Question

A garage mechanic keeps a box of good springs to use as replacements on customers cars. The box contains $5$ springs. A colleague, thinking that the springs are for scrap, tosses three faulty springs into the box. The mechanic picks two springs out of the box while servicing a car. Find the probability that the second spring drawn is faulty.

• A. $\frac{1}{8}$
• B. $\frac{2}{8}$
• C. $\frac{3}{8}$
• D. $\frac{4}{8}$

Answer 1

Answer: (C)

$\frac{3}{8}$

Let A ={first spring chosen is faulty}, B ={second spring chosen is faulty}

Given:

Total number of springs in box = 8

Number of faulty springs = 3

To find:

The probability that the second spring drawn is faulty

$P\left(B\right)=P\left(B/A\right)P\left(A\right)+P\left(B/A^{\prime }\right)P\left(A^{\prime }\right)=\frac{2}{7}\times \frac{3}{8}+\frac{3}{7}\times \frac{5}{8}=\frac{21}{56}=\frac{3}{8}$