answer：First, I need to calculate the total calories in the pizza. There are 12 slices, and each slice has 220 calories. So, I multiply 12 by 220 to get the total calories in the pizza.Next, I'll calculate the total calories in the tray of cookies. There are 8 cookies, and each cookie contains 270 calories. I multiply 8 by 270 to find the total calories in the cookies.Finally, to find out how many more calories the pizza has compared to the cookies, I subtract the total calories of the cookies from the total calories of the pizza.

answer：First, I note that Emily is currently 8 years old.Olivia is three times older than Emily, so Olivia's current age is 3 multiplied by 8, which equals 24 years.In five years, Emily will be 8 plus 5, which is 13 years old.Similarly, Olivia will be 24 plus 5, which is 29 years old.The difference between their ages in five years will be 29 minus 13, resulting in 16 years.

answer：To determine the value of ( f(-2) ), I start by recalling that ( f(x) ) is an even function. By definition, an even function satisfies the property ( f(-x) = f(x) ) for all ( x ) in its domain.Given that ( f(x) = x^2 + 1 ) when ( x > 0 ), I can use the evenness property to find ( f(-2) ). Applying the property, ( f(-2) = f(2) ).Next, I substitute ( x = 2 ) into the given function: ( f(2) = 2^2 + 1 = 4 + 1 = 5 ).Therefore, ( f(-2) = 5 ).

answer：First, I'll define variables for the number of each type of book: let ( z ) represent Zhuangzi, ( k ) Kongzi, ( m ) Mengzi, ( l ) Laozi, and ( s ) Sunzi.From the problem, I know that the total number of books is 300, so:[ z + k + m + l + s = 300 ]The total cost of the books is 4500 yuan, with their respective prices:[ 10z + 20k + 15m + 30l + 12s = 4500 ]It's given that the number of Zhuangzi books is equal to the number of Kongzi books:[ z = k ]Additionally, the number of Sunzi books is 15 more than four times the number of Laozi books:[ s = 4l + 15 ]Next, I'll substitute ( k ) with ( z ) and ( s ) with ( 4l + 15 ) in the total number of books equation:[ z + z + m + l + (4l + 15) = 300 ][ 2z + m + 5l = 285 ]I'll also substitute these into the total cost equation:[ 10z + 20z + 15m + 30l + 12(4l + 15) = 4500 ][ 30z + 15m + 78l = 4500 - 180 ][ 30z + 15m + 78l = 4320 ]To simplify, I'll divide the entire equation by 3:[ 10z + 5m + 26l = 1440 ]Now, I'll use the simplified total number of books equation:[ 2z + m + 5l = 285 ][ m = 285 - 2z - 5l ]Substituting ( m ) back into the cost equation:[ 10z + 5(285 - 2z - 5l) + 26l = 1440 ][ 10z + 1425 - 10z - 25l + 26l = 1440 ][ l = 15 ]Finally, using the value of ( l ) to find ( s ):[ s = 4 times 15 + 15 = 75 ]

answer：Okay, so I have this problem where I need to find the expression for ( f(x) ) when ( x ) is between -4 and -2. The function ( f(x) ) is even, which means ( f(x) = f(-x) ) for all real numbers ( x ). Also, it's given that ( f(2 + x) = f(2 - x) ), which suggests some kind of symmetry around the line ( x = 2 ). Additionally, when ( x ) is in the interval (0, 2), ( f(x) = frac{1}{x} ). First, let me try to understand the symmetries involved here. Since ( f(x) ) is even, it's symmetric about the y-axis. That means whatever happens on the right side of the y-axis will mirror on the left side. So, if I can figure out ( f(x) ) for positive ( x ), I can just mirror it to get the negative side.Now, the condition ( f(2 + x) = f(2 - x) ) tells me that the function is symmetric about the line ( x = 2 ). So, for any point ( x ), the value of the function at ( 2 + x ) is the same as at ( 2 - x ). This is like a reflection over the vertical line ( x = 2 ).Given that ( f(x) = frac{1}{x} ) on (0, 2), I can use these symmetries to extend ( f(x) ) beyond this interval. Let me try to visualize this. Starting with the interval (0, 2), where ( f(x) = frac{1}{x} ). Since the function is even, on the interval (-2, 0), ( f(x) = f(-x) = frac{1}{-x} = -frac{1}{x} ). Wait, hold on, that doesn't sound right because ( f(x) ) is even, so it should mirror the positive side. So, actually, ( f(x) = f(-x) ), so if ( x ) is