Question

a local high school has both male and female students. each student either plays a sport or doesnt. the two - way table summarizes a random sample of 60 students. \

$$\begin{tabular}{|c|c|c|} \\cline{2 - 3} & male & female \\\\ \\hline sport & 8 & 27 \\\\ \\hline no sport & 16 & 9 \\\\ \\hline \\end{tabular}$$

let no sport be the event that a randomly chosen student (from the table) does not play a sport. let female be the event that a randomly chosen student (from the table) is female. find the following probabilities. write your answers as decimals. (if necessary, consult a list of formulas.) (a) $p(\text{female}) = \square$ (b) $p(\text{no sport and female}) = \square$ (c) $p(\text{no sport }| \text{ female}) = \square$

Explanation:

Step1: Find total number of students

Total students = 8 + 27 + 16 + 9 = 60.

Step2: Calculate P(female)

Number of female students = 27 + 9 = 36.
$P(\text{female}) = \frac{\text{Number of female students}}{\text{Total number of students}} = \frac{36}{60} = 0.6$.

Step3: Calculate P(no sport and female)

Number of female students with no sport = 9.
$P(\text{no sport and female}) = \frac{\text{Number of female with no sport}}{\text{Total number of students}} = \frac{9}{60} = 0.15$.

Step4: Calculate P(no sport | female)

Using conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
We know $P(\text{no sport and female}) = 0.15$ and $P(\text{female}) = 0.6$.
So, $P(\text{no sport | female}) = \frac{0.15}{0.6} = 0.25$.

Answer:

s:
(a) $\boxed{0.6}$
(b) $\boxed{0.15}$
(c) $\boxed{0.25}$