Respuesta :
If the randomly selected sensor is identified as having been made by Carton Company, the probability of it being defective must be 0.06, The reason being that it is given that Carton sensors have a 6% defect rate.
Denote
- [tex] p_1[/tex] - the probability that sensors made by breakers are not defective and [tex] q_1[/tex] - the probability that sensors made by breakers are defective;
- [tex] p_2 [/tex] - the probability that sensors made by carton company are not defective and [tex] q_2[/tex] - the probability that sensors made by carton company are defective;
- [tex] p_3[/tex] - the probability that sensors made by flutes company are not defective and [tex] q_3[/tex] - the probability that sensors made by flutes company are defective.
Then
[tex] q_1=0.04,\\p_1=1-0.004=0.96,\\q_2=0.06,\\p_2=1-0.06=0.94,\\q_3=0.09,\\p_3=1-0.09=0.91.[/tex]
The probability that selected sensor was made by:
- breakers manufacturing company is 0.8;
- carton company is 0.15;
- flutes company is 0.05.
Use formula for the conditional probability:
[tex] Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr B}=\dfrac{0.8\cdot 0.04+0.15\cdot 0.06+\cdot 0.05\cdot 0.09}{0.15}= 0.3033333\approx 0.3.[/tex]
Answer: 0.3, or in percent 30%.