Confusion about conditional probability and independent events

Question

Insurance company has $85$% regular customers who in a year have an accident with probability $0.01$ and $15\%$ new customers who have an accident with probability $0.4$ Probability that given client will have an accident is invariant for him, no matter if he had an accident before or not. One can compute that probability that randomly chosen client will have an accident is equal to $0.0685$ (which I was able to do) and probability that he will have second accident if we know he had first is equal to $0.3516$ (**which I can't compute). Model assumes invariance of probability, yet first accident increases chances of second. Why?

Answer in my textbook is:

It is effect of draw. Appearance of an accident has no impact on probability, but is a hint, that randomly chosen person has higher tendency to accidents.

(**)I thought that since second accident is not depended on first the probability of second accident given that the first accident happened, using conditional probability should be:

$$P(\rm second|\rm first)=\frac{P(second \cap first)}{P(first)}=\frac{P(\rm second)P(first)}{P(first)}=P(second)=P(first)=0.0685$$

I know that my textbook probably explained why it is the case but I don't understand it.

Answer 1

Let $E_1$ denote the event that the client had at least one accident.

Let $E_2$ denote the event that the client had a second accident, given that they had at least one accident.

You are being asked to compute

$$p(E_2 ~| ~E_1) = \frac{p(E_1 \cap E_2)}{p(E_1)}. \tag1 $$

To justify the formula in (1) above, you can consult Conditional Probability. The related article Bayes' Theorem may also be helpful.

To compute $p(E_1 \cap E_2)$, you have to sum the probabilities of two mutually exclusive events:

• Old client had two accidents:
  
  Probability is $~\displaystyle (0.85) \times (0.01)^2.$

• New client had two accidents:
  
  Probability is $~\displaystyle (0.15) \times (0.4)^2.$

Therefore,

$$p(E_1 \cap E_2) = \left[ ~(0.85) \times (0.01)^2 ~\right] + \left[ ~(0.15) \times (0.4)^2 ~\right] = 0.024085.$$

Similarly, as the original poster indicated in his posting,

$$p(E_1) = \left[ ~(0.85) \times (0.01) ~\right] + \left[ ~(0.15) \times (0.4) ~\right] = 0.0685.$$

Therefore,

$$p(E_2 ~| ~E_1) = \frac{0.024085}{0.0685} \approx 0.3516.$$

---

As has already been discussed in the comments, the dramatic increase in probability, from $~0.0685~$ to $~0.3516~$ is intuitively explained by reasoning that a specific driver who has an accident is far more likely to be a new driver.

In fact, (as Lulu has already indicated in the comments):

Let $~E_3~$ denote the event that the driver had an accident.
Let $~E_4~$ denote the event that the driver is new.

Then the probability that the driver who had an accident is new is

$$p(E_4 ~| ~E_3) = \frac{p(~E_4 \cap E_3)}{p(E_3)} \\ = \frac{0.15 \times 0.4}{\left[ ~(0.85) \times (0.01) ~\right] + \left[ ~(0.15) \times (0.4) ~\right]} = \frac{0.06}{0.0685} \approx 0.8759. \tag2 $$

Another (intuitive) way of explaining the final computation in (2) above, which indicates that drivers in accidents are about $7$ times more likely to be new, rather than old is:

The approximately $6$ to $1$ ratio of old drivers to new is radically offset by the fact that one random new driver is $(40)$ times more likely to have an accident than one random old driver.

Answer 2

(**)I thought that since second accident is not depended on first the probability of second accident given that the first accident happened, using conditional probability should be:

You are dealing with conditional independence. The probability for a $S$econd accident is independent of the $F$irst for a given client. Thus you have to consider whether the client is $N$ew or $\mathcal O$ld, using the Law of Total Probability.

"... that given client will have an accident is invariant for him, no matter if he had an accident before or not."

$\qquad\begin{align}\mathsf P(S\mid F) &=\dfrac{\mathsf P(S,F)}{\mathsf P(F)} \\[1ex] &= \dfrac{\mathsf P(S, F\mid N)\,\mathsf P(N)+\mathsf P(S, F\mid\mathcal O)\,\mathsf P(\mathcal O)}{\mathsf P(F\mid N)\,\mathsf P(N)+\mathsf P(F\mid \mathcal O)\,\mathsf P(\mathcal O)}\\[1ex]&= \dfrac{\mathsf P(S\mid N)\,\mathsf P(F\mid N)\,\mathsf P(N)+\mathsf P(S\mid\mathcal O)\,\mathsf P(F\mid \mathcal O)\,\mathsf P(\mathcal O)}{\mathsf P(F\mid N)\,\mathsf P(N)+\mathsf P(F\mid \mathcal O)\,\mathsf P(\mathcal O)}\\[1ex]&= \dfrac{0.4^2\cdot0.15+0.01^2\cdot0.85}{0.4\cdot0.15+0.01\cdot 0.85}\end{align}$