Question

question 3 (5 points)
data was collected about the favorite sport and grade level of students at a school. leave all answers as fractions (do not simplify).
suppose you randomly select a student, what is the probability that...
a) the student is in 12th grade?
p(12th) =
p(not 10th) =
p(10th and basketball) =
p(11th and volleyball) =
p(12th or basketball) =

Explanation:

Step1: Probability formula

The probability \(P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)

Step2: Part (a)

The number of 12th - grade students is \(n = 120\), and the total number of students is \(N=440\). So \(P(12\text{th})=\frac{120}{440}\)

Step3: Part (b)

The number of 10th - grade students is \(n_{10}=100\). The number of non - 10th - grade students is \(N - n_{10}=440 - 100\). So \(P(\text{not }10\text{th})=\frac{440 - 100}{440}\)

Step4: Part (c)

The number of students who are in 10th grade and prefer basketball is \(n = 50\). So \(P(10\text{th and basketball})=\frac{50}{440}\)

Step5: Part (d)

The number of students who are in 11th grade and prefer volleyball is \(n = 12\). So \(P(11\text{th and volleyball})=\frac{12}{440}\)

Step6: Part (e)

Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
Let \(A\) be the event that the student is in 12th grade (\(n_A = 120\)) and \(B\) be the event that the student prefers basketball (\(n_B=146\)). The number of students who are in 12th grade and prefer basketball is \(n_{A\cap B} = 30\)
\(P(12\text{th or basketball})=\frac{120 + 146-30}{440}\)

Answer:

a) \(\frac{120}{440}\)
b) \(\frac{440 - 100}{440}=\frac{340}{440}\)
c) \(\frac{50}{440}\)
d) \(\frac{12}{440}\)
e) \(\frac{120 + 146-30}{440}=\frac{236}{440}\)