Learning Outcomes
- Compute a conditional probability for an event
- Use Baye’s theorem to compute a conditional probability
- Calculate the expected value of an event
We can use permutations and combinations to help us answer more complex probability questions.
A 4 digit PIN number is selected. What is the probability that there are no repeated digits?
In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.
In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.
The previous examples are worked in the following video.
Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.
View the following for further demonstration of these examples.
Birthday Problem
Let’s take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?
Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.
Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?
Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?
Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?
The birthday problem is examined in detail in the following.
If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.
This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.
Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?
Without considering different cases.
This is another way to get the same answer as already answerd above.
Here we start by finding the total amount of the four-digit numbers with distinct digits, then finding the amount of odd digits, filling the same criteria, and last, we subtract the odd numbers from the total, to find the even numbers filling the criteria.
We have totally seven digits. We know that the digit zero cannot be placed at the position representing the thousand position. That leaves us with six digits to choose from for this position and that can be made in
$P(6,1) = \frac{6!}{(6-1)!} = \frac{6!}{5!}=6$, different ways.
Now, we have three positions left to fill and six digits to choose from, including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in
$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}=6 \cdot5 \cdot4$, different ways.
Finally, with distinct digits, there is
$6 \cdot6 \cdot5 \cdot4 =720$
four-digit numbers to be constructed.
We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers.
The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd?
The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in
$P(3,1)=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3$, different ways.
To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in
$P(5,1)=\frac{5!}{(5-1)!}=\frac{5!}{4!}=5$, different ways.
Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in
$P(5,2)=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5 \cdot4=20$, different ways.
The total number of odd four-digit numbers, with distinct digits are
$5 \cdot5 \cdot4 \cdot3 =300$.
Now, we can answer the question how many of these 720, four-digit numbers, are even, by the subtraction
$720-300=420$.
The even numbers are 420.
Guys, does anyone know the answer?
get what is the probability of making an even number of 4 digit using 1234 from screen.
A four
A four-digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is ½. To learn more, visit BYJU’S
JEE QuestionsA Four Digit Number Is Formed By The Digits 1 2 3 4 With No Repetition
A four-digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is 1) Zero 2) ⅓ 3) ¼ 4) None of these
Solution: Option (D) None of theseThe number of ways that the four digits formed by the digits 1, 2, 3 4 without repetition = 4!
Number of favourable outcomes = 12
Therefore, the required probability is = 12/ 4!
=12/(4.3.2.1) = ½
Was this answer helpful?
2.5 (8)
Join BYJU'S Learning Program
Leave a Comment
स्रोत : byjus.com
What is the probability of making an even number of 4 digits using 1, 2, 3 and 4?
Answer (1 of 4): This isn’t really a permutations or combinations matter. The problem refers only to the singles digit, of which there are four choices. Two of them are odd and two of them are even.
Permutations and Combinations
Probability (statistics)
Numbers (mathematics)
What is the probability of making an even number of 4 digits using 1, 2, 3 and 4?
Ad by Amazon Web Services (AWS)
AWS is how.
AWS removes the complexity of building, training, and deploying machine learning models at any scale.
4 Answers Ellis Cave
40+ years as an Electrical EngineerAuthor has 5K answers and 2.5M answer views2y
Using the J programming language, brute force approach:
x:(+/%#)-.2|10#.>:perm 4
1r2
The answer is 1/2 or 50%
Related questions
A bag contains 4 strawberries and 8 grapes. What is the probability that both the fruits drawn from it are strawberries?
Using the digits 0, 1, 2, 3 and 4, four-digit numbers are made without repetition. What is the probability that the number formed is even?
How many 4 digit numbers can be formed using 1,2,3,4,5,6, and 7 with none of the digits being repeated?
What is x if 8-1*x=(-4)-1?
What is a 4-digit number formed with 1, 1, 2, 2, 3, and 4?
Al Katz
Author has 801 answers and 256.1K answer views2y
Since your choices for the least significant digit are from two odd numbers and two even numbers, the odds of the least significant digit being even is 0.5; other digits are irrelevant (assuming only integers).
