Question

suppose 3 companies x y z produce tvs x pproduces twice as many as x while y and z produce same number as.it is known that 2%of x,2%of y,4%of z are defected all the tvs are produced are put into 1 shop and then 1 tv is choosen at random what is the probability that the tv is defeceted.suppppose a tv choosen is defective what is the probability that this tv is produced by company x.

Answer 1

Let $X$ denote the event "tv is produced by company x."

Let $Y$ denote the event "tv is produced by company y."

Let $Z$ denote the event "tv is produced by company z."

Let $D$ denote the event "tv is defected".

Given

$$P(X)=0.5, P(Y)=P(Z)=0.25$$

$$P(D|X)=0.02, P(D|Y)=0.02,P(D|Z)=0.04$$

a) Theorem of total probability

$$P(D)=P(X)P(D|X)+P(Y)P(D|Y)$$

$$+P(Z)P(Z|D)$$

$$=0.5(0.02)+0.25(0.02)+0.25(0.04)=0.025$$

The probability that the tv is defected is $0.025.$

ii) By the Bayes' Rule

$$P(X|D)$$

$$=\dfrac{P(X)P(D|X)}{P(X)P(D|X)+P(Y)P(D|Y)+P(Z)P(Z|D)}$$

$$=\dfrac{0.5(0.02)}{0.5(0.02)+0.25(0.02)+0.25(0.04)}$$

$$=0.4$$

The probability that this tv is produced by company x is $0.4.$