i’ve watched other mathematics youtubers (blackpenredpen, michael penn, etc.), but you have quickly become my favorite! i can’t put it into words, but your videos give me incredible joy!!
I love the standard chalkboards. In my school they use whiteboards (i am support staff aka the electrician) and they always complain about sun shining on the board ... scratch head. From the digits shown I found the following possibilities X= any unused digit and Nz nonzero digit picked first: X X X 2 0 3! positive outcomes X X X 4 0 3! NzX X 1 2 2 ways for first digit and 2! for rest so overall 4 ways NzX X 3 2 4 X X X 0 4 3! NzX X 2 4 4 ways total 3*6+3*4=30 positive outcomes out of 5! =120 posible ones so 1/4 . Cute problem
There are 120 ways to shuffle 5 digits. Yes, 03412 is 3412, therefore no 5 digit number, but it is a possible arrangement that just does not fit one of the requirements.
Yes I believe you are correct. I think the problem he wanted to propose was "Find the probability that a random 5 digit number formed by the digits 0,1,2,3,4 (without repetitions) is divisible by 4."
This just made probability a thousand and times easier. Thanks a million sir! Ps: i love the way you just paused and wondered why 42 isnt divisible by 4 lol, i was wondering the same thing😂
Technically the phrasing of the question in the thumbnail is "what is the probability that a random arrangement of 01234 forms a 5 digit number divisible by 4", that means we look at all 120 permutations of 01234, and pick the 30 that are both divisible by 4 and are 5 digits, so the answer would be 30/120=1/4
The two last digits requirements: 1. If the first digit is odd, the second digit should be 2 or 6. 2. If the first digit is even, the second digit should be 0, 4, or 8.
So at 6:10 "Okay we're done'. I'm yelling at the screen, NO!!! What about 20?!?! As you picked each digit you didn't work back with zero!! lol I'm sure you'll catch it in a minute.
The thing is that since we can't take 0 in the first digit, so that excludes all those numbers which actually start with 0 and then rearrange with only the last 4 digits. That is why, the possibility is not 120, but 96.
@@sukhwinderchawla6037Yes, you are right. First of all with due regards to the sir, I would like to make it clear I am not trying to find fault with the solution. I just tried to interpret the phrasing of problem. In my view, for the solution given here, the phrasing of problem should have been like "five digit numbers formed by random arrangement of 0,1,2,3,4....." instead of "random arrangement of digits 0,1,2,3,4...".
@@tomtomspa Because starting with 0 will not make a 5 digit number. I will remain a 4 digit. like 1234. The question wants us to form 5 digit numbers and then among them calculate the probability of finding a number which is divisible by 4.
@@sudiptoatutube no, this is an assumption not justified. The problem is to find the number of arrangements of five digits wich are divisible by four. Five digits arrangements is not the same as arrangements of five digits numbers. 03124 is most definitely such an arrangement.
@@sudiptoatutube”What is the probability that a random arrangement of all five digits 0-1-2-3-4 forms a number divisible by 4?” is literally the first frame of this video. Otherwise it should have been: “…that a random arrangement of all five digits 0-1-2-3-4 forms a five digit number divisible by four”. It is clearly stated.
On the thumbnail it says "of all five digits", which implies each one has to be used once. Of course, it also has no explicit specification that said number has to be a 5-digit number. It just says that it must form "a number divisible by 4". 01324, for example, fulfils both requirements even if it results in a 4-digit number.
question never asked you to form 5 digit number. it says " random arrangement of number of all 5 digits " then it says that arrangement form "a number" (never says 5 digit number) divisible by 4. for ex : 1324 can be arranged like '01324'. so total number of possible ways should be 120. so probability should be 1/4 or .25 .
Solution: If it has a 1 or 3 at the end, it is not divisible by 4, so 2/5 are already out. of the other 3/5, it depends on the digit before. So we have: 10 > no 20 > yes 30 > no 40 > yes 02 > no 12 > yes 32 > yes 42 > no 04 > yes 14 > no 24 > yes 34 > no This is 6/12 or 1/2. In total, we have 1 - (2/5 + 3/5 * 1/2) = 1 - (4/10 + 3/10) = 1 - 7/10 = 3/10 = 30%
There is an even easier way. The number has to end in one of the six sequences: 04, 12, 20, 24, 32, 40. The probability of getting 04 for the last two digits is 1 ∕ 5⋅1 ∕ 4 = 1 ∕ 20. The same is true for the other five sequences as well, so the probability of getting either of them is 6⋅1 ∕ 20 = 0.3 EDIT: After watching the video I noticed that Newton didn't allow for the total sequence of 5 digits to begin with 0 (as if e.g. 01432 is not divisible by 4 LOL), which makes it a little trickier. It can still be solved using Bayes' theorem, but it's hardly any easier than the way Newton did it.
