# Conditional Probability: Verification

I'm having doubts that my answer is correct and wondered if someone here could clarify. It is a relatively basic question but I have doubts!

60% of new drivers in a particular country have had additional driving education. During their first year of driving, new drivers who have $\mathit{not}$ had additional driving education have a probability of 0.12 of having an accident, while new drivers who $\mathit{have}$ had additional driving instruction have a probability of 0.03 of having an accident. Calculate the probability that a new driver does not have an accident during their first year of driving.

My thoughts are thus;

Let $N_e$ be the event that a new driver has had additional instruction and $N_n$ be the event that the driver has $\mathit{not}$ had additional instruction and let $A$ be the event that a driver has an accident. Then,

$$P(A|N_n) = 0.12$$ and $$P(A|N_e) = 0.03.$$

Then the total probability of having an accident is

$$P(A|N_e) + P(A|N_n) = 0.15.$$

As such, the probability of a new driver $\mathit{not}$ having an accident is the compliment of this. That is,

$$P(A^c) = 1 - 0.15 = 0.85.$$

Am I completely mistaken?

Thanks!

• You need to fix one thing: the Law of Total Probability says that $P(A)=P(A|N_e)P(N_e)+P(A|N_n)P(N_n)$. Mar 23 '16 at 16:58
• I see. After having corrected this, would the $P(A^c)$ step be correct? Mar 23 '16 at 17:00
• Yes, that step is fine. Mar 23 '16 at 17:01
• Perfect, thanks for your help. Mar 23 '16 at 17:01

Just to give an answer to this question for future reference.

Using the notation given above,

\begin{align} P(A) &= P(A|N_e)P(N_e) + P(A|N_n)P(N_n)\\ &= (0.03 \times 0.6) + (0.12 \times 0.4)\\ &= \frac{33}{500}. \end{align}

Hence,

$$P(A^c) = 1 - \frac{33}{500} = \frac{467}{500} \approx 93\%.$$

$P(A|N_n) = P(A\cap N_n)/P(N_n)$, so $P(A\cap N_n) = 0.12/0.4 = 0.3$

$P(A|N_e) = P(A\cap N_e)/P(N_e)$, so $P(A\cap N_e) = 0.03/0.6 = 0.05$

So $P(A) = P(A\cap N_n) + P(A\cap N_e) = 0.35$

Hence, $P(A^c) = 0.65$

• The probability of an accident is $P(A) = P(A \mid N_e)P(N_e) + P(A \mid N_n)P(N_n)$. Mar 23 '16 at 23:39