negative, ( f(x) = f(-x) = frac{1}{-x} ) only if ( -x ) is in (0, 2). So, for ( x ) in (-2, 0), ( f(x) = frac{1}{-x} ). Hmm, but ( frac{1}{-x} ) is negative, but since ( f(x) ) is even, it should be positive. Wait, maybe I made a mistake here.Wait, no. If ( f(x) ) is even, then ( f(-x) = f(x) ). So, if ( x ) is in (-2, 0), then ( -x ) is in (0, 2), so ( f(x) = f(-x) = frac{1}{-x} ). But ( frac{1}{-x} ) is negative, but ( f(x) ) should be positive because it's even. Hmm, that doesn't make sense. Maybe I got confused.Wait, no, actually, ( f(x) = f(-x) ), so if ( x ) is negative, ( f(x) = f(-x) ), and since ( -x ) is positive, ( f(-x) = frac{1}{-x} ). So, ( f(x) = frac{1}{-x} ) when ( x ) is in (-2, 0). But ( frac{1}{-x} ) is negative, but ( f(x) ) is even, so it should be positive. Hmm, maybe I need to reconsider.Wait, perhaps I'm overcomplicating. Let me think step by step.1. ( f(x) ) is even: ( f(x) = f(-x) ) for all ( x ).2. ( f(2 + x) = f(2 - x) ) for all ( x ): so the function is symmetric about ( x = 2 ).3. ( f(x) = frac{1}{x} ) for ( x in (0, 2) ).So, starting from ( x in (0, 2) ), we can use the symmetry about ( x = 2 ) to find ( f(x) ) in another interval.Let me consider the reflection about ( x = 2 ). For any ( x ), ( f(2 + x) = f(2 - x) ). So, if I take ( x = t ), then ( f(2 + t) = f(2 - t) ). So, for ( t ) such that ( 2 + t ) is in some interval, we can relate it to ( 2 - t ).Let me try to find ( f(x) ) for ( x in (2, 4) ). Let me set ( x = 2 + t ), where ( t in (0, 2) ). Then, ( f(2 + t) = f(2 - t) ). But ( 2 - t ) is in (0, 2), so ( f(2 - t) = frac{1}{2 - t} ). Therefore, ( f(2 + t) = frac{1}{2 - t} ). So, substituting back, if ( x = 2 + t ), then ( t = x - 2 ), so ( f(x) = frac{1}{2 - (x - 2)} = frac{1}{4 - x} ). Therefore, for ( x in (2, 4) ), ( f(x) = frac{1}{4 - x} ).Okay, so now I have ( f(x) ) defined on (0, 2) as ( frac{1}{x} ) and on (2, 4) as ( frac{1}{4 - x} ). Since the function is even, I can mirror these to the negative side.So, for ( x in (-2, 0) ), ( f(x) = f(-x) ). Since ( -x in (0, 2) ), ( f(-x) = frac{1}{-x} ). Therefore, ( f(x) = frac{1}{-x} ) for ( x in (-2, 0) ). Wait, but ( frac{1}{-x} ) is negative, but since ( f(x) ) is even, it should be positive. Hmm, so maybe I need to take the absolute value? Or perhaps I made a mistake.Wait, no. Let me think again. If ( f(x) ) is even, then ( f(-x) = f(x) ). So, if ( x ) is negative, ( f(x) = f(-x) ). Since ( -x ) is positive, ( f(-x) = frac{1}{-x} ) only if ( -x in (0, 2) ). So, for ( x in (-2, 0) ), ( f(x) = frac{1}{-x} ). But ( frac{1}{-x} ) is negative, but ( f(x) ) is even, so it should be positive. Hmm, that seems contradictory.Wait, maybe I need to reconsider the definition. Since ( f(x) ) is even, ( f(-x) = f(x) ), so ( f(x) = f(-x) ). If ( x ) is negative, ( -x ) is positive, so ( f(-x) = frac{1}{-x} ) only if ( -x in (0, 2) ). So, for ( x in (-2, 0) ), ( f(x) = frac{1}{-x} ). But since ( f(x) ) is even, it should be equal to ( f(-x) ), which is ( frac{1}{-x} ). Wait, but ( f(-x) ) is ( frac{1}{-x} ), but ( f(x) ) is ( f(-x) ), so ( f(x) = frac{1}{-x} ). But that would make ( f(x) ) negative for ( x in (-2, 0) ), which contradicts the fact that ( f(x) ) is even, because if ( f(x) ) is even, it should take the same value as ( f(-x) ), which is positive.Wait, maybe I'm confusing something here. Let me take a specific example. Let ( x = 1 ), which is in (0, 2). Then ( f(1) = 1/1 = 1 ). Since ( f(x) ) is even, ( f(-1) = f(1) = 1 ). But according to the formula ( f(x) = frac{1}{-x} ), when ( x = -1 ), ( f(-1) = frac{1}{-(-1)} = 1 ). Oh, wait, that's correct. So, actually, ( f(x) = frac{1}{-x} ) when ( x in (-2, 0) ) simplifies to ( f(x) = -frac{1}{x} ), but since ( x ) is negative, ( -frac{1}{x} ) is positive. So, for ( x = -1 ), ( f(-1) = -frac{1}{-1} = 1 ), which is correct.So, in general, for ( x in (-2, 0) ), ( f(x) = frac{1}{-x} ), which is the same as ( -frac{1}{x} ), but since ( x ) is negative, this is positive. So, that works out.Now, moving on. I need to find ( f(x) ) for ( x in (-4, -2) ). Let me think about how to use the