0

Guest

Courses New

A Company has two plants to manufacture cycles. The first plant manufactures 60% of the cycles and the second plant manufactures 40%. 80% of the cycles of the first plant are rated of standard quality and 90% of the cycles of the second plant are rated of standard quality. A cycle is picked up at random and it is found to be of standard quality. Find the probability that it comes from the second plant.

A Company has two plants to manufacture cycles. The first plant manufactures 60% of the
cycles and the second plant manufactures 40%. 80% of the cycles of the first plant are rated
of standard quality and 90% of the cycles of the second plant are rated of standard quality. A
cycle is picked up at random and it is found to be of standard quality. Find the probability
that it comes from the second plant.

Grade:12

5 Answers

yaswanth
36 Points
9 years ago
answer=2/5.9/10
              2/5.9/10+3/5.4/5
  =18/42
=3/7
 
ARUN KAUSHIK N R
18 Points
9 years ago
let f be event that the cycle is from first plant.
let s be event that the cycle is from second plant.
let g be event that the cycle is  of standard quality.
from sum,           P(f)=0.6,                 P(s)=0.4,
                           P(g/f)=0.8,              P(g/s)=0.9
required probability=P(s/g)
                               =P(g/s) .P(s) / { P(g/f).P(f) +P(g/s) .P(s)} (by bayes theorem)
                               = 0.4*0.9 / {0.6*0.8 + 0.4*0.9}
                               = 0.36/0.84
                               = 3/7
                               = 0.42857
                               
ARUN KAUSHIK N R
18 Points
9 years ago
let f be event that the cycle is from first plant.
let s be event that the cycle is from second plant.
let g be event that the cycle is  of standard quality.
from sum,           P(f)=0.6,                 P(s)=0.4,
                           P(g/f)=0.8,              P(g/s)=0.9
required probability=P(s/g)
                               =P(g/s) .P(s) / { P(g/f).P(f) +P(g/s) .P(s)} (by bayes theorem)
                               = 0.4*0.9 / {0.6*0.8 + 0.4*0.9}
                               = 0.36/0.84
                               = 3/7
                               = 0.42857
                               
ARUN KAUSHIK N R
18 Points
9 years ago
let f be event that the cycle is from first plant.
let s be event that the cycle is from second plant.
let g be event that the cycle is  of standard quality.
from sum,           P(f)=0.6,                 P(s)=0.4,
                           P(g/f)=0.8,              P(g/s)=0.9
required probability=P(s/g)
                               =P(g/s) .P(s) / { P(g/f).P(f) +P(g/s) .P(s)} (by bayes theorem)
                               = 0.4*0.9 / {0.6*0.8 + 0.4*0.9}
                               = 0.36/0.84
                               = 3/7
                               = 0.42857
                               
Aryakanya Amin
23 Points
9 years ago
A= standered quality cycles 
P(E1) = cycles frm 1st plant= 60/100
P(E2) = cycles frm 2nd plant= 40/100
P(A/E1) = 80/100
P(A/E2) = 90/100
 
P(E2/A)= P(E2)*P(A/E2)
               P(E1)*P(A/E1)
            =40/100*90/100                                     
               40/100*90/100 + 60/100*80/100
            =  36/100                        
                    36/100 +48/100
             = 36/84 = 6/14  = 3/7 :) 
               
 

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More