i am not underage, i'm just inept at probabilitywe have a pin code made up of 4 *distinct* numbers, guessing the first digit has a probability of 1/10, guessing the next should have a probability of 1/9 and so on (because the numbers are distinct), therefore the probability of someone guessing the code by inputting random digits is 1/(10*9*8*7), is this correct?moreover the probability of someone guessing all digits but not necessarily in the correct position should be the previous probability multiplied by 4! because that's all the possible combinations of the digits one can try, correct?finally someone guessing at least 2 numbers in the correct position should be (1/(10*9) * (1/8*7)*2!) because it's the probability of guessing 2 digits in 2 correct positions * all possible combinations of the remaining digits, is this correct?
That makes no sense though. Is it not possible for a pin to be the same 4 digits? Wouldn't it be 10x10x10x10 and not 10x9x8x7 as there is a chance that the numbers don't vary within the code
>>1460096sorry i was perhaps a bit unclear, the problem states that the digits are distinct
>>1460092I think there’s a mistake in your third part, the numerator should be (4 choose 2) * 8 * 7 because that would be the number of combinations that share two digits with the correct code. (4 choose 2 comes from choosing which digits are correct and 8*7 comes from the remaining digits).
>>1460130so i can't treat those as separate digits? how do i properly recognize when to use (x choose y)? i understand it basically means "the number of groups of y elements i can make from x elements" but sometimes i don't understand which numbers to use in, for example, urn problems
>>1460133In the case of this particular problem, saying first and second digits are correct is no different from saying second and first digits are correct. I don’t know if this helps very much, but one way I think about it is seeing if the order i choose the elements in matters to what I’m looking for. (E.g. if I want to see the probability that i draw two different color balls from an urn is, the order doesn’t matter for this event, but for the probability that I draw a red ball first and a blue ball second, the order does matter).
>>1460133>>1460150Also for probability specifically, the idea of sample space and event are helpful. In the case of this problem, the sample space is all the valid pass codes and an event would be something like passcode contains the digit nine. Since this event doesn’t specify where the nine occurs in the passcode, the numerator of the probability would involve a combination (because your event includes multiple orderings of digits), and then the denominator would be the size of your sample space.
lol the password is 1 2 3 4
>>1460151i think i get what you're sayingso let's take the event "someone guesses the first digit correctly and in the correct position"theres a probability of 1/10 to guess it because since its the first position it means we havent chosen any other digits yet (if it was second position it would be 1/9), however once guessed the other digits are bound by that choice, for example if the first digit is correctly guessed and it is 9 it means every combination WOULD need to be calculated as 1/9*8*7 however since they're bound by the first choice i only have to consider the groups that start with 9 and those groups are all of size 3 so 4 choose 3 multiplied by however many digits i can choose from (9*8*7 in this case), does this make sense?also when i normally write 4 choose 3 i'd be considering combinations that contain another 9 in it but since im dividing by 9*8*7 i am, in a way, only considering the probability of every other combination not counting the probability of the first digit, also simplifying the fraction only leaves me with 4 choose 3if an event were "someone guesses one digit" without specifying in which position and which digit it is does that make the problem incomplete or do i have to add up the probability of every possible correct guess? (so 1/10 + 1/9 + 1/8 + 1/7)
>>14600921. Yes that's correct.2. No that's incorrect. The probability of such a thing is 24/5040.3. 365/5040
>>1460297Hmm, I think I didn't explain myself clearly.>if it was second position it would be 1/9The event "someone guesses the first digit correctly" has the same probability as the event "someone guesses the second digit correctly".I'll try to explain the ideas of event and sample space in more detail that makes it easier to see why this is not true. So like I said earlier, sample space is the set of all possibilities we are considering (in this case all of the valid guesses for passcodes). Events have to be subsets of that sample space. In the case of this problem, this means that events are sets containing valid passcode guesses, so the event you describe at the bottom is not actually well defined. Thinking about the event "someone guesses the first digit correctly" can be rephrases as "all passcodes that share the first digit with the correct passcode". In most cases where you deal with discrete events, the probability is defined as # of things in event / # of things in sample space. The probability of the event mentioned above is thus (1*9*8*7)/(10*9*8*7) = 1/10, which is what you would expect.Now, intuitively, the number of passcodes that have the second digit as the correct passcode should be equal to the number of passcodes that have the same first digit. This suggests that the event "someone guesses the first digit correctly" has the same probability as the event "someone guesses the second digit correctly". To actually calculate this, you have to be a bit careful. Let's say the correct passcode 1 2 3 4. Then any passcode that has the same second digit can't start with 2. So the event has a size of (9*1*8*7), which is the same as the size of the event "someone guesses the first digit correctly". I think this is why thinking systematically in terms of events is helpful because it can help you catch subtleties like this.
