conitional probability-probability this given this

Question

According to U.S. Department of Labor employment reports,in 2010, 39.3% of jobs were classified as management, professional, and related occupations (Mgt), 14.5% were Service occupations (Svc), 23.2% were Sales and office occupations (SO), 9.9% were Natural resources, construction, and maintenance occupations (NCM), and 13.1% were Production, transportation, and material moving occupations (PTM).

Among Mgt, 48.6% were held by male workers, and 51.4% were held by female workers.

The other categories were:

• Svc, 50.6% male, 49.4% female
• SO, 38.3% male, 61.7% female
• NCM, 95.9% male, 4.1% female
• PTM, 80.2% male, 19.8% female

D=NCM, F=female

I want to calculate $P(D|F).$

I know that $P(D|F)=\frac{P(D∩F)}{P(F)}=\frac{P(D)+P(F)−P(D∪F)}{P(F)}$ but how would I find $P(D∪F)$? (I think this is what you're supposed to do? Correct me if I'm wrong) I do know that the answer is about $0.009$ by the way but I'm trying to figure out how to get that

Answer 1

$$P(D|F)=\frac{P(DF)}{P(F)}=\frac{P(F|D)P(D)}{P(F)}\frac{0.041~ X~0.099}{P(F)}$$ To find $P(F)$ you can use the law of total probability which states:

$$ P(F)=\sum_n P(F \cap A_n)=\sum_nP(F|A_n)P(A_n) $$ So you would have: $$ P(F)=P(F|D)P(D)+P(F|Mgt)P(Mgt)+... $$ Hope this helps

Answer 2

$$P\left(D\mid F\right)=P\left(F\mid D\right)\times\frac{P\left(D\right)}{P\left(F\right)}$$ and:

$$P\left(F\right)=P\left(F\mid D\right)P\left(D\right)+\dots+P\left(F\mid Svc\right)P\left(Svc\right)$$