Question

midterm cumulative test
of 50 students going on a class - trip, 35 are student athletes and 5 are left - handed. of the student athletes, 3 are left - handed. which is the probability that one of the students on the trip is an athlete or is left - handed?
athletes left - handed
0.8
0.74
0.2
0.5

Explanation:

Step1: Identify the number of athletes, left - handed, and both

Let \(A\) be the set of athletes and \(L\) be the set of left - handed students. We know \(n(A)=35\), \(n(L) = 5\), and \(n(A\cap L)=3\), and the total number of students \(n(T)=50\).

Step2: Use the formula for the probability of the union of two events

The formula for \(P(A\cup L)\) is \(P(A\cup L)=P(A)+P(L)-P(A\cap L)\). Since \(P(A)=\frac{n(A)}{n(T)}\), \(P(L)=\frac{n(L)}{n(T)}\) and \(P(A\cap L)=\frac{n(A\cap L)}{n(T)}\), we have \(P(A)=\frac{35}{50}\), \(P(L)=\frac{5}{50}\) and \(P(A\cap L)=\frac{3}{50}\).

Step3: Calculate \(P(A\cup L)\)

\[

$$\begin{align*} P(A\cup L)&=\frac{35}{50}+\frac{5}{50}-\frac{3}{50}\\ &=\frac{35 + 5-3}{50}\\ &=\frac{37}{50}\\ &=0.74 \end{align*}$$

\]

Answer:

0.74