Question

The probability that a trainee will remain with a company is 0.8. The probability that an employee earns more than 20,000 per month is 0.4. The probability that an employee, who was a trainee and remained with the company or who earns more than 20,000 per month is 0.9. what is the probability that an employee earns more than 20,000 per month given that he is a trainee, who stayed with the company?​

Answer 1

Answer: The probability that an employee earns more than 20,000 per month given that he is a trainee, who stayed with the company is 0.3

Step-by-step explanation:

P(A) ⇒ Probability that a trainee will remain with a company = 0.8

P(B) ⇒ Probability that employee earns more than 20000 per month = 0.4

P(A∪B) ⇒ Probability that an employee, who was a trainee and remained with the company or who earns more than 20,000 per month = 0.9

P(A∩B) ⇒ Probability that an employee earns more than 20,000 per month given that he is a trainee, who stayed with the company

P(A∪B) = P(A) + P(B) - P(A∩B)

0.9        =  0.8  + 0.4  - P(A∩B)

P(A∩B)  = 0.3

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Answer 2

Answer:

The probability is $\frac{3}{8}$

If a trainee stayed with the company, there is a $\frac{3}{8}$ chance that he will earn more than 20,000 per month.

step-by-step explanation:

Consider two scenarios: A, where the trainee stays with the business, and B, where the employee earns more than 20,000 annually.

P(A) = chance of a trainee staying with an organization

P(B) = possibility that an employee will make more than 20,000 per month

P(A∪B) = Probability that a worker who joined the company as a learner and has stayed with it or who makes more than 20,000 each monthly

P(A∩B) = Possibility that a worker, who was a learner and remained with the organization, makes more than 20,000 per monthly

$P(A) = 0.8,$

$P(B) = 0.4 and$

$P(A{\displaystyle \cup }B)=0.9$

$P(A{\displaystyle \cup }B)=P(A)+P(B)-P(A{\displaystyle \cap }B)$

$P(A{\displaystyle \cap }B)=P(A) + P(B) - P (A{\displaystyle \cup }B)$

              $= 0.8+0.4-0.9$

              $= 0.3$

$P(\frac{B}{A} )=\frac{P(A{\displaystyle \cap }B)}{P(A)}$

         $=\frac{3}{8}$

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