Ronald Deep
Worked at The University of DaytonAuthor has 1.6K answers and 767K answer views2y
(2 * 3 * 2 * 1) / 256 = 12/256 = 0.0468 without repetition
or (4 * 4 * 4 * 2) / 256 = 1/2 with repetition
Sponsored by Aspose
What is the Aspose.Diagram for the .NET library?
Easy to use and reliable API capable of generating, editing, manipulating Visio diagrams and more.
Malcolm MacMartin 2y
This isn’t really a permutations or combinations matter. The problem refers only to the singles digit, of which there are four choices. Two of them are odd and two of them are even.
Related questions
A bag contains 4 strawberries and 8 grapes. What is the probability that both the fruits drawn from it are strawberries?
Using the digits 0, 1, 2, 3 and 4, four-digit numbers are made without repetition. What is the probability that the number formed is even?
How many 4 digit numbers can be formed using 1,2,3,4,5,6, and 7 with none of the digits being repeated?
What is x if 8-1*x=(-4)-1?
What is a 4-digit number formed with 1, 1, 2, 2, 3, and 4?
What is the value of (a,y) in (13-20- a* 13y) = 168* 13-22?
What is the probability that the selected number is "an even number " if a five digit number is selected using digits 1, 2, 3, 4, 5, and 6 (each digit is used once)?
If both 11^2 and 3^3 are factors of the number a * 4^3 * 6^2 * 13^11, then what is the smallest possible value of 'a'?
What is the next digit in the sequence 1, 2, 2, 3, 2, 4, 2, 4, 3, 4?
What 3 1/2 - 4 1/4 = 3 1/2 +?
What is the probability of getting 2 odd and 2 even numbers from 1 to 9?
How many 5 digit numbers from 1, 2, 3, 4?
What is 3/4 का 1/2 [2/3-3/4+ {1/2-2/ 3+ (1/4-1/5)}]?
Four digit numbers are formed by using 1, 2, 3, 4, and 5 without repeating any digit. Can you find the probability that anumber, chosen at random, is an odd number?
Let S = {0, 1, 2 , 4, 5, 6, 9}. In how many ways can 3 digits be arranged such that they are all even and find the probability of each event?
स्रोत : www.quora.com
A four
Given numbers are 1,2,3 and 4. Possibilities for unit's place digit (either 1 or 3) = 2 Possibilities for ten's place digit = 3 Possibilities for hundred's place digit = 2 Possibilities for thousand's place digit = 1 therefore Number of favourable outcomes =2xx3xx2xx1=12 Number of four digit numbers formed by 1,2,3,4 (without repetition) = 4! therefore Required probability =(12)/(4xx3xx2)=(1)/(2)
Home > English > Class 12 > Maths > Chapter > Probability >
A four-digit number is formed ...
A four-digit number is formed by the digits 1,2,3,4 with no repetition. The probability that the number is odd, is
Updated On: 27-06-2022
Get Answer to any question, just click a photo and upload the photo and get the answer completely free,
UPLOAD PHOTO AND GET THE ANSWER NOW!
Text Solution Open Answer in App A zero B 1 3 13 C 1 4 14 D None of these Answer
The correct Answer is D
Solution
Given numbers are 1,2,3 and 4.
Possibilities for unit's place digit (either 1 or 3) = 2
Possibilities for ten's place digit = 3
Possibilities for hundred's place digit = 2
Possibilities for thousand's place digit = 1
∴ ∴
Number of favourable outcomes
=2×3×2×1=12 =2×3×2×1=12
Number of four digit numbers formed by 1,2,3,4 (without repetition) = 4!
