Question

A random sample of 88 U.S. 11th- and 12th-graders was selected. The two-way table summarizes the gender of the students and their response to the question "Do you have allergies?" Suppose we choose a student from this group at random.

\(\begin{array}{c|cc|c} & \text { Female } & \text { Male } & \text { Total } \\ \hline \text{ Yes } & 19 & 15 & 34 \\ \text{ No } & 24 & 30 & 54 \\ \hline \text{ Total } & 43 & 45 & 88\\ \end{array}\)
What is the probability that the student is female or has allergies?
\((a)\frac{19}{88}\)
(b)\(\frac{39}{88}\)
(c)\(\frac{58}{88}\)
(d)\(\frac{77}{88}\)

Answer 1

DEFINITIONS
Complement rule
\(P(A^{c})=P(\text{not}\ A)=1-P(A)\)
General addition rule for any two events:
\(P(A\ or\ B)=P(A)+P(B)-P(A\text{ and } B)\)
SOLUTION
\(\begin{array}{c|cc|c} & \text { Female } & \text { Male } & \text { Total } \\ \hline \text{ Yes } & 19 & 15 & 34 \\ \text{ No } & 24 & 30 & 54 \\ \hline \text{ Total } & 43 & 45 & 88\\ \end{array}\)
We note that the table contains information about 88 peoples (given in the bottom right corner of the table).
Moreover, 34 of the 88 people have allergies, becatise 34 is mentioned in the row Yes” and in the column *Total” of the table.
The probability is the number of favorable outcomes divided by the number of possible outcomes:
\(P(Allergies)=\frac{\# \text{ of favorable outcomes}}{\# \text{ of possible outcomes}}=\frac{34}{88}\) We note that 43 of the 88 people are female, becatise 43 is mentioned in the row * Total” and in the column *Female” of the given table.
\(P(Female)=\frac{\# \text{ of favorable outcomes}}{\# \text{ of possible outcomes}}=\frac{43}{88}\) We note that 19 of the 88 people are females and have allergies, because 19 is mentioned in the row ” Yes” and in the column *Female” of the given table.
\(P(\text{ Allergies and Female})=\frac{\# \text{ of favorable outcomes}}{\# \text{ of possible outcomes}}=\frac{19}{88}\)
Use the general addition rule:
\(P(\text{ Allergies and Female})=P(\text{Allergies})+P(\text{Female})-P(\text{ Allergies and Female})\)
\(=\frac{34}{88}+\frac{43}{88}-\frac{19}{88}\)
\(=\frac{34+43-19}{88}\)
\(=\frac{58}{88}\)
Result: \((c)\frac{58}{88}\)