Question

A certain item is manufactured by machine $ M_1 $ , and $ M_2 $ . It is known that machine $ M_1 $ turns out twice as many items as machine $ M_2 $ . It is also known that $ 4 \%$ of the items produced by machine $ M_1 $ , and $ 3 \ % $ of the items produced by machine $ M_2 $ are defective. All the items produced are put into one stock pile and then one item is selected at random. The probability that the selected item is defective is equal to

• A. $ \frac {10}{300} $
• B. $ \frac {11}{300} $
• C. $ \frac {10}{200} $
• D. $ \frac {11}{200} $

Answer 1

Answer: (B)

Let number of item manufactured by machine $M_1 =200$
According to the question, number of items manufactured by $M_1$ is twice of number of item manufactured by machine $M_2$.
$\therefore $ Number of item manufactured by machine
$M_2 = \frac{200}{2} = 100$
$\therefore 4 \%$ of the items produced by machine $M_1$ and $3\%$ of the items produced by machine $M_2$ are defective.
$\therefore $ Number of defective items produced by machine
$ M_1 = 200 \times 4\% = 200 \times \frac{4}{100} = 8$
and number of defective items produced by machine
$M_2 = 100 \times 3\% = 100 \times \frac{3}{100} = 3$
$\therefore $ Total number of items produced by machine $M_1$ and $M_2$
$=200 + 100 = 300$
and total number of defective items produced by $M_1$ and $M_2$
$ = 8 + 3 = 11$
$\therefore $ Probability that the selected item is defective
$ = \frac{\text{ Total number of defective items produced by machine $M_1$ and $M_2$}}{\text{Total number of items produced by machine $M_1$ and $M_2$}}$
$ =\frac{11}{300}$