given symmetries to extend beyond (-2, 0).Since the function is symmetric about ( x = 2 ), perhaps I can use that to find ( f(x) ) for ( x ) beyond 4 or less than 0. Wait, but I need to find ( f(x) ) for ( x in (-4, -2) ). Let me see.First, let's consider the reflection about ( x = 2 ). For any ( x ), ( f(2 + x) = f(2 - x) ). So, if I take ( x = t ), then ( f(2 + t) = f(2 - t) ). So, if ( t ) is such that ( 2 + t ) is in some interval, we can relate it to ( 2 - t ).But I need to find ( f(x) ) for ( x in (-4, -2) ). Let me set ( x = -2 - t ), where ( t in (0, 2) ). Then, ( x = -2 - t ), so ( t = -2 - x ). Hmm, not sure if that helps.Alternatively, maybe I can use the evenness of the function. Since ( f(x) ) is even, ( f(x) = f(-x) ). So, if I can find ( f(-x) ) for ( x in (-4, -2) ), which would be ( f(-x) ) for ( -x in (2, 4) ). But I already found that ( f(x) = frac{1}{4 - x} ) for ( x in (2, 4) ). Therefore, ( f(-x) = frac{1}{4 - (-x)} = frac{1}{4 + x} ). Therefore, ( f(x) = f(-x) = frac{1}{4 + x} ) for ( x in (-4, -2) ).Wait, let me verify that. If ( x in (-4, -2) ), then ( -x in (2, 4) ). Since ( f(-x) = frac{1}{4 - (-x)} = frac{1}{4 + x} ). But since ( f(x) = f(-x) ), then ( f(x) = frac{1}{4 + x} ) for ( x in (-4, -2) ).Let me test this with a specific value. Let me take ( x = -3 ), which is in (-4, -2). Then, ( f(-3) = f(3) ) because ( f(x) ) is even. Now, ( f(3) ) is in (2, 4), so ( f(3) = frac{1}{4 - 3} = 1 ). Therefore, ( f(-3) = 1 ). According to the expression I found, ( f(-3) = frac{1}{4 + (-3)} = frac{1}{1} = 1 ). So, that works.Another test: ( x = -2.5 ). Then, ( f(-2.5) = f(2.5) ). Since ( 2.5 in (2, 4) ), ( f(2.5) = frac{1}{4 - 2.5} = frac{1}{1.5} = frac{2}{3} ). According to my expression, ( f(-2.5) = frac{1}{4 + (-2.5)} = frac{1}{1.5} = frac{2}{3} ). That also works.So, it seems that for ( x in (-4, -2) ), ( f(x) = frac{1}{4 + x} ).Wait, but let me think again about the reflection about ( x = 2 ). If I take ( x = -2 ), which is the boundary, then ( f(2 + (-4)) = f(2 - (-4)) ), which would be ( f(-2) = f(6) ). But I don't know what ( f(6) ) is yet. Hmm, but maybe I don't need to go that far.Alternatively, maybe I can use the reflection about ( x = 2 ) to find ( f(x) ) for ( x in (-4, -2) ). Let me try that.Let me consider ( x in (-4, -2) ). Let me set ( x = 2 + t ), but wait, that would make ( t = x - 2 ), which for ( x in (-4, -2) ), ( t in (-6, -4) ). Hmm, not sure if that helps.Alternatively, maybe I can use the reflection about ( x = 2 ) twice. Since ( f(2 + x) = f(2 - x) ), and ( f(x) ) is even, perhaps combining these symmetries can help.Let me take ( x in (-4, -2) ). Let me set ( x = -4 + t ), where ( t in (0, 2) ). So, ( x = -4 + t ), which is in (-4, -2) when ( t in (0, 2) ). Now, I want to express ( f(x) ) in terms of ( f ) evaluated at some other point.Since ( f(x) ) is even, ( f(x) = f(-x) ). So, ( f(-4 + t) = f(4 - t) ). Now, ( 4 - t ) is in (2, 4) because ( t in (0, 2) ). And for ( x in (2, 4) ), we have ( f(x) = frac{1}{4 - x} ). So, ( f(4 - t) = frac{1}{4 - (4 - t)} = frac{1}{t} ). Therefore, ( f(-4 + t) = frac{1}{t} ). But ( t = x + 4 ), so ( f(x) = frac{1}{x + 4} ).Therefore, for ( x in (-4, -2) ), ( f(x) = frac{1}{x + 4} ).Wait, that's the same result as before. So, whether I use the evenness or the reflection about ( x = 2 ), I end up with ( f(x) = frac{1}{x + 4} ) for ( x in (-4, -2) ).Let me test this with another value. Let me take ( x = -3.5 ). Then, ( f(-3.5) = f(3.5) ) because ( f(x) ) is even. ( f(3.5) ) is in (2, 4), so ( f(3.5) = frac{1}{4 - 3.5} = frac{1}{0.5} = 2 ). According to my expression, ( f(-3.5) = frac{1}{-3.5 + 4} = frac{1}{0.5} = 2 ). Perfect, that works.Another test: ( x = -2.5 ). ( f(-2.5) = f(2.5) = frac{1}{4 - 2.5} = frac{2}{3} ). Using the expression ( f(x) = frac{1}{x + 4} ), ( f(-2.5) = frac{1}{-2.5 + 4} = frac{1}{1.5} = frac{2}{3} ). Correct again.So, it seems consistent. Therefore, the expression for ( f(x) ) when ( x in (-4, -2) ) is ( frac{1}{x + 4} ).But just to make sure I haven't missed anything, let me recap:1. ( f(x) ) is even: ( f(x) = f(-x) ).2. ( f(2 + x) = f(2 - x) ): symmetric about ( x = 2 ).3. ( f(x) = frac{1}{x} ) for ( x in (0, 2) ).From 2 and 3, we found that ( f(x) = frac{1}{4 - x} ) for ( x in (2, 4) ).Using the evenness, ( f(x) = f(-x) ), so for ( x in (-4, -2) ), ( f(x) = f(-x) ), and ( -x in (2, 4) ), so ( f(-x) = frac{1}{4 - (-x)} = frac{1}{4 + x} ). Therefore, ( f(x) = frac{1}{4 + x} ) for ( x in (-4, -2) ).Yes, that makes sense. So, the final expression is ( frac{1}{x + 4} ) for ( x in (-4, -2) ).

