Question

consider your statistics 213 lecture section. there are 120 students registered. of these 120 students, 53 are haskayne school of business students (the remaining students belong to some other faculty). moreover, 69 students are classified as 1st - year students and 20 are 1st - year students and who are not haskayne students. you are to randomly choose one student from the 120 in your lecture section. let hs represent the event that the chosen student is a haskayne - student and 1st represent the event that this chosen student is a 1st - year student. part (a) complete the table below. use four decimals in each of your answers.

	hs	hs^c	row probabilities
1st^c
column probabilities

part (b) find the probability that the student you randomly chose is a haskayne student or in their 1st - year. enter your answer to four decimals.
part (c) find the probability that the student you randomly chose is neither a haskayne student nor in their 1st - year. (use four decimals)
part (d) what proportion of all the students in your statistics 213 lecture section are 1st - year students who are not in the haskayne school of business? (use four decimals)
part (e) are the events hs and 1st mutually exclusive events? select the most appropriate reason below.
a. hs and 1st are not mutually exclusive events, because p(hs ∩ 1st) ≠ p(hs)p(1st).
b. hs and 1st are not mutually exclusive events, because p(hs ∩ 1st) ≠ 0.
c. hs and 1st are mutually exclusive events, because p(hs ∩ 1st) = 0.
d. hs and 1st are mutually exclusive events, because p(hs ∩ 1st) = p(hs)p(1st).
e. hs and 1st are mutually exclusive events, because p(hs ∪ 1st) = 0.

Explanation:

Step1: Calculate the number of students in each category

Total students \(n = 120\). Number of Haskayne students \(n(HS)=53\), so number of non - Haskayne students \(n(HS^c)=120 - 53=67\). Number of 1st - year students \(n(1st)=69\), and number of 1st - year non - Haskayne students \(n(1st\cap HS^c)=20\). Then number of 1st - year Haskayne students \(n(1st\cap HS)=69 - 20 = 49\). Number of non - 1st - year Haskayne students \(n(1st^c\cap HS)=53 - 49 = 4\). Number of non - 1st - year non - Haskayne students \(n(1st^c\cap HS^c)=67 - 20=47\).

Step2: Calculate the probabilities for the table

• \(P(1st\cap HS)=\frac{n(1st\cap HS)}{n}=\frac{49}{120}\approx0.4083\)
• \(P(1st\cap HS^c)=\frac{n(1st\cap HS^c)}{n}=\frac{20}{120}\approx0.1667\)
• \(P(1st)=\frac{n(1st)}{n}=\frac{69}{120}=0.5750\)
• \(P(1st^c\cap HS)=\frac{n(1st^c\cap HS)}{n}=\frac{4}{120}\approx0.0333\)
• \(P(1st^c\cap HS^c)=\frac{n(1st^c\cap HS^c)}{n}=\frac{47}{120}\approx0.3917\)
• \(P(1st^c)=\frac{n(1st^c)}{n}=\frac{120 - 69}{120}=\frac{51}{120}=0.4250\)
• \(P(HS)=\frac{n(HS)}{n}=\frac{53}{120}\approx0.4417\)
• \(P(HS^c)=\frac{n(HS^c)}{n}=\frac{67}{120}\approx0.5583\)
• \(P(1)=1\)

The completed table:

	\(HS\)	\(HS^c\)	Row Probabilities
\(1st^c\)	\(0.0333\)	\(0.3917\)	\(0.4250\)
Column Probabilities	\(0.4417\)	\(0.5583\)	\(1.0000\)

Step3: Calculate \(P(HS\cup1st)\) for part (b)

Using the formula \(P(HS\cup1st)=P(HS)+P(1st)-P(HS\cap1st)\)
\(P(HS)=\frac{53}{120}\), \(P(1st)=\frac{69}{120}\), \(P(HS\cap1st)=\frac{49}{120}\)
\(P(HS\cup1st)=\frac{53 + 69-49}{120}=\frac{73}{120}\approx0.6083\)

Step4: Calculate \(P(HS^c\cap1st^c)\) for part (c)

We know that \(P(HS^c\cap1st^c)=\frac{n(HS^c\cap1st^c)}{n}=\frac{47}{120}\approx0.3917\)

Step5: Calculate the proportion for part (d)

The proportion of 1st - year non - Haskayne students is \(\frac{n(1st\cap HS^c)}{n}=\frac{20}{120}\approx0.1667\)

Step6: Determine mutual - exclusivity for part (e)

Two events \(A\) and \(B\) are mutually exclusive if \(P(A\cap B) = 0\). Since \(P(HS\cap1st)=\frac{49}{120}
eq0\), \(HS\) and \(1st\) are not mutually exclusive events. The reason is \(P(HS\cap1st)
eq0\).

Answer:

Part (a):

	\(HS\)	\(HS^c\)	Row Probabilities
\(1st^c\)	\(0.0333\)	\(0.3917\)	\(0.4250\)
Column Probabilities	\(0.4417\)	\(0.5583\)	\(1.0000\)

Part (b): \(0.6083\)
Part (c): \(0.3917\)
Part (d): \(0.1667\)
Part (e): B. HS and 1st are not mutually exclusive events, because \(P(HS\cap1st)
eq0\).