thepillow

haha:). here's that q5 explanation: We are choosing a pair of 2 papers from the 4 papers, and repeating this over and over until we have selected all of the possible pairs of 2. Because it doesn't matter in which order we pull out the 2 pieces of paper, only 6 unique pairs exist (4c2). These are: [ 0 ] and [ 1 ] [ 0 ] and [ 2 ] [ 0 ] and [ 3 ] [ 1 ] and [ 2 ] [ 1 ] and [ 3 ] [ 2 ] and [ 3 ] Now, there is a way to avoid any further computation on this problem if we notice a certain feature of our group of 4 papers: there is no way to get a sum of 3 with different colored papers!! In other words, if the numbers don't add to 3 then the papers must be different colors. If we select two papers then (A) the numbers will always add to 3 or (B) the papers will be different colors. So there is a 100% chance that (A) or (B) occurs. So all 6 of the 6 possible unique 2-paper combinations will have a sum of 3 or the papers will be different colors. This seems really obvious once you know the answer but it is definitely not obvious initially. It's quite tricky. I think drawing the scenario really helps, but even doing that doesn't make it completely apparent.

That's the million dollar question! Well, I think the first step should always be to "translate" the question into more intuitive and colloquial language. This might seem trivial, but unless it is immediately apparent to you what the question is asking, this little trick can make a huge difference. If you can express the same question in different words, then you have almost certainly truly understood what it is asking. I'll use one of your original questions as an example: "A fair die (sides having numbers 1-6) is to be rolled until the first number greater than 4 appear.What is the probability that first number greater than 4 will appear on 3rd or 4th roll?" The language here isn't too convoluted, but we can take the question apart and rephrase it so it is more obvious what's going on. The first part - "a fair die (sides having numbers 1-6) is to be rolled" - is easily understood, but let's just change it to "we roll a die". We can assume it's fair unless told otherwise. Then we have "until a number greater than 4 appears". This can be restated as "until getting a 5 or a 6". The next part - "the first number greater than 4 will appear on the 3rd or 4th roll" - can be "translated" as "this takes us 3 or 4 rolls". Now the whole question is: We roll a die until getting a 5 or a 6. What is the probability that this takes 3 or 4 rolls? Again, the original question was: "A fair die (sides having numbers 1-6) is to be rolled until the first number greater than 4 appear.What is the probability that first number greater than 4 will appear on 3rd or 4th roll?" In my opinion our new version is much easier to deal with. But they both say the same thing! It might seem like this type of translation is a slow process and that it will waste time on the test. But with a bit of practice this can become second nature and can end up saving you time.

1. On any given roll the odds of getting a number greater than 4 is 2/6 = 1/3. (Only 5 and 6 are greater than 4, so this is 2 out of the 6 numbers.) This means that the probability of getting a number that is 4 or less is 2/3. We'll use these numbers in a little bit. We want to consider 2 cases: Case 1: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is greater than 4. We can write this as: P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (1/3) Case 2: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is less than or equal to 4 and our fourth roll is greater than 4. We can write this as: P(4 or less) x P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (2/3) x (1/3) We want to know about either case happening so we can add up these probabilities: (2/3)(2/3)(1/3) + (2/3)(2/3)(2/3)(1/3) = 4/27 + 8/81 = 12/81 + 8/81 = 20/81 2. In order for there to be a range of 7 we need the largest number minus the smallest number to equal 7. From the numbers between 1 and 10, we can make 3 different pairs that satisfy this: 8 and 1; 9 and 2; 10 and 3 Let's take 8 and 1 as an example: along with our 8 and 1, we need 2 other discs out of the 6 numbers between 8 and 1 (we can't get a 9 or a 10 because that would change the range). So there are 6c2 = 15 different pairs we could choose for our other 2 discs. This works out the exact same way for 9 and 2, as well as 10 and 3. So in each of the 3 cases there are 15 combinations for a total of 45. The number of total groups of 4 that we can get from the 10 discs is 10c4 = 210. So 45 of the 210 groups meet our requirements. 45/210 can be reduced to 3/14. Hope that helps! Let me know if I need to explain something better.