answer：Okay, so I have this problem about the probability of passing an exam in Anchuria by guessing. Let me try to figure this out step by step. First, in 2011, the exam required answering 3 out of 20 questions correctly. Each question has a 0.25 chance of being answered correctly since there are probably four options, right? So, guessing each question gives a 25% chance of getting it right. In 2012, they changed the requirements. Now, you need to answer 6 out of 40 questions correctly. So, the number of correct answers needed doubled, and the total number of questions also doubled. Hmm, I wonder if this makes it harder or easier? Intuitively, I might think it's harder because you need more correct answers, but maybe the increase in total questions affects the probability in a different way.I think I need to calculate the probability of getting exactly 3 correct answers in 2011 and exactly 6 correct answers in 2012, and then compare them. But wait, actually, to pass, you just need at least 3 correct in 2011 and at least 6 correct in 2012. So, I should calculate the cumulative probability of getting 3 or more correct in 2011 and 6 or more correct in 2012.This sounds like a binomial probability problem. The binomial formula is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where:- P(k) is the probability of getting exactly k successes,- C(n, k) is the combination of n things taken k at a time,- p is the probability of success on a single trial.But since we need the probability of getting at least a certain number, we have to sum the probabilities from k to n. That could be a bit tedious, especially for 20 and 40 questions, but maybe there's a way to approximate it or use a calculator.Alternatively, since the number of trials is large (20 and 40), maybe we can use the normal approximation to the binomial distribution. That might make the calculations easier.Let me recall the conditions for using the normal approximation: np and n(1-p) should both be greater than 5. Let's check:For 2011:n = 20, p = 0.25np = 20 * 0.25 = 5n(1-p) = 20 * 0.75 = 15So, np is exactly 5, which is the lower bound. Some sources say np and n(1-p) should be at least 10, so maybe the approximation isn't great here, but it might still be usable.For 2012:n = 40, p = 0.25np = 40 * 0.25 = 10n(1-p) = 40 * 0.75 = 30Here, both are above 10, so the normal approximation should work better.But I'm not sure if I should use the approximation or calculate the exact probabilities. Since the numbers aren't too large, maybe I can compute the exact probabilities using the binomial formula.Starting with 2011:We need P(X >= 3) where X ~ Binomial(n=20, p=0.25). So, this is 1 - P(X <= 2). Let me compute P(X=0), P(X=1), P(X=2), and subtract their sum from 1.Calculating each term:P(X=0) = C(20, 0) * (0.25)^0 * (0.75)^20 = 1 * 1 * (0.75)^20I can compute (0.75)^20. Let me see, 0.75^2 = 0.5625, 0.75^4 = ~0.3164, 0.75^8 = ~0.1001, 0.75^16 = ~0.01003. Then, 0.75^20 = 0.75^16 * 0.75^4 ≈ 0.01003 * 0.3164 ≈ 0.00317.So, P(X=0) ≈ 0.00317.P(X=1) = C(20, 1) * (0.25)^1 * (0.75)^19C(20,1) = 20(0.25)^1 = 0.25(0.75)^19 = (0.75)^20 / 0.75 ≈ 0.00317 / 0.75 ≈ 0.00423So, P(X=1) = 20 * 0.25 * 0.00423 ≈ 20 * 0.0010575 ≈ 0.02115.P(X=2) = C(20, 2) * (0.25)^2 * (0.75)^18C(20,2) = 190(0.25)^2 = 0.0625(0.75)^18 = (0.75)^20 / (0.75)^2 ≈ 0.00317 / 0.5625 ≈ 0.00563So, P(X=2) = 190 * 0.0625 * 0.00563 ≈ 190 * 0.0003519 ≈ 0.06686.Adding these up: P(X=0) + P(X=1) + P(X=2) ≈ 0.00317 + 0.02115 + 0.06686 ≈ 0.09118.Therefore, P(X >= 3) = 1 - 0.09118 ≈ 0.90882, or about 90.88%.Wait, that seems