A package contains 11 resistors, 2of which are defective. If 4 are selected find the probability of getting the following results.P(1 defective)=

Question

asked Mar 10, 2023 214k views

1 Answer

NicolasR · Answer 1 · 2023-03-16T04:39:11+0000

From the given values, we know there are 2 defective resistors and 9 not defective. then, we have

$P(1\text{ defective)=}(C^9_3* C^2_1)/(C^(11)_4)$

where

denote the combinations of 9 elements taking in 3, that is 3 good resistor from the 9 which are not defective. Similarly,

denotes the combinations of 2 elements taking in 1, that is, 1 defective resistor form the 2 defective resistors.

Then, we have

$P(1\text{ defective)=}(84*2)/(330)$

then, the answer is

$P(1\text{ defective) =}0.509$