Question

In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. Then the probability that the bolt drawn is defective is

• A. 0.0345
• B. 0.345
• C. 3.45
• D. 0.0034

Answer 1

Answer: (A)

0.0345

Answer 2

Let $E _ { 1 } , E _ { 2 } , E _ { 3 }$ and A be the events defined as follows:

$E _ { 1 } =$ the bolts is manufactured by machine A; $E _ { 2 } =$ the

bolts is manufactured by machine B; $E _ { 3 } =$ the bolts is

manufactured by machine C, and A = the bolt is defective.

Then $P \left( E _ { 1 } \right) = \frac { 25 } { 100 } = \frac { 1 } { 4 } , P \left( E _ { 2 } \right) = \frac { 35 } { 100 } , P \left( E _ { 3 } \right) = \frac { 40 } { 100 }$ .

$P \left( A / E _ { 1 } \right) =$ Probability that the bolt drawn is defective given the condition that it is manufactured by machine A = 5/100.

Similarly $P \left( A / E _ { 2 } \right) = \frac { 4 } { 100 }$ and $P \left( A / E _ { 3 } \right) = \frac { 2 } { 100 }$ .

Using the law of total probability, we have

$= \frac { 25 } { 100 } \times \frac { 5 } { 100 } + \frac { 35 } { 100 } \times \frac { 4 } { 100 } + \frac { 40 } { 100 } \times \frac { 2 } { 100 } = 0.0345$.