high. Let me double-check my calculations because 90% seems quite high for guessing.Wait, actually, 0.25 is a low probability, so getting 3 out of 20 might not be that high. Maybe I made a mistake in computing (0.75)^20.Let me recalculate (0.75)^20 more accurately.0.75^2 = 0.56250.75^4 = (0.5625)^2 ≈ 0.316406250.75^8 = (0.31640625)^2 ≈ 0.1001129150.75^16 = (0.100112915)^2 ≈ 0.0100225880.75^20 = 0.75^16 * 0.75^4 ≈ 0.010022588 * 0.31640625 ≈ 0.003171212So, P(X=0) ≈ 0.003171212P(X=1): 20 * 0.25 * (0.75)^19(0.75)^19 = (0.75)^20 / 0.75 ≈ 0.003171212 / 0.75 ≈ 0.004228283So, P(X=1) ≈ 20 * 0.25 * 0.004228283 ≈ 20 * 0.00105707 ≈ 0.0211414P(X=2): C(20,2) = 190(0.25)^2 = 0.0625(0.75)^18 = (0.75)^20 / (0.75)^2 ≈ 0.003171212 / 0.5625 ≈ 0.00563616So, P(X=2) ≈ 190 * 0.0625 * 0.00563616 ≈ 190 * 0.00035226 ≈ 0.0669294Adding up: 0.003171212 + 0.0211414 + 0.0669294 ≈ 0.091242So, P(X >= 3) ≈ 1 - 0.091242 ≈ 0.908758, or about 90.88%. Hmm, that seems correct. So, in 2011, the probability of passing by guessing is approximately 90.88%.Now, moving on to 2012. The exam now requires 6 correct answers out of 40. So, we need P(Y >= 6) where Y ~ Binomial(n=40, p=0.25). Again, this is 1 - P(Y <= 5). Let's compute P(Y=0) through P(Y=5) and subtract their sum from 1.Calculating each term:P(Y=0) = C(40, 0) * (0.25)^0 * (0.75)^40 = 1 * 1 * (0.75)^40Calculating (0.75)^40. Let's see:We know that (0.75)^20 ≈ 0.003171212, so (0.75)^40 = (0.75)^20 squared ≈ (0.003171212)^2 ≈ 0.00001005.So, P(Y=0) ≈ 0.00001005.P(Y=1) = C(40, 1) * (0.25)^1 * (0.75)^39C(40,1) = 40(0.25)^1 = 0.25(0.75)^39 = (0.75)^40 / 0.75 ≈ 0.00001005 / 0.75 ≈ 0.0000134So, P(Y=1) ≈ 40 * 0.25 * 0.0000134 ≈ 40 * 0.00000335 ≈ 0.000134P(Y=2) = C(40, 2) * (0.25)^2 * (0.75)^38C(40,2) = 780(0.25)^2 = 0.0625(0.75)^38 = (0.75)^40 / (0.75)^2 ≈ 0.00001005 / 0.5625 ≈ 0.00001786So, P(Y=2) ≈ 780 * 0.0625 * 0.00001786 ≈ 780 * 0.000001116 ≈ 0.000870P(Y=3) = C(40, 3) * (0.25)^3 * (0.75)^37C(40,3) = 9880(0.25)^3 = 0.015625(0.75)^37 = (0.75)^40 / (0.75)^3 ≈ 0.00001005 / 0.421875 ≈ 0.00002382So, P(Y=3) ≈ 9880 * 0.015625 * 0.00002382 ≈ 9880 * 0.000000372 ≈ 0.00368Wait, let me compute that again:9880 * 0.015625 = 9880 * (1/64) ≈ 9880 / 64 ≈ 154.375Then, 154.375 * 0.00002382 ≈ 0.00368Yes, that seems right.P(Y=4) = C(40,4) * (0.25)^4 * (0.75)^36C(40,4) = 91390(0.25)^4 = 0.00390625(0.75)^36 = (0.75)^40 / (0.75)^4 ≈ 0.00001005 / 0.31640625 ≈ 0.00003176So, P(Y=4) ≈ 91390 * 0.00390625 * 0.00003176First, 91390 * 0.00390625 ≈ 91390 * (1/256) ≈ 91390 / 256 ≈ 357.0Then, 357.0 * 0.00003176 ≈ 0.01135So, P(Y=4) ≈ 0.01135P(Y=5) = C(40,5) * (0.25)^5 * (0.75)^35C(40,5) = 658,008(0.25)^5 = 0.0009765625(0.75)^35 = (0.75)^40 / (0.75)^5 ≈ 0.00001005 / 0.2373046875 ≈ 0.00004235So, P(Y=5) ≈ 658,008 * 0.0009765625 * 0.00004235First, 658,008 * 0.0009765625 ≈ 658,008 * (1/1024) ≈ 658,008 / 1024 ≈ 642.3Then, 642.3 * 0.00004235 ≈ 0.02718So, P(Y=5) ≈ 0.02718Now, adding up all these probabilities:P(Y=0) ≈ 0.00001005P(Y=1) ≈ 0.000134P(Y=2) ≈ 0.000870P(Y=3) ≈ 0.00368P(Y=4) ≈ 0.01135P(Y=5) ≈ 0.02718Total ≈ 0.00001005 + 0.000134 + 0.000870 + 0.00368 + 0.01135 + 0.02718 ≈Let's add them step by step:Start with 0.00001005 + 0.000134 ≈ 0.000144050.00014405 + 0.000870 ≈ 0.001014050.00101405 + 0.00368 ≈ 0.004694050.00469405 + 0.01135 ≈ 0.016044050.01604405 + 0.02718 ≈ 0.04322405So, the total P(Y <= 5) ≈ 0.04322405Therefore, P(Y >= 6) = 1 - 0.04322405 ≈ 0.95677595, or about 95.68%.Wait, that can't be right. If in 2011, the probability was ~90.88%, and in 2012 it's ~95.68%, that would mean it's easier to pass in 2012 by guessing. But that seems counterintuitive because you need to answer twice as many questions correctly. Maybe I made a mistake in my calculations.Let me check the calculations again, especially for P(Y=5). Wait, when I calculated P(Y=5), I had:C(40,5) = 658,008(0.25)^5 = 0.0009765625(0.75)^35 ≈ 0.00004235So, 658,008 * 0.0009765625 ≈ 658,008 * (1/1024) ≈ 642.3Then, 642.3 * 0.00004235 ≈ 0.02718Hmm, that seems correct. But let me verify the combination numbers because 40 choose 5 is actually 658,008, which is correct.Wait, but