>>1460297>>1460313Sorry, I had to split the post up.n choose k and n permute k are kind of like shortcuts you can use in calculating the size of events. For the third question in your OP, this question is tricky because the position of the two correct digits aren't specified. However, it's easy to reason that if I give you two correct digits and their positions, then there are 8*7 (8 permute 2) possible passcodes. Note that we use the permute operation because the position of the digits matter (1 X 2 X is not the same as 2 X 1 X). So now all we need to know is the number of possibilities for what the positions of the correct digits are. Naively, we might guess 4*3 (4 permute 2), reasoning that there are 4 possible positions for the first correct digit and 3 for the second. But first correct digit and second correct digit are completely arbitrary, as this adds an order to a situation where one doesn't exist, so we should divide by 4!, which gives 4 choose 2. The size of this event is thus 4 choose 2 * 8 permute 2. This might be super redundant, but a lot of people use this terminology, which could help when looking at other people's explanations.>>1460150Also, I realized this example is kind of misleading. The event "the color of the two balls I drew are different" is a situation where the fact that the order doesnt matter actually increases the size of the event, because the different orderings are individual elements of this event (this is called undercounting). In the passcode scenario, it's not valid to consider first and second digits are correct as distinct from second and first digits are correct. This is why 4*3 would not give the correct answer (we overcounted the number of possibilities by using 4 permute 2).
>>1460303where do 24 and 365 come from?>>1460318i'm still having a bit of trouble understanding, like i understand where you say given 2 correct digits there are 8*7 possibilities left, but i dont get where the 4*3 comes from (im ESL so i may have just misunderstood what you mean)my professor does write all the events in a rather verbose way in assignments so maybe if i do that it will help me think about problems the correct wayi will try some easy excercises and work my way up to harder ones, thank you for the explanations
>>1460357>where do 24 and 365 come from?24 is 4! which gives the same answer as in your op. Also, I’m not entirely confident in my own answer, but 365 = 5*73, so I think that anon is just mistaken.>4*3A correct digit can be in the first, second, third, or fourth position. Once we fix the position of the first correct digit, this leaves three remaining positions for the second correct digit. Then we have to divide by 2! so we dont overcount.I’m not exactly a master at this stuff, so I hope my explanations gave some help, good luck in your class.
>>1460374Oops, I realized there’s a mistake in my answer to the third problem, here’s why.Let’s say I guess the correct passcode, then all of the digits are correct. My method of counting considers cases where the first and second digits are correct, the first and third digits are correct, and so on. But the correct passcode fits all of these cases, so I have overcounted the number of passcodes in my event. So the correct approach would involve splitting this event into specific cases, exactly four digits are correct, exactly three digits are correct, and exactly two digits are correct.
>>1460404this makes my brain hurt, the professor isnt even good at explaining eitherand this is supposed to be one of the easier assignments, well i will look up some problems with an attached explanationthanks for your time
>>1460357Yeah I was mistaken about you being incorrect on the second one. I thought the exclamation point was excitement, not a factorial. My bad.365 is the addition of all instances of only 2 numbers being correct, only 3 numbers being correct, and all 4 numbers being correct. 8*7*6 + 7*4 + 1