hey Rain, 1. Useful fact: a right triangle maximizes the area. So if you know 2 sides of a triangle, then you can figure out the maximum possible area by treating the triangle as a right triangle. With sides of 12 and 18, the largest right triangle will be if these 2 sides are the legs (that way the hypotenuse is bigger than both of them). The area would be (1/2)(12)(18) = 108. So 112 is larger than the maximum possible area of a triangle with the 2 sides we're given. 2. We have: AB + BC = 22 AB + AC = 13 This tells us that the 2 equal sides can't be AC and BC (otherwise AB + BC and AB + AC should be equal). So AB is either equal to BC or to AC. If AB = BC, then using the equations we can determine that they both are 11 and AC is 2. The perimeter of the triangle would be 24. If AB = AC, then we can determine that they are both 6.5 and BC is 15.5. The perimeter of the triangle would be 28.5. However,can this triangle exist? The third side of a triangle must be less than the sum of the other 2, and 15 > 6.5 + 6.5, so this triangle is not possible. Therefore the perimeter can't be 28.5. We're left with 24, which is less than 28. 3. Standard deviation is a measure of how spread out a set of numbers/data is. Adding the 11th number will increase the average distance from the numbers to the mean, so this set must have a slightly higher standard deviation. 4. Let's say, for example, that we're dealing with saltwater solutions. Let's let A = amount we need of the solution with 20% salt and B = amount we need of the solution with 50% salt. The total amount of liquid we will have is therefore A + B. Out of this A + B how much will be salt? In A, 20% will be salt. We can write this as (0.2)A. In B, 50% will be salt. We can write this as (0.5)B. The total amount of salt is therefore (0.2)A + (0.5)B. So in our new solution the % of salt will be the amount of salt divided by the total amount of liquid. We want this to be 40%, so we can write: ((0.2)A + (0.5)B) / (A+B) = 0.4 Simplifying we get: (0.2)A + (0.5)B = (0.4)A + (0.4)B (0.1)B = (0.2)A (0.1)/(0.2) = A/B 1/2 = A/B So the ratio of A:B should be 1:2 Hope that helps! Let me know if I didn't explain something clearly enough.

Hey Kamakshi, I'm glad you like my posts! Hopefully they are helpful. Your updated answer to question 1 is correct (it's 90). For question 2 you actually had the right answer the first time. We we do care about order of flavors but not order of toppings. So for the flavors is 15p3 and the toppings 5c3. This works out to (15p3)(5c3) = 15*14*13*10 = 27,300. Your answer for question 3 is also correct (it's 142) The answers to questions 4 and 5 need a bit of tweaking. Let me give you a few hints and then see what you come up with: For question 4, you can visualize the situation like this: _P_R_V_D_ The five blanks are the places where the three vowels could potentially go. And then we also need to take into account the fact that PRVD don't need to appear in this order. For question 5, try visualizing the 4 pieces of colored and numbered paper like this: [ 0 ] - [ 1 ] - [ 2 ] - [ 3 ] It's also important that the question asks for a sum of 3 or different colors. Hope that helps!

