Question

question 2 34% of ua students are known to use macbooks. 72% are known to use iphones. assume the probability of intersection is 29%. a student is selected at random and questioned about their technology usage. what is the probability of the student using neither device? 0 77% 0 72% 0 29% 0 23% 1 points save answer

Explanation:

Step1: Use the principle of inclusion - exclusion

Let \(A\) be the set of students using Mac - Books and \(B\) be the set of students using iPhones. The formula for \(P(A\cup B)\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Given \(P(A) = 0.34\), \(P(B)=0.72\) and \(P(A\cap B)=0.29\).
\[P(A\cup B)=0.34 + 0.72-0.29\]

Step2: Calculate \(P(A\cup B)\)

\[P(A\cup B)=0.77\]

Step3: Calculate the probability of using neither

The probability of using neither is \(1 - P(A\cup B)\).
\[1 - 0.77=0.23\]

Answer:

0.23 or 23%