Question

1. a survey was sent to a random sample of individuals to learn about their traveling preferences. of the 140 individuals who responded, 45 individuals reported united as their favorite airline, and 102 individuals said they fly only with a carry - on, rather than checking a bag. let a represent the event of an individual reporting united as their favorite airline and b represent the event of an individual flying with only a carry - on. which of the following statements is true?

a) ( p(a or b)=p(a)+p(b) )
b) ( p(a and b)>p(a or b) )
c) ( p(a)>p(b^{c}) )
d) ( p(a^{c})=p(b) )
e) ( p(a or b)>p(a)+p(b) )

Explanation:

Step1: Calculate \(P(A)\), \(P(B)\), \(P(A^{C})\), \(P(B^{C})\)

• \(n = 140\) (total number of respondents)
• \(P(A)=\frac{45}{140}=\frac{9}{28}\approx0.321\)
• \(P(B)=\frac{102}{140}=\frac{51}{70}\approx0.729\)
• \(P(A^{C}) = 1 - P(A)=1-\frac{45}{140}=\frac{140 - 45}{140}=\frac{95}{140}=\frac{19}{28}\approx0.679\)
• \(P(B^{C})=1 - P(B)=1-\frac{102}{140}=\frac{140-102}{140}=\frac{38}{140}=\frac{19}{70}\approx0.271\)

Step2: Analyze each option

• Option A: \(P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)\). Since \(P(A\ and\ B)\geq0\), \(P(A\ or\ B)

eq P(A)+P(B)\) (unless \(P(A\ and\ B) = 0\), which we have no information about).

• Option B: \(P(A\ and\ B)\leq P(A)\) and \(P(A\ or\ B)\geq P(A)\) and \(P(A\ or\ B)\geq P(B)\). So \(P(A\ and\ B)>P(A\ or\ B)\) is false.
• Option C: \(P(A)=\frac{9}{28}\approx0.321\) and \(P(B^{C})=\frac{19}{70}\approx0.271\). Since \(\frac{9}{28}=\frac{9\times5}{28\times5}=\frac{45}{140}\) and \(\frac{19}{70}=\frac{38}{140}\), \(P(A)>P(B^{C})\)
• Option D: \(P(A^{C})=\frac{95}{140}\) and \(P(B)=\frac{102}{140}\), so \(P(A^{C})

eq P(B)\)

• Option E: \(P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)\leq P(A)+P(B)\)

Answer:

C. \(P(A)>P(B^{C})\)