Question

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.

a. what’s the probability that this person is a graduate student or mainly uses a mac?
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.
find (p(g^{c}cap m^{c})). interpret this value in context.

Explanation:

Step1: Identify given probabilities

Let $G$ be the event of being a graduate student and $M$ be the event of mainly using a Mac. We know that $P(G^c)=0.6$ (60% undergraduates, so 40% graduates $P(G) = 0.4$), and among respondents, 23% used Macs, so $P(M)=0.23$, and 67% of respondents were graduates among those who used Macs. Let's first find $P(G\cap M)$. If we assume the number of Mac - using students is the reference group for the 67% statistic, and $P(M) = 0.23$, then $P(G\cap M)=0.67\times0.23 = 0.1541$.

Step2: Use the formula for $P(G\cup M)$

The formula for $P(G\cup M)$ is $P(G\cup M)=P(G)+P(M)-P(G\cap M)$. Substitute $P(G) = 0.4$, $P(M)=0.23$ and $P(G\cap M)=0.1541$ into the formula. So $P(G\cup M)=0.4 + 0.23-0.1541=0.4759$.

Step3: Use De - Morgan's law to find $P(G^c\cap M^c)$

By De - Morgan's law, $P(G^c\cap M^c)=1 - P(G\cup M)$. Substitute $P(G\cup M)=0.4759$ into the formula. So $P(G^c\cap M^c)=1 - 0.4759 = 0.5241$.

For part b:
Draw two overlapping circles. Label one circle $G$ (graduate students) and the other circle $M$ (Mac - using students). The overlapping region represents $G\cap M$ with probability $0.1541$. The non - overlapping part of the $G$ circle represents $P(G\cap M^c)=P(G)-P(G\cap M)=0.4 - 0.1541 = 0.2459$. The non - overlapping part of the $M$ circle represents $P(G^c\cap M)=P(M)-P(G\cap M)=0.23 - 0.1541 = 0.0759$. The region outside both circles represents $P(G^c\cap M^c)=0.5241$.

Interpretation of $P(G^c\cap M^c)$: The probability that a randomly selected student is an undergraduate and does not mainly use a Mac is 0.5241 or 52.41%.

Answer:

a. $P(G\cup M)=0.4759$, $P(G^c\cap M^c)=0.5241$
b. Draw two overlapping circles labeled $G$ and $M$. The overlapping region $G\cap M$ has probability $0.1541$, non - overlapping part of $G$ is $0.2459$, non - overlapping part of $M$ is $0.0759$ and the region outside both circles is $0.5241$.