∴ ∴
Required probability
= 12 4×3×2 = 1 2 =124×3×2=12 Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6
CHAAYA PRAKASHINI (BENGALI)
WB JEE PREVIOUS YEAR PAPER BANGALI
A DAS GUPTA
IA MARON
JEE MAINS PREVIOUS YEAR ENGLISH
RD SHARMA ENGLISH CENGAGE ENGLISH
XII BOARD PREVIOUS YEAR PAPER ENGLISH
NCERT EXEMPLAR ENGLISH
MCGROW HILL PUBLICATION
KUMAR PRAKASHAN OSWAAL PUBLICATION MTG-WBJEE DISHA PUBLICATION
IPUCET PREVIOUS YEAR PAPERS
MODERN PUBLICATION MODERN PUBLICATION FULL MARKS
KCET PREVIOUS YEAR PAPERS
CBSE COMPLEMENTARY MATERIAL
TS EAMCET PREVIOUS YEAR PAPERS
NTA MOCK TESTS SUBHASH PUBLICATION SURA PUBLICATION VK JAISWAL
VIKAS GUPTA (BLACK BOOK)
VIKAS GUPTA (BLACK BOOK)
VIKAS GUPTA (BLACK BOOK)
CHHAYA PUBLICATION MARVEL PUBLICATION
VIKRAM PUBLICATION ( ANDHRA PUBLICATION)
ARIHANT MATHS
XII BOARDS PREVIOUS YEAR
XII BOARDS PREVIOUS YEAR
JEE MAINS PREVIOUS YEAR
RD SHARMA CENGAGE
JEE ADVANCED PREVIOUS YEAR
RS AGGARWAL KC SINHA NCERT EXEMPLAR NAGEEN PRAKASHAN OBJECTIVE RD SHARMA OBJECTIVE RD SHARMA NDA PREVIOUS YEARS KVPY PREVIOUS YEAR
MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS
SUNSTAR PUBLICATION USHA PUBLICATION
VIKAS GUPTA (BLACK BOOK) ENGLISH
ARIHANT MATHS ENGLISH
HIMALAYA PUBLICATION
ICSE CHETANA PUBLICATION MBD PUBLICATION
ARIHANT PUBLICATION BIHAR
ML KHANNA CENGAGE KC SINHA ENGLISH
OBJECTIVE RD SHARMA ENGLISH
MODERN PUBLICATION V PUBLICATION UNITED BOOK HOUSE
A N EXCEL PUBLICATION
BODY BOOKS PUBLICATION
MAXIMUM PUBLICATION PSEB
VGS PUBLICATION-BRILLIANT
BITSAT GUIDE BANSAL NCERT ENGLISH NCERT
AKSHRANAND PUBLICATION
VIRAJ PUBLICATION NCERT GUJARATI NCERT ARIHANT HINDI ERRORLESS HINDI
BIHAR BOARD- PREVIOUS YEAR PAPER
TRIPUTI PUBLICATION
UPTU PREVIOUS YEAR PAPER
CGPET PREVIOUS YEAR PAPERS
JEE Main & Advanced (Hindi Medium)
SCIENCE PUBLICATION MITTAL PUBLICATION SANJEEV PUBLICATION YUGBODH
SHIVLAAL PUBLICATION
NAVBODH DAS GUPTA RAJEEV BANSAL RAJEEV BANSAL
MP BOARD - PREVIOUS YEAR PAPERS
HARYANA BOARD - PREVIOUS YEAR PAPERS
PRABODH PUBLICATION
RAJASTHAN BOARD PREVIOUS YEAR
CHATTISGARH BOARD PREVIOUS YEAR
KC SINHA MANOHAR RAY NOOTAN HINDI BALAJI PUBLICATION NDA PATHFINDER
BIHAR BOARD- PREVIOUS YEAR PAPER
ARIHANT PUBLICATION CHHATISGARH
ARIHANT PUBLICATION BIHAR
ARIHANT PUBLICATION UP
NCERT EXEMPLAR HINDI
SBPD PUBLICATION
AGRAWAL EXAM CART BIHAR
NTA MOCK TESTS
ARIHANT PUBLICATION MP
S CHAND
JHARKHAND BOARD PREVIOUS YEAR PAPERS
NDA PREVIOUS YEAR HINDI
BANSAL HINDI NCERT HINDI
PSEB (PUNJABI MEDIUM)
SURA PUBLICATION(TAMIL)
NCERT TAMIL NCERT TELUGU Answer
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Click here to get PDF DOWNLOAD for all questions and answers of this chapter - MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS Class 12 PROBABILITY
Click here to get PDF DOWNLOAD for all questions and answers of this Book - MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS Class 12 MATHS
Related Videos
648233370
स्रोत : www.doubtnut.com