Question

A box contains 5 radio tubes of which 2 are defective.The tubes are tested one after the other until the 2 defective tubes are discovered .
Find the probability that the process stopped on the (i) Second test; (ii) third test, find the probability that the first tube is non-defective.

Answer 1

Solution -

P(Defected) $=\frac{2}{5}$ P(not defected)$=\frac{3}{5}$

P(Process stopped on second test) $=\frac{2}{5}\times \frac{2}{5}=\frac{4}{25}$

P(Process stopped on third test) $=\frac{3}{5}\times \frac{2}{5}\times \frac{2}{5}+\frac{2}{5}\times \frac{3}{5}\times \frac{2}{5}+\frac{2}{5}\times \frac{2}{5}\times \frac{3}{5}$

$=\frac{36}{125}$

P(first tube defective) $=\frac{2}{5}$