Case Based Questions (MCQ)

Chapter 13 Class 12 Probability
Serial order wise

Β

## (D) 1/30

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Question A factory has 3 machines X, Y and Z, producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and it is found to be defective. Let, E1 = event of drawing a bolt produced by machine X. E2 = event of drawing a bolt produced by machine Y. E3 = event of drawing a bolt produced by machine Z. E = event of drawing a defective bolt. Based on the above information answer the following questions:Question 1 What is the value of P(E2)? (A) 1/6 (B) 1/3 (C) 1/2 (D) 1/4 Now, P(E2) = (ππ’ππππ ππ ππππ‘π  πππππ’πππ ππ¦ πππβπππ π)/(πππ‘ππ ππ’ππππ ππ ππππ‘π  πππππ’πππ ππ¦ πππ 3 πππβππππ ) Number of bolts produced by machine Y = 2000 Total number of defective bolts = 1000 + 2000 + 3000 = 6000 Thus, P(E2) = 2000/6000= π/π So, the correct answer is (B) Question 2 Find the value of P(E|E1)? (A) 1/100 (B) 3/200 (C) 1/50 (D) 1/10 P(E | E1) = P(defective bolt | produced by machine X) = P( defective bolt is produced by machine X) Since Machine X produces 1% defective bolts = 1% = π/πππ So, the correct answer is (a) Question 3 Find the value of P(E | E2)? (A) 1/100 (B) 3/200 (C) 1/50 (D) 1/10 P(E | E1) = P(defective bolt | produced by machine Y) = P( defective bolt is produced by machine Y) Since Machine Y produces 1.5% defective bolts = 1.5 % = 1.5/100 = π/πππ So, the correct answer is (B) Question 4 Find the value of P(E|E3)? (A) 1/100 (B) 3/200 (C) 1/50 (D) 1/10 P(E | E3) = P(defective bolt | produced by machine Z) = P( defective bolt is produced by machine Z) Since Machine Z produces 2% defective bolts = 2 % = π/πππ = π/ππ So, the correct answer is (c) Question 5 What is the probability that the drawn bolt has been produced by the machine X? (A) 1/10 (B) 3/200 (C) 1/20 (D) 1/30 We need to find Probability that the drawn defective bolt is produced by machine X i.e. P(π¬_π "|E") So, "P(" π¬_π "|E) = " (π(πΈ_1 ). π(πΈ|πΈ_1))/(π(πΈ_1 ). π(πΈ|πΈ_1 ) + π(πΈ_2 ). π(πΈ|πΈ_2 )+π(πΈ_3 ). π(πΈβ€| πΈ_3) ) "P(" π¬_π ")" = Probability that the bolt is made by machine X = 1000/6000 = π/π π·("E|" π¬_π) This is calculated in Question 2 π("E|" πΈ_1) = π/πππ "P(" π¬_π ")" = Probability that the bolt is made by machine Y = 2000/6000 = π/π π·("E|" π¬_π) This is calculated in Question 3 π("E|" πΈ_2) = π/πππ "P(" π¬_π ")" = Probability that the bolt is made by machine Z = 3000/6000 = π/π π·("E|" π¬_π) This is calculated in Question 4 π("E|" πΈ_3) = π/ππ Putting values in formula, "P(" πΈ_1 "|E) = " (1/6 Γ 1/100)/(1/6 Γ 1/100 + 1/3 Γ 3/200 + 1/2 Γ 1/50) = ( 1/600)/(1/600 + 1/200 + 1/100) = ( 1/600)/((1 + 3 + 6)/600 ) = ( 1/600)/(10/600 ) Putting values in formula, "P(" πΈ_1 "|E) = " (1/6 Γ 1/100)/(1/6 Γ 1/100 + 1/3 Γ 3/200 + 1/2 Γ 1/50) = ( 1/600)/(1/600 + 1/200 + 1/100) = ( 1/600)/((1 + 3 + 6)/600 ) = ( 1/600)/(10/600 ) = 1/600Γ600/10 = π/ππ So, the correct answer is (a)

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.