Answered

A company categorizes its employees in three groups: I, II, and III, which constitute 20%, 60%, and 20% of the workforce respectively. Furthermore, 10%, 20%, 5% employees of group I, II, III respectively are considered certified. Given that a randomly picked employee is not certified, what is the probability that this employee belongs to group I?

Answer :

nuhulawal20

Answer:

the probability that this employee belongs to group I is 0.2118

Step-by-step explanation:

Given the data in the question;

P( group 1 )  = 0.20

P( group 11 ) = 0.60

P( group 111 ) = 0.20

now

[tex]P( \frac{certified}{Group1} )[/tex]         = 0.10

[tex]P( \frac{Not-certified}{Group1} )[/tex] = 0.90

[tex]P( \frac{certified}{Group11} )[/tex]        = 0.20

[tex]P( \frac{Not-certified}{Group11} )[/tex] = 0.80

[tex]P( \frac{certified}{Group111} )[/tex]        = 0.05

[tex]P( \frac{Not-certified}{Group111} )[/tex] = 0.95

so

[tex]P( \frac{group1}{Not-certisfied} )[/tex] = [ [tex]P( \frac{Not-certified}{Group1} )[/tex] × P( group 1 ) ] / [ [tex]P( \frac{Not-certified}{Group1} )[/tex] × P( group 1 ) + [tex]P( \frac{Not-certified}{Group11} )[/tex] × P( group 11 ) + [tex]P( \frac{Not-certified}{Group111} )[/tex] × P( group 111 ) ]

we substitute

[tex]P( \frac{group1}{Not-certisfied} )[/tex] = [ 0.90×0.20 ] / [ 0.90×0.20 + 0.80×0.60 + 0.95×0.20

[tex]P( \frac{group1}{Not-certisfied} )[/tex] = 0.18 / [ 0.18 + 0.48 + 0.19 ]

[tex]P( \frac{group1}{Not-certisfied} )[/tex] = 0.18 / 0.85

[tex]P( \frac{group1}{Not-certisfied} )[/tex] = 0.2118

Therefore, the probability that this employee belongs to group I is 0.2118

Other Questions