let me think about the overall probabilities. If the number of questions doubled, but the required correct answers also doubled, does that make it easier or harder? Intuitively, with more questions, the law of large numbers would make the distribution more concentrated around the mean. The mean number of correct answers is n*p, so in 2011, it was 5, and in 2012, it's 10. So, in 2011, you needed 3, which is below the mean, so the probability is high. In 2012, you need 6, which is below the mean of 10, so the probability should still be high, but maybe higher or lower?Wait, in 2011, the required correct answers were 3, which is 60% below the mean (since mean is 5). In 2012, required is 6, which is 40% below the mean (since mean is 10). So, in 2012, you're closer to the mean, so the probability should be higher. So, 95.68% in 2012 vs 90.88% in 2011. So, actually, it's easier to pass in 2012 by guessing.But that seems counterintuitive because you have more questions. But mathematically, it's because the required number is proportionally closer to the mean in 2012.Wait, but let me check my calculations again because 95.68% seems very high for 6 out of 40. Maybe I made a mistake in calculating P(Y <=5).Alternatively, maybe I should use the normal approximation to double-check.For 2012:n = 40, p = 0.25, so mean μ = np = 10, variance σ² = np(1-p) = 40 * 0.25 * 0.75 = 7.5, so σ ≈ 2.7386.We want P(Y >= 6). Using continuity correction, we can approximate P(Y >= 6) ≈ P(Z >= (5.5 - 10)/2.7386) = P(Z >= -1.64). Looking up Z = -1.64 in the standard normal table, the area to the right is 1 - Φ(-1.64) = 1 - (1 - Φ(1.64)) = Φ(1.64). Φ(1.64) is approximately 0.9495. So, P(Y >=6) ≈ 0.9495, which is about 94.95%, which is close to my exact calculation of ~95.68%. So, that seems consistent.For 2011:n = 20, p = 0.25, μ = 5, σ² = 20 * 0.25 * 0.75 = 3.75, σ ≈ 1.9365.We want P(X >=3). Using continuity correction, P(X >=3) ≈ P(Z >= (2.5 - 5)/1.9365) = P(Z >= -1.29). Φ(-1.29) is the area to the left of -1.29, which is about 0.0985. So, P(X >=3) ≈ 1 - 0.0985 = 0.9015, or 90.15%, which is close to my exact calculation of ~90.88%. So, again, consistent.So, according to both exact calculations and normal approximation, the probability of passing by guessing is higher in 2012 (~95.68%) than in 2011 (~90.88%).Wait, but that seems surprising because in 2012, you have more questions and need more correct answers. But the required number is proportionally closer to the mean, making it easier. So, the answer is that in 2012, the probability is higher.But let me just think again: in 2011, you needed 3 out of 20, which is 15%, while the expected correct answers are 5, so 3 is 40% of the expected. In 2012, you need 6 out of 40, which is 15%, same proportion as 3 out of 20. Wait, no, 3/20 is 15%, 6/40 is also 15%. So, the required proportion is the same. So, why is the probability higher in 2012?Because with more trials, the distribution is more concentrated around the mean. So, even though the required number is the same proportion, the variance is higher in 2012, but the required number is further from the mean in 2011. Wait, no, in 2011, the required is 3, mean is 5, so 2 below. In 2012, required is 6, mean is 10, so 4 below. So, in terms of standard deviations:In 2011: (3 - 5)/1.9365 ≈ -1.03 standard deviations.In 2012: (6 - 10)/2.7386 ≈ -1.46 standard deviations.So, in 2011, you're only 1.03 SD below the mean, while in 2012, you're 1.46 SD below. Wait, that would mean that 2011 is closer to the mean, so the probability should be higher. But according to my calculations, 2012 has a higher probability. That contradicts.Wait, no, actually, in 2011, the required is 3, which is 2 below the mean of 5, so z-score is (3 - 5)/1.9365 ≈ -1.03.In 2012, required is 6, which is 4 below the mean of 10, so z-score is (6 - 10)/2.7386 ≈ -1.46.So, in 2011, you're closer to the mean, so the probability of being above 3 should be higher than in 2012, where you're further below the mean. But according to my calculations, 2012 has a higher probability. That