Question

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

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

Answer 1

Answer: (C)

$8/25$

$P\left(D/A\right)$ denotes that a defective articles is from $A$

$P\left(D/A\right)=\frac{2}{5},P\left(D/B\right)=\frac{1}{5},P\left(A\right)=\frac{3}{5},P\left(B\right)=\frac{2}{5}$

$P\left(D\right)=P\left(A\right)P\left(D/A\right)+P\left(B\right)P\left(D/B\right)$

$P\left(D\right)=\frac{3}{5}\times \frac{2}{5}+\frac{2}{5}\times \frac{1}{5}$

$P\left(D\right)=\frac{6}{25}+\frac{2}{25}=\frac{8}{25}$