Very Tough Probability Question?

Suppose a hi-tech company's workforce displays the following characteristics:


20% of the workforce have 3 yrs or less of university education


51% of the workforce have 4 yrs of university education


29% of the workforce have over 4 yrs of university education





An internal company survey showed that the likelihood of an employee earning over $60,000/yr is related to yrs of education in the following manner:


-4% having 3 yrs or less of university earn over $60,000/yr


-10% having 4 yrs of university earn over $60,000/yr


-12% having over 4 yrs of university earn over $60,000/yr





1. An employee earns over $60,000. What is the probability that they have 3 yrs or less of university? (Round to 3 decimals-0.5555 would be 0.556)





2. An employee earns over $60,000. What is the probability that they have 4 yrs of university? (Round to three decimals)





3. An employee earns over $60,000. What is the probability that they have over 4 yrs of university? (Round to 3 decimals)





Thank you for anything.

Very Tough Probability Question?
X = event that employee has 3 or less years univ


Y = event that employee has 3 or less years univ


Z = event that employee has 3 or less years univ





W = event that employee makes over 60k





Given: P(X) = 0.2, P(Y) = 0.51, P(Z) = 0.29


P(W|X) = 0.04


P(W|Y) = 0.10


P(W|Z) = 0.12





RTF: P(X|W), P(Y|W), P(Z|W)





1) P(X|W) = P(X n W) / P(W) ...(1)





P(W|X) = P(X n W) / P(X)


thus P(X n W) = P(W|X) * P(X)


= 0.04 * 0.2 = 0.008





P(W) = P(W|X)*P(X) + P(W|Y)*P(Y) + P(W|Z)*P(Z)


= 0.008 + 0.1*0.51 + 0.12*0.29


= 0.0938





So back to eqn 1, P(X|W) = 0.008 / 0.0938


= 0.085





Similarly P(Y|W) = 0.544


P(Z|W) = 0.371