Example — Driving test

Example — Driving test

The probability of passing the driving test is $30\%$ for those who have taken at least $10$ lessons, and $10\%$ for those who have taken fewer than $10$. Among those who passed the test, those who took at least $10$ lessons are $50\%$. What is the percentage of people who take few lessons?

Let us define $A$ = “passing the test”, $B$ = “having taken at least $10$ lessons”. Data:

$P(A\mid B) = 0,3, \quad P(A\mid \overline{B}) = 0,1, \quad P(B\mid A) = 0,5.$

Requested: $P(\overline{B}) = 1-P(B)$.

By the law of compound probability, $P(A\cap B)$ can be written in two ways:

$P(A\cap B) = P(B)\cdot P(A\mid B) = 0,3\,P(B),$

$P(A\cap B) = P(A)\cdot P(B\mid A) = 0,5\,P(A).$

Equating: $0,3\,P(B) = 0,5\,P(A)$, that is $P(A) = 0,6\,P(B)$.

On the other hand, using the decomposition of $P(A)$ over the two cases (lessons $\ge 10$ and lessons $<10$):

$P(A) = P(A\cap B) + P(A\cap\overline{B}) = 0,3\,P(B) + 0,1\,(1-P(B)) = 0,2\,P(B) + 0,1.$

Combining with $P(A) = 0,6\,P(B)$:

$0,6\,P(B) = 0,2\,P(B) + 0,1 \implies 0,4\,P(B) = 0,1 \implies P(B) = \boxed{0,25}.$

So $25\%$ of the people take at least $10$ lessons, and $\boxed{75\%}$ take fewer than $10$.

Check via table: $P(A) = 0,6\cdot 0,25 = 0,15$ (that is, $15\%$ pass the test, a value intermediate between $30\%$ and $10\%$, reasonable). The contingency table is filled in as:

	$A$ (passed)	$\overline{A}$ (failed)	Total
$B$ ($\ge 10$ lessons)	$0,075$	$0,175$	$0,25$
$\overline{B}$ ($<10$ lessons)	$0,075$	$0,675$	$0,75$
Total	$0,15$	$0,85$	$1$

All the constraints are satisfied ✓.

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Topics: Probability
Concepts: Product formula · Conditional probability · Total probability
Methods: Conditional probability · Total probability
Skills: Probability calculation · Solving equations