Question

b) the probability that none of the 3 randomly selected math 164 students will transfer is
c) the probability that at least one of the 3 randomly selected math 164 students will transfer is

Explanation:

1. Assume the probability of a single - student not transferring is \(p\) (not given in the problem, let's assume the probability of a single student transferring is \(q\) and \(p = 1 - q\)):

• Since the selection of each student is an independent event, for independent events \(A\), \(B\), and \(C\), the probability that \(A\) and \(B\) and \(C\) occur is \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\).
• For the three randomly - selected students, the probability that none of them will transfer is \(P(\text{none transfer})=p\times p\times p=p^{3}\).

1. For part (c):

• The probability that at least one of the students will transfer is the complement of the event that none of them will transfer.
• Let \(P(X\geq1)\) be the probability that at least one student will transfer and \(P(X = 0)\) be the probability that none of the students will transfer.
• We know that \(P(X\geq1)=1 - P(X = 0)\). Since \(P(X = 0)=p^{3}\), then \(P(X\geq1)=1 - p^{3}\).

Step1: Define probability for a single - student

Let the probability of a single student not transferring be \(p\). For independent events, the probability that none of the 3 students transfer is the product of the probabilities of each student not transferring.

$$P(\text{none transfer})=p\times p\times p=p^{3}$$

Step2: Use the complement rule

The probability that at least one student transfers is the complement of the event that none of them transfer.

$$P(\text{at least one transfer})=1 - P(\text{none transfer})=1 - p^{3}$$

Answer:

b) \(p^{3}\) (where \(p\) is the probability of a single student not transferring)
c) \(1 - p^{3}\) (where \(p\) is the probability of a single student not transferring)