1. mac or pc? a recent census at a major university revealed that 60% of its students were undergraduates…

1. mac or pc? a recent census at a major university revealed that 60% of its students were undergraduates. the rest mainly used pcs. at the time of the census, 67% of the school’s students who said that they used macs as their main computers. suppose we select a student at random from among those who were part of the census.\n a. what’s the probability that this person is a graduate student or mainly uses a mac?\n b. construct a venn diagram to represent the outcomes of this chance process using the events g: is a graduate student and m: mainly uses a mac.\n c. find p(g^c ∩ m^c). interpret this value in context.

1. mac or pc? a recent census at a major university revealed that 60% of its students were undergraduates. the rest mainly used pcs. at the time of the census, 67% of the school’s students who said that they used macs as their main computers. suppose we select a student at random from among those who were part of the census.\n a. what’s the probability that this person is a graduate student or mainly uses a mac?\n b. construct a venn diagram to represent the outcomes of this chance process using the events g: is a graduate student and m: mainly uses a mac.\n c. find p(g^c ∩ m^c). interpret this value in context.

Answer

Explanation:

Step1: Calculate probabilities of events

Let $P(G)$ be the probability of being a graduate student. Since 60% are undergraduates, $P(G)=1 - 0.6=0.4$. Let $P(M)$ be the probability of mainly using a Mac. Given that 67% mainly used PCs, $P(M)=1 - 0.67 = 0.33$. Also, 23% of respondents who were graduate students said they used Macs. Let's assume we need to use the formula for the union of two - events $P(G\cup M)=P(G)+P(M)-P(G\cap M)$. We know that the proportion of graduate - Mac users is part of the information we can use to find $P(G\cap M)$. But we first answer part a. The probability that a person is a graduate student or mainly uses a Mac is given by the formula $P(G\cup M)=P(G)+P(M)-P(G\cap M)$. We know $P(G) = 0.4$, $P(M)=0.33$. The proportion of graduate students who use Macs is 23% of the graduate students. So $P(G\cap M)=0.4\times0.23 = 0.092$. Then $P(G\cup M)=0.4 + 0.33-0.092=0.638$.

Step2: Construct Venn - diagram for part b

Draw two overlapping circles. Label one circle $G$ (graduate students) and the other circle $M$ (Mac - users). The non - overlapping part of $G$ represents graduate students who do not use Macs, which is $P(G)-P(G\cap M)=0.4 - 0.092 = 0.308$. The non - overlapping part of $M$ represents non - graduate students who use Macs, which is $P(M)-P(G\cap M)=0.33 - 0.092 = 0.238$. The overlapping part $G\cap M$ has a value of 0.092. The area outside both circles represents non - graduate students who do not use Macs.

Step3: Calculate $P(G^{c}\cap M^{c})$ for part c

By De - Morgan's law, $P(G^{c}\cap M^{c})=1 - P(G\cup M)$. Since $P(G\cup M)=0.638$, then $P(G^{c}\cap M^{c})=1 - 0.638 = 0.362$. In context, this is the probability that a randomly selected student is an undergraduate and mainly uses a PC.

Answer:

a. $0.638$ b. (Describe the Venn - diagram as above: two overlapping circles $G$ and $M$ with non - overlapping part of $G$ as 0.308, non - overlapping part of $M$ as 0.238, overlapping part as 0.092 and the outside area representing non - graduate PC users) c. $0.362$, the probability that a randomly selected student is an undergraduate and mainly uses a PC.