Question

Suppose 70% of all companies are classified as small companies and the rest as large companies. Suppose further, 82% of large companies provide training to employees, but only 18% of small companies provide training. A company is randomly selected without knowing if it is a large or small company; however, it is determined that the company provides training to employees.

What are the prior probabilities that the company is a large company or a small company? (Round the answers to 2 decimal place.)
p(small) =
p(large) =
What are the revised probabilities that the company is large or small? (Round your answers to 4 decimal places.) p(small) =
p(large) =
Based on your analysis, what is the overall percentage of companies that offer training? (Round your answer to 1 decimal place.)
Percentage = %

Answer 1

a) p(small) = 0.126

p(large) = 0.246

b) p(small) = 0.6613

p(large) = 0.3387

c) 37.2%

Step-by-step explanation:

A) determine that the company picked is a large company or small company

condition : the company provides training to its employees

Given data:

p( small ) = 0.7, p( large ) = 0.3, p( training ∩ small ) = 0.18, p( training ∩ large ) = 0.82 , p( No-training ∩ small ) = 0.82 , p( no-training ∩ large ) = 0.18

A) hence the probability of picking a small company that provides training

P( small | training ) = P(Training ∩ Small)* P(Small) = 0.18 * 0.7 = 0.126

Probability of picking a large company that provides training

P( large | training ) = P(training ∩ Large) *P(Large) = 0.82 * 0.3 = 0.246

B) Determine the revised probabilities that company picked is large or small

Revised probability for a large company; P( large | training )

P(Large | training) = P(Large ∩ training) / P(training)

= 0.246 / ( 0.126 + 0.246 ) = 0.6613

P( small | training ) = P( small ∩ training ) / P(training )

= 0.126 / ( 0.126 + 0.246 ) = 0.3387

C) Overall percentage of companies that offer training

p( training ) = 0.126 + 0.246 = 0.372 = 37.2%