Question

A box contains $15$ transistors, $5$ of which are defective. An inspector takes out one transistor at random, examines it for defects, and replace it. After it has been replaced another inspector does the same thing, and then so does a third inspector. The probability that at least one of the inspector 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 $\frac{19}{27}$

Let $A_i$ be the event that $i^{th}$ inspector finds a defective transistor.

Hence required probability is

$P(A_1\cup A_2\cup A_3)=1-P{(A_1\cup A_2\cup A_3)}^{\prime }$

$=1-P(A_1^{\prime }\cup A_2^{\prime }\cup A_2^{\prime })=1-P\left(A_1^{\prime }\right).P\left(A_2^{\prime }\right)P\left(A_3^{\prime }\right)$

$($ As $A_1,A_2,A_3$ are mutually independent events $)$

$=1-\frac{10}{15}.\frac{10}{15}.\frac{10}{15}=\frac{19}{27}(\because P\left(\overline{A_i}\right)=\frac{10}{15}=\frac{2}{3})$