2
$\begingroup$

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

  1. Roll the dice
  2. If the outcome is green, say yes if the driver had a car accident last year and say no if no accidents
  3. 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

$\endgroup$
2
  • $\begingroup$ The question is ambiguous. I suspect it is intended that the driver will respond "yes" if and only if they had an accident and the outcome is green. The ambiguity arises from the absence of any indication here of what the response should be if the driver had no accident and the outcome is green. $\endgroup$
    – whuber
    Sep 28, 2022 at 17:37
  • 1
    $\begingroup$ @whuber there was a typo in the previous edition, I corrected it $\endgroup$
    – Simple
    Sep 28, 2022 at 17:44

2 Answers 2

1
$\begingroup$

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: $$P(A)\approx\frac1{40}=2.5\%$$

$\endgroup$
0
$\begingroup$

One third of drivers got red, and all of them said "yes", so that accounts for 66.6 of the "yes". That leaves 3.3 people, who must then be drivers who got green and answered "yes", out of a total of 133.3 (two thirds of the 200 got green). So there were 133.3 people who got green, and 3.3 of them said "yes". 3.3/133.3 = 2.5%.

This make various assumptions of independence and that observed frequencies represent underlying probabilities, and of course it ignores the fact there are fractional people.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.