seems contradictory.Wait, maybe I'm confusing the direction. Let me think again.In 2011, the probability of getting >=3 is the area to the right of 3, which is 1 - P(X <=2). Since the mean is 5, 3 is to the left of the mean, so the area to the right is larger.In 2012, the probability of getting >=6 is the area to the right of 6, which is 1 - P(Y <=5). The mean is 10, so 6 is to the left of the mean, so the area to the right is larger than the area to the left of 6.But in 2011, the z-score is -1.03, so the area to the right is 1 - Φ(-1.03) = Φ(1.03) ≈ 0.8485.In 2012, z-score is -1.46, so the area to the right is 1 - Φ(-1.46) = Φ(1.46) ≈ 0.9279.Wait, so according to this, in 2011, the probability is ~84.85%, and in 2012, it's ~92.79%, which is higher. But my exact calculations gave 90.88% and 95.68%. So, the normal approximation suggests that 2012 has a higher probability, which aligns with my exact calculations.Wait, but earlier I thought that being closer to the mean would make the probability higher, but in reality, the probability of being above a certain point is higher when that point is closer to the mean. So, in 2011, 3 is closer to the mean of 5 than 6 is to the mean of 10. Wait, no, 3 is 2 below 5, and 6 is 4 below 10. So, in terms of distance from the mean, 3 is closer. But in terms of z-scores, 3 is -1.03, and 6 is -1.46. So, 3 is closer in z-score terms as well.Wait, but the area to the right of 3 is larger than the area to the right of 6 because 3 is closer to the mean. So, why does the normal approximation suggest that 2012 has a higher probability?Wait, no, actually, the area to the right of 3 is larger than the area to the right of 6 because 3 is closer to the mean. So, in 2011, the probability should be higher than in 2012. But according to my exact calculations and the normal approximation, 2012 has a higher probability. There's a contradiction here.Wait, let me clarify:In 2011:- Mean = 5- Required = 3- Distance from mean = 2- Z-score = (3 - 5)/1.9365 ≈ -1.03- Area to the right (probability of passing) ≈ Φ(1.03) ≈ 0.8485In 2012:- Mean = 10- Required = 6- Distance from mean = 4- Z-score = (6 - 10)/2.7386 ≈ -1.46- Area to the right (probability of passing) ≈ Φ(1.46) ≈ 0.9279So, according to this, 2012 has a higher probability, which contradicts the intuition that being closer to the mean should have a higher probability. Wait, no, actually, the area to the right of 3 is larger than the area to the right of 6 because 3 is closer to the mean. But according to the z-scores, 2012 has a higher probability. That seems contradictory.Wait, no, actually, the z-score in 2011 is -1.03, so the area to the right is Φ(1.03) ≈ 0.8485.In 2012, z-score is -1.46, so area to the right is Φ(1.46) ≈ 0.9279.So, 0.9279 > 0.8485, meaning 2012 has a higher probability.But wait, 3 is closer to the mean than 6 is to its mean, so why is the probability higher in 2012? Because the required number is proportionally closer to the mean in 2012? Wait, 3 is 60% of the mean (3/5=0.6), and 6 is 60% of the mean (6/10=0.6). So, the required proportion is the same. So, why is the probability higher in 2012?Because with more trials, the distribution is more concentrated, so the probability of being above a certain proportion is higher. Wait, no, actually, the variance increases with n, but the required number is also increasing. Hmm, I'm getting confused.Wait, let me think in terms of the exact probabilities. In 2011, the probability of passing is ~90.88%, and in 2012, it's ~95.68%. So, 2012 is higher. So, despite needing more correct answers, the probability is higher because the required proportion is the same, but the distribution is more concentrated, making it more likely to reach that proportion.Wait, but actually, the required number is the same proportion (15%), but the variance in 2012 is higher because n is larger. So, the standard deviation is higher, which would make the distribution more spread out, but the required number is also further in absolute terms. Hmm, I'm not sure.Wait, maybe it's better to think