@@jumpman8282 Not really easier, just the inverse of my calculation. I calculated the probability of not being divisible by 4 and subtracted it from 1.
I think total outcomes should be 5! = 120. Because in the question it is given: What is the probability that a random arrangement of the digits 0,1,2,3,4, forms a 5 digit number divisible by 4? Which means favourable cases are the numbers which are 5 digit and also divisible by 4. And total arrangements of 5 digits, i.e , sample space is 120. So, probability = 30/120 = 1/4.
@@maburwanemokoena7117 yes you are right but here the question is if we arrange the digits 0,1,2,3,4 randomly then what is the probability that we get a 5 digit number divisible by 4. So, sample space is number of arrangement of 0,1,2,3,4 which are possible. And favorable cases are those arrangements which are 5 digit numbers and also divisible by 4. So, Sample space= 5! = 120 And, favourable cases= number of 5 digit number divisible by 4 , which is = 30. So, Probability= favorable cases/ total possible cases = 30/120 = 1/4
@@shivnathbanerjee5868 I do hear you, but normally in such cases we assume the first digit cannot be a zero. How do you think the question should’ve been phrased such that the above solution is true?
Like this question. The 0 being the first digits could create confusion but I understand and agree with your reasoning! My personal preference for getting the 96 is get the total number including the ones starting zero which is 120 the subtract 24 the number starting 0. Warning for any answering any probability question - it is very easy 5o correctly answer a different question!
I didn't understand why four digit numbers divisible by 4 were calculated instead of five digit numbers divisible by 4 as the question asks. Would someone please answer my question. I am very weak at probability. I would very much appreciate it. Thanks.
That second part of the video was a bit inexact in that regard, but the answer is still correct. If the last 2 digits are fixed in each case, then you're just trying to arrange the other 3 digits. Once you've figured out the number of possibilities for the first 2, there's only 1 possibility for the last one. So each of his cases should have been multiplied by another (x1), to be technically correct, but that doesn't change the actual answer.
It is never stated that it has to be a five digit number, only that it has to be an arrangement of those five digits, so your answer is wrong. Why not counting 01234 as a possible arrangement?
Well the total for out probability is 5! Unless it means any arrangement 5 numbers or less. Then, because Prime is a tactical genius, he will break down the different cases. Because I am a gorilla, I will smash my head against the wall trying to brute force it. Lets see whose method is better!
Let x=abcde, a≠0 x=10⁴a+10³b+10²c+10d+10e 4 | x --> 4 | 10d+e (d,e)={(0,4),(1,2),(2,0),(2,4),(3,2),(4,0)} Number of digits left for a, b, and c correspondingly is 3, 2, 3, 2, 3, 2. If the number left for a, b, and c is 3 there are 6 possible arrangements, but only for if the number left for a, b, and c is 2. Thus the total number of arrangement is 6x3+4x2=30 out of 4x4!=96. The probability is 5/16.
10324 10432, 12340, 13240. 13420. 14032. 14320. 20124. the firat 3 digits don.t mater they can within in peramters 12, 20, 24, 32 , 40 have to be the last 2 digitd any random combo of leftover digits except zero as first digit
So, the rules are: 1. There are 120 unique arrangements of the digits 0, 1, 2, 3, and 4 if no digits are repeated. 2. "0" cannot be the first digit. 3. The 5th digit must be even. 4. If the 5th digit is 0 or 4, the 4th digit must be even. 5. If the 5th digit is 2, the 4th digit must be odd. So, a table is helpful, even if I have to go to extreme lengths to make one in UA-cam. (Anyone know how to type a comment in monospaced font?) The first row shows the first digit. The first column shows the 5th digit, and the entries show the permissible combinations of the 4th and 5th digits. Numbers are represented by the hours on a clock, except for zero (goose egg). Pluses, minuses, and exclamation points are gridlines, and blanks are for spacing. "🚫🚫" means the combination that _would_ be in that spot is impermissible because one of the digits is already used for the first digit. ◻❗🕐◻❗🕑◻❗🕒◻❗🕓◻ ➖➕➖➖➕➖➖➕➖➖➕➖➖ 🥚❗🕑🥚❗🚫🚫❗🕑🥚❗🕑🥚 ◻❗🕓🥚❗🕓🥚❗🕓🥚❗🚫🚫 ➖➕➖➖➕➖➖➕➖➖➕➖➖ 🕑❗🚫🚫❗🚫🚫❗🕐🕑❗🕐🕑 ◻❗🕒🕑❗🚫🚫❗🚫🚫❗🕒🕑 ➖➕➖➖➕➖➖➕➖➖➕➖➖ 🕓❗🥚🕓❗🥚🕓❗🥚🕓❗🚫🚫 ◻❗🕑🕓❗🚫🚫❗🕑🕓❗🚫🚫 So, we can count 15 permissible combinations for the 1st, 4th, and 5th digits. For each permissible combination, the remaining 2 digits can be either in order A-B or B-A, so there are 2 x 15 = 30 possible combinations out of 120 total, and 30/120 = 25%. Anyone want to try to draw out a decision tree?
