Question

The items produced by a firm are supposed to contain 5% defective items. The probability that a sample of 8 items will contain less than 2 defective items, is

• A. $\frac { 27 } { 20 } \left( \frac { 19 } { 20 } \right) ^ { 7 }$
• B. $\frac { 533 } { 400 } \left( \frac { 19 } { 20 } \right) ^ { 6 }$
• C. $\frac { 153 } { 20 } \left( \frac { 1 } { 20 } \right) ^ { 7 }$
• D. $\frac { 35 } { 16 } \left( \frac { 1 } { 20 } \right) ^ { 6 }$

Answer 1

Answer: (A)

$\frac { 27 } { 20 } \left( \frac { 19 } { 20 } \right) ^ { 7 }$

Answer 2

Required probability

= ${ } ^ { 8 } \mathrm { C } _ { 1 } \left( \frac { 1 } { 20 } \right) ^ { 1 } \left( \frac { 19 } { 20 } \right) ^ { 7 } + { } ^ { 8 } \mathrm { C } _ { \mathrm { o } } \left( \frac { 1 } { 20 } \right) ^ { 0 } \left( \frac { 19 } { 20 } \right) ^ { 8 }$ = $\frac { 27 } { 20 } \left( \frac { 19 } { 20 } \right) ^ { 7 }$