in terms of the exact calculations. Since in 2012, the probability is higher, despite needing more correct answers, because the required number is proportionally the same, but the distribution is such that the probability of exceeding that number is higher.Alternatively, maybe my exact calculations are wrong. Let me check the exact probability for 2012 again.Wait, when I calculated P(Y <=5), I got approximately 0.0432, so P(Y >=6) ≈ 0.9568. But according to the normal approximation, it's ~0.9279. So, there's a discrepancy. Which one is correct?Wait, the exact calculation is more accurate, but for n=40, the normal approximation should be pretty good. So, why is there a difference?Wait, maybe I made a mistake in the exact calculation. Let me recalculate P(Y=5) again.P(Y=5) = C(40,5) * (0.25)^5 * (0.75)^35C(40,5) = 658,008(0.25)^5 = 0.0009765625(0.75)^35 ≈ ?Wait, I approximated (0.75)^35 as 0.00004235, but let me compute it more accurately.We know that (0.75)^40 ≈ 0.00001005So, (0.75)^35 = (0.75)^40 / (0.75)^5 ≈ 0.00001005 / (0.2373046875) ≈ 0.00004235So, that seems correct.So, P(Y=5) ≈ 658,008 * 0.0009765625 * 0.00004235First, 658,008 * 0.0009765625 ≈ 658,008 * (1/1024) ≈ 642.3Then, 642.3 * 0.00004235 ≈ 0.02718So, that seems correct.Adding up P(Y=0) to P(Y=5):0.00001005 + 0.000134 + 0.000870 + 0.00368 + 0.01135 + 0.02718 ≈ 0.04322405So, P(Y >=6) ≈ 0.95677595But according to the normal approximation, it's ~0.9279. So, the exact calculation is higher. That suggests that the exact probability is indeed higher in 2012.Therefore, the conclusion is that in 2012, the probability of passing by guessing is higher than in 2011.Wait, but that contradicts the intuition that more questions would make it harder. But mathematically, it's because the required number is proportionally the same, but with more questions, the distribution is such that the probability of reaching that proportion is higher.Alternatively, maybe I'm misunderstanding the problem. Let me read it again."In 2011, to receive a certificate, one needed to answer correctly 3 questions out of 20. In 2012, ... one needs to correctly answer 6 questions out of 40."So, in both cases, the required proportion is 15%. So, the question is, is it easier to get 15% correct in 2011 or 2012 by guessing.But with more questions, the variance increases, but the required number is also higher. However, the exact calculations show that the probability is higher in 2012.Wait, maybe it's because in 2012, the required number is 6, which is closer to the mean of 10, but in 2011, 3 is closer to the mean of 5. Wait, no, 3 is 60% of 5, and 6 is 60% of 10. So, the required proportion is the same.But in 2012, the mean is higher, so the required number is further in absolute terms, but proportionally the same. So, the probability of getting at least 60% of the mean is higher in 2012 because the distribution is more concentrated around the mean.Wait, no, actually, the variance increases with n, so the distribution is more spread out. Hmm, I'm confused.Wait, let me think of it in terms of the binomial distribution. For a binomial distribution, as n increases, the distribution becomes more bell-shaped and concentrated around the mean. So, even though the required number is higher, the probability of reaching that number might be higher because the distribution is more concentrated.Wait, but in 2011, the required number is 3, which is 15% of 20, and in 2012, it's 6, which is 15% of 40. So, the required proportion is the same. So, why is the probability higher in 2012?Because with more trials, the probability of being above a certain proportion is higher? Or is it the opposite?Wait, actually, the probability of being above a certain proportion might be lower because the variance increases, but the required number is also higher. It's a bit tricky.But according to the exact calculations, 2012 has a higher probability. So, the answer is that in 2012, the probability is higher.Therefore, the probability of receiving an Anchurian certificate by guessing is higher in 2012 than in 2011.