i’ve watched other mathematics youtubers (blackpenredpen, michael penn, etc.), but you have quickly become my favorite! i can’t put it into words, but your videos give me incredible joy!!
I Agree ......he makes math so tangible....
And not afraid to make mistakes. Very refreshing.
Same lol
I love your channel!I I’m also impressed by your delivery! It may sound stupid, but your chalkboard is the best!
I like the sound!
I love the standard chalkboards. In my school they use whiteboards (i am support staff aka the electrician) and they always complain about sun shining on the board ... scratch head.
From the digits shown I found the following possibilities X= any unused digit and Nz nonzero digit picked first:
X X X 2 0 3! positive outcomes
X X X 4 0 3!
NzX X 1 2 2 ways for first digit and 2! for rest so overall 4 ways
NzX X 3 2 4
X X X 0 4 3!
NzX X 2 4 4 ways
total 3*6+3*4=30 positive outcomes out of 5! =120 posible ones so 1/4 . Cute problem
Prime Newtons is probably the best math teacher on UA-cam! 😊
“You can’t mess this up.” Proceeds to almost mess this up…😂
The ever-present human factor
There are 120 ways to shuffle 5 digits. Yes, 03412 is 3412, therefore no 5 digit number, but it is a possible arrangement that just does not fit one of the requirements.
Yes I believe you are correct. I think the problem he wanted to propose was "Find the probability that a random 5 digit number formed by the digits 0,1,2,3,4 (without repetitions) is divisible by 4."
I took the time to calculate the answer in my head, only to find out that the problem was incorrectly stated. 😐
You are the best maths teacher on utube. Hope you can separate A level maths and university maths
This just made probability a thousand and times easier. Thanks a million sir!
Ps: i love the way you just paused and wondered why 42 isnt divisible by 4 lol, i was wondering the same thing😂
6:25 hehehe 🙂
A very nice explanation.
Technically the phrasing of the question in the thumbnail is "what is the probability that a random arrangement of 01234 forms a 5 digit number divisible by 4", that means we look at all 120 permutations of 01234, and pick the 30 that are both divisible by 4 and are 5 digits, so the answer would be 30/120=1/4
The two last digits requirements:
1. If the first digit is odd, the second digit should be 2 or 6.
2. If the first digit is even, the second digit should be 0, 4, or 8.
Cool! Make more of these probability videos
This is great, I didn't have any paper on me so I gave it a go in my head, took me several minutes but I did get there. Great question 😊
So at 6:10 "Okay we're done'. I'm yelling at the screen, NO!!! What about 20?!?! As you picked each digit you didn't work back with zero!! lol I'm sure you'll catch it in a minute.
42 is the answer to the life, universe and everything except for thr divisibility by 4! Great explanation based on common sense.
As per problem, sample space is "random arrangement of all five digits" and that should be 120.
I see. That was not my intention. I'd phrase better next time.
The thing is that since we can't take 0 in the first digit, so that excludes all those numbers which actually start with 0 and then rearrange with only the last 4 digits. That is why, the possibility is not 120, but 96.
@@sukhwinderchawla6037Yes, you are right. First of all with due regards to the sir, I would like to make it clear I am not trying to find fault with the solution. I just tried to interpret the phrasing of problem. In my view, for the solution given here, the phrasing of problem should have been like "five digit numbers formed by random arrangement of 0,1,2,3,4....." instead of "random arrangement of digits 0,1,2,3,4...".
I'm curious why you write the closed version of the digit 4, did you pick it up while learning to do block lettering? 🤔
I solved it myself without seeing your video! And it matched your answer 30/96. Now I will see how you did it.
How come you ignored all the arrangements starting with zero?
@@tomtomspa Because starting with 0 will not make a 5 digit number. I will remain a 4 digit. like 1234. The question wants us to form 5 digit numbers and then among them calculate the probability of finding a number which is divisible by 4.
