Question

the venn diagram represents the types of devices a group of students own. student devices a total of 619 students own a smartwatch. a total of 572 students own a smartwatch and a cell phone. a total of 157 students own a smartwatch and a tablet. how many students own a smartwatch, cell phone, and tablet? number of students who own a smartwatch, cell phone, and tablet =

Explanation:

Step1: Recall the principle of triple - set intersection

We use the principle of inclusion - exclusion for three sets \(A\), \(B\), and \(C\): \(n(A\cap B\cap C)=n(A\cap B)+n(A\cap C)+n(B\cap C)-n(A)-n(B)-n(C)+n(A\cup B\cup C)\). Here, let \(A\) be the set of students with smart - watches, \(B\) be the set of students with cell phones, and \(C\) be the set of students with tablets. But we can also think in terms of the given information directly. We know that the number of students who own a smart - watch and a cell phone and a tablet is what we need to find. We are given the number of students who own a smart - watch (\(n(A) = 619\)), the number of students who own a smart - watch and a cell phone (\(n(A\cap B)=572\)), the number of students who own a smart - watch and a tablet (\(n(A\cap C) = 157\)), and the number of students who own only a smart - watch (\(12\)).

Step2: Calculate the number of students who own all three devices

We know that \(n(A)=n(\text{only }A)+n(A\cap B)+n(A\cap C)-n(A\cap B\cap C)\). Rearranging the formula to solve for \(n(A\cap B\cap C)\), we get \(n(A\cap B\cap C)=n(\text{only }A)+n(A\cap B)+n(A\cap C)-n(A)\). Substitute the values: \(n(\text{only }A) = 12\), \(n(A\cap B)=572\), \(n(A\cap C)=157\), and \(n(A)=619\). Then \(n(A\cap B\cap C)=12 + 572+157 - 619\).
\[

$$\begin{align*} n(A\cap B\cap C)&=(12 + 572+157)-619\\ &=(584 + 157)-619\\ &=741-619\\ &=122 \end{align*}$$

\]

Answer:

122