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In a office, 30% are supervisors, 40% are male employees, 60% of supervisors are male. Probability that a random chosen employee is a male or a female?
Answer:
I would say probability of randomly chosen a male employee is 22% while that of female is 48%.
Base on my understanding of the question,
let 'x' be the total employee (supervisors and ordinary employees)
supervisor: 30% of 'x' = 0.3x
male supervisors: 60% of 0.3x = 0.18x
and female supervisor: 40% of 0.3x = 0.12x
total male employees: 40% of 'x' = 0.4x
therefore, number of ordinary male employee = 0.4x - 0.18x = 0.22x
=> P(male employee) = 0.22x/x = 0.22
i.e. probability of male employee is 22%
total female employees: 60% of 'x' = 0.6x
therefore, number of ordinary female employee = 0.6x - 0.12x = 0.48x
=> P(female employee) = 0.48x/x = 0.48
i.e. probability of female employee is 48%
I hope I've interpreted the question correctly.
Base on my understanding of the question,
let 'x' be the total employee (supervisors and ordinary employees)
supervisor: 30% of 'x' = 0.3x
male supervisors: 60% of 0.3x = 0.18x
and female supervisor: 40% of 0.3x = 0.12x
total male employees: 40% of 'x' = 0.4x
therefore, number of ordinary male employee = 0.4x - 0.18x = 0.22x
=> P(male employee) = 0.22x/x = 0.22
i.e. probability of male employee is 22%
total female employees: 60% of 'x' = 0.6x
therefore, number of ordinary female employee = 0.6x - 0.12x = 0.48x
=> P(female employee) = 0.48x/x = 0.48
i.e. probability of female employee is 48%
I hope I've interpreted the question correctly.
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