- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Make the sample space of tossing two coins simultaneously and find the probability of:
1. Exactly two heads
2. At least one head
3. At least one tail
4. At most two tails
Given:
Two coins are tossed simultaneously.
To do:
We have to find the probability of getting:
1. Exactly two heads
2. At least one head
3. At least one tail
4. At most two tails
Solution:
The sample space of the experiment is $(HH,\ HT,\ TH,\ TT)$.
Total number of outcomes $=\ 4$
1. Outcomes in favour of getting exactly two heads $=\ ( HH)$
Number of outcomes in favour of getting exactly two heads $=\ 1$
Probability of getting exactly heads on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
$=\frac{1}{4}$
2. Outcomes in favour of getting at least one head on tossing the two coins $=\ ( HT,\ TH,\ HH)$
Number of outcomes in favour of getting at least one head $=\ 3$
Probability of getting at least one head on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
$=\frac{3}{4}$
3. Outcomes in favour of getting at least one tail on tossing the two coins $=\ ( HT,\ TH,\ TT)$
Number of outcomes in favour of getting at least one tail $=\ 3$
Probability of getting at least one tail on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
$=\frac{3}{4}$
4. Outcomes in favour of getting at most two tails on tossing the two coins $=\ ( HT,\ TH,\ TT, HH)$
Number of outcomes in favour of getting at most two tails $=\ 4$
Probability of getting at most two tails on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
$=\frac{4}{4}$
$=1$
85 Views
