Probability

Question

An electronic gadget has 5 components of which two are of type-1 and other three are of type-2. Unless all components are functional the gadget will not be operative. During some interval probability of failure of each of components of type - 1 is 0.04 and the probability of failure of each of the components of type-2 is 0.03 what is the corresponding probability of failure of the gadget?

Answer 1

As mentioned in the question , the gadget will be defective if even one component is not working properly so its complementary event would be that the gadget is not defective , meaning that all components are working fine.Hence , 

P(defective) = 1 - P(not defective) 

                   = 1 - P(all components are fine)

So for all components are fine , we have to use multiplication principle , since here choice is not there and each of the components have to be fine and taken into consideration.

Hence p(not defective) = (1 - 0.04)² * (1 - 0.03)³ = 0.841(Since there are 2 balls of type having failure of each = 0.04 and 3 balls of type 2 having failure of each = 0.03)

Hence p(defective) = 1 - 0.841 = 0.159 = 0.16(approx) 

Answer 2

Find the probability of success first:

$P(\text{all 5 components are ok}) = (0.96)^2*(0.97)^3 = 0.841$

=> probability of failure = 1 -  $P(\text{all 5 components are ok})$ = $0.159$

Comments

QS is why not direct counting ? because it involves lots of combinations.