Student's Age in Fictional Stats Class Problem (Probability) Problem:
In a fictional stats class, 40% of students are female, and the rest are male. Of the female students, 30% are less than 20 years old and 90% are less than 30 years old. Of the male students, half are less than 20 years old and 70% are less than 30 years old.
(a) Make a contingency table to describe these two variables
(b) Find the probability that a randomly selected studet is 30 years or older
(c) If a student is 20 years or older, what is the probability that the student is female?
(d) If a student is less than 30 years old, what is the probability that the student is 20 years or older?

My Thoughts:
(b) P(<30 years) = 1 - 0.78 = 0.22
(c) What I first did was find P(S2 given 'not A1'), but the answer doesn't make sense because the denominator ended up being smaller than the nominator.
(d) Do I solve this problem by doing 'not 20 years'?

Let's follow @BGM's suggestion in the comments. 
Let $F$ denote female; let $M$ denote male; let $A$ denote age.
Since $40\%$ of the students are female and $30\%$ of them are less than $20$ years old, the probability that a student is female and less than $20$ years old is 
$$P(F~\cap A < 20) = P(F)P(A < 20 \mid F) = 0.40 \cdot 0.30 = 0.12$$
Since $90\%$ of the female students are less than $30$ years old, the probability that a student is female and less than $30$ years old is 
$$P(F~\cap A < 30) = P(F)P(A < 30 \mid F) = 0.40 \cdot 0.90 = 0.36$$
The probability that a student is female, at least $20$ years old, and less than $30$ years old can be found by subtracting the probability that she is less than $20$ years old from the probability that she is less than $30$ years old, which yields
$$P(F~\cap 20 \leq A \leq 30) = P(F~\cap A < 30) - P(F~\cap A < 20) = 0.36 - 0.12 = 0.24$$
Finally, the probability that a student is female and at least $30$ years old is found by subtracting the probability that a student is female and less than $30$ years old from the probability that a student is female, which yields
$$P(F~\cap A \geq 30) = P(F) - P(F~\cap A < 30) = 0.40 - 0.36 = 0.04$$
By using similar reasoning, we can fill in the table for the male students.
$$
\begin{array}{l | c | c | c | c}
  & A < 20 & 20 \leq A < 30 & A \geq 30 & Total\\ \hline
F & 0.12 & 0.24 & 0.04 & 0.40\\
M & 0.30 & 0.12 & 0.18 & 0.60\\ \hline
Total & 0.42 & 0.36 & 0.22 & 1 
\end{array}
$$
The probability that a student is at least $30$ years old is stated in the contingency table.
To find the probability that a student who is at least $20$ years old is female, divide the probability that a female student is at least $20$ years old by the probability that a student is at least $20$ years old, both of which can be found by adding the appropriate columns in the table.
The probability that a student who is less than $30$ years old is at least $20$ years old can be found by subtracting the probability that the student is less than $20$ years old from the probability the student is less than $30$ years old.  To find the probability that a student is less than $30$ years old, you can subtract the probability that a student is greater than $30$ years old from $1$.
