Question

data were collected from a survey given to graduating college seniors on the number of times they had changed majors. from that data, a probability distribution was constructed. the random variable x is defined as the number of times a graduating senior changed majors. it is shown below:

x	p(x = x)
1	0.165
2	0.319
3	0.16
4	0.061
5	0.033
6	0.01
7	0.003
8	0.002

a. what is the probability that a randomly selected student changed his or her major at least once?
b. what is the probability that a randomly selected student changed his or her major at most twice?
c. given that a randomly selected person did change majors, what is the probability that he or she changed majors more than three times?

Explanation:

Step1: Recall probability - sum rule

The sum of all probabilities in a probability - distribution is 1. For a discrete random variable \(X\) with probability mass function \(P(X = x)\), \(\sum_{x}P(X = x)=1\).

Step2: Calculate the probability for part a

The probability that a student changed his or her major at least once is \(P(X\geq1)\). We know that \(P(X\geq1)=1 - P(X = 0)\). Given \(P(X = 0)=0.247\), then \(P(X\geq1)=1 - 0.247 = 0.753\).

Step3: Calculate the probability for part b

The probability that a student changed his or her major at most twice is \(P(X\leq2)\). We use the formula \(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\). Given \(P(X = 0)=0.247\), \(P(X = 1)=0.165\), and \(P(X = 2)=0.319\), then \(P(X\leq2)=0.247 + 0.165+0.319 = 0.731\).

Step4: Calculate the probability for part c

The probability that a student changed his or her major more than three times is \(P(X>3)\). We use the formula \(P(X>3)=P(X = 4)+P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)\). Given \(P(X = 4)=0.061\), \(P(X = 5)=0.033\), \(P(X = 6)=0.01\), \(P(X = 7)=0.003\), \(P(X = 8)=0.002\), then \(P(X>3)=0.061 + 0.033+0.01+0.003+0.002=0.11\).

Answer:

a. \(0.753\)
b. \(0.731\)
c. \(0.11\)