a. Probability that the student got a B = 24/54 = 0.444444
= 0.4444 (4 decimal places)
b. Probability that the student was female AND got an "A":
[tex]\begin{gathered} \text{ = }\frac{3}{54} \\ =0.055555 \\ =0.0556\text{ (4 decimal places)} \end{gathered}[/tex]
c.
Probability that the student was male OR got a "B"
[tex]\begin{gathered} \text{ }\frac{33}{54}\text{ + }\frac{24}{54}-\frac{14}{54}\text{ = }\frac{43}{54}\text{ } \\ =0.796296 \\ =0.7963\text{ (4 decimal places)} \end{gathered}[/tex]
d. Probability that the student got an 'B' GIVEN they are female
[tex]\begin{gathered} \text{ }\frac{p(B\text{ n Female)}}{p(\text{Female)}} \\ =\frac{10}{21} \\ =0.47619048 \\ =0.4762\text{ (4 decimal places)} \end{gathered}[/tex]