   Chapter 14.CT, Problem 13CT ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

#### Solutions

Chapter
Section ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem

# 13. In a group of four students, what is the probability that at least two have the same astrological sign?

To determine

To find:

The probability of getting at least two students of same astrological sign from a group of four students.

Explanation

Approach:

The formula of probability of an event is given by,

P(E)=n(E)n(S)

Here, n(E) is the number of favorable outcomes for event E and n(S) is the number of total possible outcomes.

The formula of compliment of probability of an event is given by,

P(E)=1P(E)

Calculation:

There are 12 astrological signs.

So,

n(S)=12

The ways by which a student can have a astrological sign is also 12.

So,

n(E1)=12

Substitute 12 for n(S) and 12 for n(E1) in above mentioned formula of probability in order to find the probability of student can have a astrological sign.

P(E1)=1212=1

The ways by which second student has a different astrological sign is 11.

So,

n(E2)=11

Substitute 12 for n(S) and 11 for n(E2) in above mentioned formula of probability in order to find the probability that second student has a different astrological sign.

P(E2)=1112

The ways by which third student has a different astrological sign is 10.

So,

n(E3)=10

Substitute 12 for n(S) and 10 for n(E3) in above mentioned formula of probability in order to find the probability that third student has a different astrological sign

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 