@@sudiptoatutube no, this is an assumption not justified. The problem is to find the number of arrangements of five digits wich are divisible by four. Five digits arrangements is not the same as arrangements of five digits numbers.
03124 is most definitely such an arrangement.
@@tomtomspa Review his question again carefully.
@@sudiptoatutube”What is the probability that a random arrangement of all five digits 0-1-2-3-4 forms a number divisible by 4?” is literally the first frame of this video. Otherwise it should have been: “…that a random arrangement of all five digits 0-1-2-3-4 forms a five digit number divisible by four”.
It is clearly stated.
Did I miss the part of the video where you specified this should be done without replacement?
On the thumbnail it says "of all five digits", which implies each one has to be used once.
Of course, it also has no explicit specification that said number has to be a 5-digit number. It just says that it must form "a number divisible by 4". 01324, for example, fulfils both requirements even if it results in a 4-digit number.
I have a doubt. Why did you take 4 blocks instead of 5 while you checking the possible last two digits ?
Treat the last 2 digits as a block
question never asked you to form 5 digit number. it says " random arrangement of number of all 5 digits " then it says that arrangement form "a number" (never says 5 digit number) divisible by 4. for ex : 1324 can be arranged like '01324'. so total number of possible ways should be 120. so probability should be 1/4 or .25 .
Sir y u have taken 4 in the sample space?
Nice video
❤❤❤
If 0 can't be the first digit, this restriction should be clearly stated. Afterall, random means random.
My mans way too wholesome ❤
Expected the explanation: due too 100 being 4*25, we can ignore the 100-unit (3th digit) and above, and just look at the last 2 digts.
Solution:
If it has a 1 or 3 at the end, it is not divisible by 4, so 2/5 are already out.
of the other 3/5, it depends on the digit before. So we have:
10 > no
20 > yes
30 > no
40 > yes
02 > no
12 > yes
32 > yes
42 > no
04 > yes
14 > no
24 > yes
34 > no
This is 6/12 or 1/2.
In total, we have 1 - (2/5 + 3/5 * 1/2) = 1 - (4/10 + 3/10) = 1 - 7/10 = 3/10 = 30%
There is an even easier way. The number has to end in one of the six sequences: 04, 12, 20, 24, 32, 40.
The probability of getting 04 for the last two digits is 1 ∕ 5⋅1 ∕ 4 = 1 ∕ 20.
The same is true for the other five sequences as well, so the probability of getting either of them is 6⋅1 ∕ 20 = 0.3
EDIT: After watching the video I noticed that Newton didn't allow for the total sequence of 5 digits to begin with 0 (as if e.g. 01432 is not divisible by 4 LOL), which makes it a little trickier. It can still be solved using Bayes' theorem, but it's hardly any easier than the way Newton did it.
@@jumpman8282 Not really easier, just the inverse of my calculation. I calculated the probability of not being divisible by 4 and subtracted it from 1.
I think total outcomes should be 5! = 120.
Because in the question it is given: What is the probability that a random arrangement of the digits 0,1,2,3,4, forms a 5 digit number divisible by 4?
Which means favourable cases are the numbers which are 5 digit and also divisible by 4. And total arrangements of 5 digits, i.e , sample space is 120. So, probability = 30/120 = 1/4.
You are wrong. A 5 digit number cannot start with a zero.
@@maburwanemokoena7117 yes you are right but here the question is if we arrange the digits 0,1,2,3,4 randomly then what is the probability that we get a 5 digit number divisible by 4.
So, sample space is number of arrangement of 0,1,2,3,4 which are possible. And favorable cases are those arrangements which are 5 digit numbers and also divisible by 4.
So, Sample space= 5! = 120
And, favourable cases= number of 5 digit number divisible by 4 , which is = 30.
So, Probability= favorable cases/ total possible cases = 30/120 = 1/4
@@shivnathbanerjee5868 I do hear you, but normally in such cases we assume the first digit cannot be a zero. How do you think the question should’ve been phrased such that the above solution is true?
@@maburwanemokoena7117 it require you to form a number not 5 digit number, sir. please read again and also my comment (not this reply)
@@maburwanemokoena7117 it should have been phrased ' five digit number' instead of 'a number' in the tail end of the question
Like this question. The 0 being the first digits could create confusion but I understand and agree with your reasoning! My personal preference for getting the 96 is get the total number including the ones starting zero which is 120 the subtract 24 the number starting 0.
Warning for any answering any probability question - it is very easy 5o correctly answer a different question!
