Question

Lot $A$ consists of $3G$ and $2D$ articles. Lot $B$ consists of $4G$ $1D$ article. A new lot $C$ is formed by taking $3$ articles from $A$ articles from $A$ and $2$ form $B$ . The probability that an article chosen at random from $C$ is

• A. $1/3$
• B. $2/5$
• C. $8/25$
• D. none

Answer 1

Answer: (C)

$8/25$

A] A : event it is chosen from lot A

$P\left(a\right)=\frac{3}{5}$

$P\left(B\right)=\frac{2}{5}$

P (Defective) $=P\left(A\right)\times P\left(\frac{D}{A}\right)+P\left(B\right)\times P\left(\frac{D}{B}\right)$

$=\frac{3}{5}\times \frac{2}{5}+\frac{2}{5}\times \frac{1}{5}$

$=\frac{6}{25}+\frac{2}{25}$

$=\frac{8}{25}$

$\therefore P\left(D\right)=\frac{8}{25}$