Question

fill in the blank 5 points
how many participants own exactly 1 cat? 11
how many participants own exactly 1 dog? 13
how many participants own exactly 1 cat and exactly 1 dog? 3
according to the union rule, how many participants own exactly 1 cat or exactly 1 dog? type your answer...
use the union rule idea and your answers above to find the the probability that a participant owns exactly 1 cat or exactly 1 dog? (enter as fraction, out of 34) type your answer...
multiple choice 1 point
if a participant is selected at random from this group, what is the probability that they know no cats and no dogs?
20/34
33/34
1/34

Explanation:

Step1: Recall the union - rule formula

The union - rule for two events \(A\) and \(B\) is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), where \(n(A)\) is the number of elements in event \(A\), \(n(B)\) is the number of elements in event \(B\), and \(n(A\cap B)\) is the number of elements in both \(A\) and \(B\). Let \(A\) be the event that a participant owns exactly 1 cat and \(B\) be the event that a participant owns exactly 1 dog. Given \(n(A) = 11\), \(n(B)=13\), and \(n(A\cap B)=3\).

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

Substitute the values into the formula: \(n(A\cup B)=11 + 13-3\).
\[n(A\cup B)=21\]

Step3: Calculate the probability

The probability \(P(A\cup B)\) that a participant owns exactly 1 cat or exactly 1 dog out of 34 participants is given by the formula \(P(A\cup B)=\frac{n(A\cup B)}{34}\). Since \(n(A\cup B) = 21\), then \(P(A\cup B)=\frac{21}{34}\).

Answer:

How many participants own exactly 1 cat or exactly 1 dog: 21
The probability that a participant owns exactly 1 cat or exactly 1 dog (out of 34): \(\frac{21}{34}\)