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	0	1	2	3	4	5	6	7	8
p(x = x)	0.247	0.165	0.319	0.16	0.061	0.033	0.01	0.003	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: Define the probability for at - least once

The probability of changing major at least once is $P(X\geq1)$. We know that $P(X\geq1)=1 - P(X = 0)$.

Step2: Calculate $P(X\geq1)$

Given $P(X = 0)=0.247$, then $P(X\geq1)=1 - 0.247=0.753$.

Step3: Define the probability for at - most twice

The probability of changing major at most twice is $P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)$.

Step4: Calculate $P(X\leq2)$

$P(X = 0)=0.247$, $P(X = 1)=0.165$, $P(X = 2)=0.319$. So $P(X\leq2)=0.247 + 0.165+0.319 = 0.731$.

Step5: Define the probability for more than three times given changed majors

First, the probability of changing majors is $P(X\gt0)=1 - P(X = 0)=0.753$. The probability of changing majors more than three times is $P(X\gt3)=P(X = 4)+P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)$.
$P(X = 4)=0.061$, $P(X = 5)=0.033$, $P(X = 6)=0.01$, $P(X = 7)=0.003$, $P(X = 8)=0.002$. So $P(X\gt3)=0.061 + 0.033+0.01+0.003+0.002=0.11$.
Then, by the formula of conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$, here $A$ is $X\gt3$ and $B$ is $X\gt0$, and $A\cap B = A$ since $A\subseteq B$. So $P(X\gt3|X\gt0)=\frac{P(X\gt3)}{P(X\gt0)}=\frac{0.11}{0.753}\approx0.146$.

Answer:

a. $0.753$
b. $0.731$
c. $0.146$