There are four machines and it is known that exactly
two of them are faulty. They are tested, one by one,
in a random order till both the faulty machines are
identified. Find the probability that exactly $3$ tests will
be required to identify the $2$ faulty machines.
$a.)\ \dfrac{1}{2} \\
b.)\ 1 \\
\color{green}{c.)\ \dfrac{1}{3}} \\
d.)\ \dfrac{2}{3} $
I did $\dbinom{3}{2} \times \dfrac{2}{4} \times \dfrac{1}{3}= \dfrac{1}{2} $
But the answer given is option $c.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
Best Answer
The probability for faulty-not faulty-faulty is
$\frac{1}{2}\times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The probability for not faulty-faulty-faulty is
$\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The sum is $\frac{1}{3}$.