Question

A box contains 15 transistors, 5 of which are defective. An inspector takes out one transistor at random, examines it for defects, and replaces it. After it has been replaced another inspector does the same things, and then so does a third inspector. The probability that atleast one of the inspectors finds a defective transistor, is equal to

• A. $\frac{1}{27}$
• B. $\frac{8}{27}$
• C. $\frac{19}{27}$
• D. $\frac{26}{27}$

Answer 1

The correct option is C $\frac{19}{27}$

Probability of defective transistor =$\frac{5}{15}=\frac{1}{3}$ and probability of non defective transistor = $1-\frac{1}{3}$
$=\frac{2}{3}$
Probability that the inspectors find a non defective transistor
$=\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}$
Hence, probability that atleast one of the inspectors finds a defective transistor
$=1-\frac{8}{27}=\frac{19}{27}$