Ok, here are 5 counting problems that definitely require a bit of careful consideration. Definitely feel free to post your responses even if you doubt they're correct. No one is getting graded here :). (Pemdas, if you want you can PM me your answers, but I think you've got these under control. I think we should leave some time for anyone to give them a try and respond before posting the solutions)​ Problem 1 How many positive and even 3-digit integers contain at least one digit that is a 7? Problem 2 At the ice cream shop there are 15 available flavors and 5 available toppings. I want to make a cone with 3 scoops of ice cream (each a different flavor) and 3 different toppings. How many ways can I do this if I don't care about the order the toppings go on, but I do care about the order of the flavors? Problem 3 At a certain hospital, every doctor has had extra training in dermatology, radiology, cardiology, or some combination of the three. 60 of the doctors have extra training in dermatology, 75 in radiology, and 100 in cardiology. If 34 doctors have extra training in both radiology and dermatology, 25 have extra training in both cardiology and dermatology, 44 have extra training in both cardiology and radiology, and 10 doctors have had extra training in all three specialties, how many doctors work at the hospital? Problem 4 How many ways can we arrange the letters in the word PROVIDE so that no two vowels are adjacent? Problem 5 A box contains four pieces of paper. They are labeled 0, 1, 2 and 3. The paper labeled 0 and the paper labeled 3 are both colored blue. The paper labeled 1 and the paper labeled 2 are both colored green. We choose 2 pieces of paper at random and record the sum of the numbers on their labels and we record their colors. If we repeat this process until we have selected each of the different possible pairs of 2 pieces of paper, then in what fraction of the unique outcomes was the sum of the two numbers equal to 3 or the colors of the two papers different?

Hi Fahad, Thanks for your nice words. I'm happy to do whatever you think will help everyone the most. You said that you have some questions already. If you want, send me them in a PM and I'll take a look. Hope all is well

Thanks. I didn't exactly understand what he was referring to before, but yes it can definitely be computed like this. Hadn't even thought of it.

Oh I see what you mean. I misunderstood what you where saying in your earlier post.

Hey Fahad, Consider the numbers: 1, 2, 128, 378 (i just picked these randomly) The average of these 4 numbers is 127.25. Of the 4 numbers, 1 and 2 are below the mean, 128 and 378 are above the mean. But the difference between 1 and 2 is clearly not equal to the difference between 128 and 378. In our example, in order for the average of the terms in Set G to be 100, the only thing we need to make sure of is that the sum of the 55 terms divided by 55 = 100. To achieve this, we need: g1 + g2 + g3 + ... + g54 + g55 = 5500 A set of 55 terms that adds to 5500 will have an average of 100. If we have: {g1, g2, ..., g27} = {1, 2, ..., 27} {g28, g29, ..., g54} = {100, 101, ..., 126} g55 = 2071 then we have 55 terms with a total sum = Sum(1 to 27) + Sum(100 to 126) + 2071 = 378 + 3051 + 2071 = 5500 We when divide by the number of terms we get: Average = 5500/55 = 100 Does that help?

Great question.Here are a few general guidelines: If we say that a set of numbers has mean µ and standard deviation σ (you can use any symbols, these are just conventions), then: 1. If some constant a is added to each number in a set, then the new set will have mean µ + abut the standard deviation will remain σ (because the numbers changed but they are no more or less spread out than before). This is true for subtraction as well. •• Example: for {1, 2, 3} µ = 2 and σ = 1. If we add 2 to each number we get {3, 4, 5} with µ = 2 + 2 = 4 and σ = 1. 2. If every number in a set is multiplied by some constant a, then the new set will have mean a∙µ and standard deviation a∙σ (because the numbers in the new set will be more spread out). This is also true for division by a constant, because division by a= multiplication by 1/a. •• Example: for {1, 2, 3} µ = 2 and σ = 1. If we multiply each number by 2 we get {2, 4, 6} with µ = 2∙2 = 4 and σ = 1∙2 = 2. 3. If the numbers in a set are all identical, then the set has standard deviation σ = 0 and it will remain 0 after performing the same operation (multiply, divide, add, subtract, etc) on all of the numbers in the set (because they will still be identical to each other). The mean will just become the new value of all the identical terms (i.e. we perform the same operations on the mean as we do on each term). •• Example: for {5, 5, 5} µ = 5 and σ = 0. If we add 2 to each number and then multiply each resulting number by 3 we get {3(5+2), 3(5+2), 3(5+2)} = {21, 21, 21} with µ = 3(5 + 2) = 21 and σ = 0. A brief summary, if done to all terms then: addition/subtraction changes the mean but not standard deviation. Multiplication/division changes both. Hope that helps!

