Question

select the correct answer
a school choir has 80 students; 36 of them are in seventh grade. also, 40 students in the choir are girls; 18 of them are in seventh grade. what is the probability that a student picked at random from the choir is either a girl or is in seventh grade?
a. \\(\frac{27}{40}\\)
b. \\(\frac{29}{40}\\)
c. \\(\frac{13}{40}\\)
d. \\(\frac{33}{40}\\)

Explanation:

Step1: Recall the principle of inclusion - exclusion for probability

The formula for \( P(A \cup B) \) (probability of \( A \) or \( B \)) is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \). Let \( A \) be the event that a student is a girl and \( B \) be the event that a student is in seventh grade.

Step2: Calculate \( P(A) \), \( P(B) \) and \( P(A \cap B) \)

• Total number of students \( n = 80 \).
• Number of girls \( n(A)=40 \), so \( P(A)=\frac{n(A)}{n}=\frac{40}{80}=\frac{1}{2} \).
• Number of seventh - grade students \( n(B) = 36 \), so \( P(B)=\frac{n(B)}{n}=\frac{36}{80}=\frac{9}{20} \).
• Number of students who are both girls and in seventh grade \( n(A\cap B)=18 \), so \( P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{18}{80}=\frac{9}{40} \).

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

\[

$$\begin{align*} P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\ &=\frac{40}{80}+\frac{36}{80}-\frac{18}{80}\\ &=\frac{40 + 36-18}{80}\\ &=\frac{58}{80}\\ &=\frac{29}{40} \end{align*}$$

\]

Answer:

B. \(\frac{29}{40}\)