I didn't understand why four digit numbers divisible by 4 were calculated instead of five digit numbers divisible by 4 as the question asks. Would someone please answer my question. I am very weak at probability. I would very much appreciate it. Thanks.
But 5 digits numbers divisible by 4 were calculated, just as the question asked
That second part of the video was a bit inexact in that regard, but the answer is still correct. If the last 2 digits are fixed in each case, then you're just trying to arrange the other 3 digits. Once you've figured out the number of possibilities for the first 2, there's only 1 possibility for the last one. So each of his cases should have been multiplied by another (x1), to be technically correct, but that doesn't change the actual answer.
@@jeffreywood4759 Thank you very much.
😂 this guy can take maths exams for me. 😁😁😁😁😁😁
It is never stated that it has to be a five digit number, only that it has to be an arrangement of those five digits, so your answer is wrong. Why not counting 01234 as a possible arrangement?
Ok
I used the exact same logic as you to solve this
Well the total for out probability is 5! Unless it means any arrangement 5 numbers or less. Then, because Prime is a tactical genius, he will break down the different cases. Because I am a gorilla, I will smash my head against the wall trying to brute force it. Lets see whose method is better!
Once again, I overlook a detail of the problem that leads me to the wrong answe 😢
Let x=abcde, a≠0
x=10⁴a+10³b+10²c+10d+10e
4 | x --> 4 | 10d+e
(d,e)={(0,4),(1,2),(2,0),(2,4),(3,2),(4,0)}
Number of digits left for a, b, and c correspondingly is 3, 2, 3, 2, 3, 2.
If the number left for a, b, and c is 3 there are 6 possible arrangements, but only for if the number left for a, b, and c is 2. Thus the total number of arrangement is 6x3+4x2=30 out of 4x4!=96. The probability is 5/16.
10324 10432, 12340, 13240. 13420. 14032. 14320. 20124. the firat 3 digits don.t mater they can within in peramters 12, 20, 24, 32 , 40 have to be the last 2 digitd any random combo of leftover digits except zero as first digit
6/(4×4!)
10:35 you have forgotten the last two digits 44, sir.
Actually he didn't.
We are working with rearrangements, and as he states at 0:55, the numbers cannot repeat themselves, so 44 is not possible.
20 is divisible by 4
6:21 and also 44 sir.
nope, every number must be different and 44, 4 duplicates
I fail to see why, for example, 01432 should not be divisible by 4, but it makes the puzzle more interesting so I forgive you :)
1432 sure is divisible by 4. But it's not a five digit number
@@twsupersieben "...a random arrangement of all five digits forms a number...". It doesn't say that it has to be a 5-digit number.
I added this additional information to the thumbnail
So, the rules are:
1. There are 120 unique arrangements of the digits 0, 1, 2, 3, and 4 if no digits are repeated.
2. "0" cannot be the first digit.
3. The 5th digit must be even.
4. If the 5th digit is 0 or 4, the 4th digit must be even.
5. If the 5th digit is 2, the 4th digit must be odd.
So, a table is helpful, even if I have to go to extreme lengths to make one in UA-cam. (Anyone know how to type a comment in monospaced font?)
The first row shows the first digit. The first column shows the 5th digit, and the entries show the permissible combinations of the 4th and 5th digits. Numbers are represented by the hours on a clock, except for zero (goose egg). Pluses, minuses, and exclamation points are gridlines, and blanks are for spacing. "🚫🚫" means the combination that _would_ be in that spot is impermissible because one of the digits is already used for the first digit.
◻❗🕐◻❗🕑◻❗🕒◻❗🕓◻
➖➕➖➖➕➖➖➕➖➖➕➖➖
🥚❗🕑🥚❗🚫🚫❗🕑🥚❗🕑🥚
◻❗🕓🥚❗🕓🥚❗🕓🥚❗🚫🚫
➖➕➖➖➕➖➖➕➖➖➕➖➖
🕑❗🚫🚫❗🚫🚫❗🕐🕑❗🕐🕑
◻❗🕒🕑❗🚫🚫❗🚫🚫❗🕒🕑
➖➕➖➖➕➖➖➕➖➖➕➖➖
🕓❗🥚🕓❗🥚🕓❗🥚🕓❗🚫🚫
◻❗🕑🕓❗🚫🚫❗🕑🕓❗🚫🚫
So, we can count 15 permissible combinations for the 1st, 4th, and 5th digits. For each permissible combination, the remaining 2 digits can be either in order A-B or B-A, so there are 2 x 15 = 30 possible combinations out of 120 total, and 30/120 = 25%.
Anyone want to try to draw out a decision tree?
That's a lot of work you did there 👏