Hey Fahad, For this question here are my 2 preferred methods for solving it: Method 1: The probability of picking Eva is 1/5. Then there are 4 names left and so the probability of picking Adrian is 1/4. The probability of Eva & then Adrian is (1/5)(1/4) = 1/20 = .05. But in this calculation the order in which we've chosen them matter. So we also need to account for picking Adrian & then Eva, which is also a probability of .05. So the total probability is .05 + .05 = .10. Method 2: We're choosing 2 names from 5. There are 5c2 = 10 ways to do this. Only 1 of these ways includes both Eva and Adrian. So this is 1/10 = .10. Pemdas, was using a slightly different version of Method 2, but he considered the permutations. Choose 2 from 5 when order matters = 5p3 = 5*4 = 20. This is just as valid, but in this case we've counted Adrian and Eva as different from Eva and Adrian. So 2 out of the 20 give us Eva and Adrian. 2/10 = 1/10 = 0.1. Does that help?

Hey Fahad, the reason I multiply by 24 is because this is the number of times that each of our 5 numbers will appear in the units place, the tens place, hundreds place, etc. Maybe this will make it a bit clearer: Here's an abridged list of the numbers we get when rearranging our digits: 13579 19537 17395 ... 31597 39157 ... 59317 57931 ... 79315 73159 ... 93517 97531 These are just a few of the numbers, but if you listed all of them out you'd see that you would have 120 numbers. 24 of them would start with 1, 24 would start with 3, 24 would start with 5, etc. 24 would also have 1 as the second digit, 24 would have 3 as the second digit, etc. This is true for all the number and all the digits. This is because if we take any of our 5 numbers and pick which digit it will be, then there are 4! = 24 ways to rearrange the other numbers around it (regardless of which position it's in). Does that make sense? I'm not entirely sure if I've answered your question.

I might not have explained it clearly, but after round 50 we still are flipping cards, it's just that it works out to be only 1 card each time. Every nth card would be every card corresponding to a multiple of n so, just like every 2nd would be card 2, card 4, card 6, etc, when we get to every 51st card we would flip card 51, card 102, etc. Since we have only 100 cards, we just flip card 51. So, when we get to the 76th round of flipping, for example, we're just flipping card 76 because we don't have a card 152 or a card 228, etc. After this entire process, the only way a card can have its red side facing up is if it has been flipped an odd number of times. They start as blue so 1 flip makes a card red, a 2nd flip turns it back to blue, 3rd red, 4th blue, 5th red, etc... So if we can figure out how many cards are flipped an odd number of times, we'll know how many are red. There's actually a bit of a trick to this, which is the fact that all numbers have an even number of factors except for perfect squares. This is because factors generally come in pairs, but perfect squares have a factor that doesn't have a pair (the one that gets multiplied by itself). When we're flipping every 2nd card, every 3rd card, every 4th card etc, we're essentially flipping the multiples of 2, of 3, of 4, etc. So, for example, card #12 will be flipped in the first round (when all cards are flipped), the 2nd round, the 3rd round, the 4th round, the 6th round and the 12th round (all the factors of 12). Likewise, card 13 would only be flipped when we flip all the cards and when we flip every 13th card. There are 10 perfect squares between 1 and 100 (inclusive). So cards 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be flipped an odd number of time and therefor have their red sides up at the end. Taking card 49 as an example, we can see why it will be red: it's flipped in the 1st round (every card is), and the 7th round (when every 7th card is flipped) and then only one more time in the 49th round (when only cards 49 and 98 are flipped). So these three flips will leave card 49 red-side up.

There are 100 cards on a table. Each of the cards is blue on one side and red on the other side. At first, all the cards are positioned so that the blue side is facing up. Next, you perform these actions: First, you flip every card over so red is up. Then you flip every second card back on its other side. Then you take every third card and, if red is facing up you flip it to blue, and if blue is facing up you flip it to red. You repeat this with every fourth card, then every fifth card, then every sixth card, and continue this pattern (7th, 8th, 9th...98th, 99th) until you finally are just flipping only the 100th card. So there will be 100 different "rounds" of flipping. At the end of these 100 rounds, how many cards have their red side facing up?