Learning Outcomes
We can use permutations and combinations to help us answer more complex probability questions. Show
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 ProblemLet’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 fourA 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 theseSolution: 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 ProgramLeave a Commentस्रोत : 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 fourGiven 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 Videos648233370 स्रोत : www.doubtnut.com |
What is the Probability of making in even number of 4 digits using 1 2 3 4 without any digit being repeated?
Postagens relacionadas
direito autoral © 2024 mesadeestudo Inc.
