7{30}$$
Then: $$P(A\mid Y)=\frac{P(A\cap Y)}{P(Y)}=\frac{P(A\cap Y)}{P(A\cap Y)+P(A^{\complement}\cap Y)}=\frac{\frac3{10}}{\frac3{10}+\frac7{30}}=\frac9{16}$$

on part b)
From:
$$\frac{70}{200}\approx P(Y)=P(A\cap Y)+P(A^{\complement}\cap Y)=P(A)+P(A^{\complement}\cap R)=P(A)+P(A^{\complement})P(R)=$$$$P(A)+(1-P(A))\frac13=\frac13+\frac23P(A)$$
we conclude that: $$P(A)\approx\frac1{40}=2.5\%$$
