EduFarming
Your list · 0हिं

Type at least two letters. Press Esc to close.

↑↓ navigate↵ openTAB cycle filter

Conditional Probability & Bayes' Theorem · Question 117 of 125

Question 117

HardOriginal question

Machine M1M_1 makes 60% of a factory's pens and machine M2M_2 the rest; 2% of M1M_1's pens and 3% of M2M_2's are defective. A pen picked at random is defective. What is the probability that M2M_2 made it?

Options for question 117

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. P(D)=0.6×0.02+0.4×0.03=0.012+0.012=0.024P(D) = 0.6 \times 0.02 + 0.4 \times 0.03 = 0.012 + 0.012 = 0.024.

  2. P(M2∣D)=0.0120.024=0.5P(M_2 \mid D) = \frac{0.012}{0.024} = 0.5.

Worth noting

0.4 is M2M_2's share of all pens; its higher defect rate lifts its share of defective pens to a half.

Concept
Bayes' theorem
Typical time
120 s
Your time
—
The rule this testsBayes' theorem
Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.