answer：Okay, so I have this problem about contrapositive statements, and I need to figure out which option is correct. Let me start by recalling what a contrapositive is. From what I remember, the contrapositive of a statement "If P, then Q" is "If not Q, then not P." It's like flipping the statement and negating both parts. The original proposition here is: "If a and b are both odd, then a + b is even." So, in logical terms, P is "a and b are both odd," and Q is "a + b is even." Therefore, the contrapositive should be "If a + b is not even, then it is not the case that both a and b are odd." Hmm, let me break that down. The contrapositive swaps the hypothesis and conclusion and negates both. So, starting with P implies Q, the contrapositive is not Q implies not P. Looking at the options given:A: If a + b is not even, then neither a nor b are odd.B: If a + b is not even, then a and b are not both odd.C: If a + b is even, then a and b are both odd.D: If a + b is even, then a and b are not both odd.So, let's analyze each option.Option A says that if a + b is not even (which is not Q), then neither a nor b are odd. Wait, so that would mean both a and b are even? Because neither being odd implies both are even. But is that necessarily true? If a + b is not even, meaning it's odd, does that mean both a and b have to be even? No, actually, if a + b is odd, one of them has to be odd and the other even. So, saying neither a nor b are odd would mean both are even, which isn't necessarily the case. So, Option A might not be correct.Option B says that if a + b is not even, then a and b are not both odd. So, not both odd means at least one of them is even. Which aligns with what I was thinking earlier—if a + b is odd, then one is odd and one is even. So, they can't both be odd. Therefore, Option B seems correct.Option C says that if a + b is even, then a and b are both odd. Wait, is that true? If a + b is even, does that mean both a and b are odd? No, because two even numbers added together also give an even number. So, a + b being even could mean both are even or both are odd. Therefore, Option C is not necessarily true.Option D says that if a + b is even, then a and b are not both odd. Wait, that seems contradictory. If a + b is even, it could be because both are odd or both are even. So, saying they are not both odd would mean they could be both even or one odd and one even. But since a + b is even, it's possible for both to be odd. So, Option D is not correct either.So, going back, Option B is the contrapositive because it correctly negates both parts and swaps them. The original statement was about both being odd leading to a + b being even. The contrapositive flips it to if a + b isn't even, then they aren't both odd, which makes sense because if a + b is odd, at least one of them has to be even.I think I might have confused myself a bit with Option A because it's similar but not exactly the contrapositive. Option A is too strong because it says neither a nor b are odd, which isn't necessary—it's enough that they aren't both odd. So, Option B is the right one.Just to double-check, let's think about the logic again. Original: P → Q. Contrapositive: ¬Q → ¬P. So, P is "both a and b are odd," so ¬P is "it is not the case that both a and b are odd," which is the same as "at least one is even." Q is "a + b is even," so ¬Q is "a + b is not even," which is "a + b is odd." So, contrapositive is "If a + b is odd, then at least one of a or b is even." Which is exactly what Option B is saying: "If a + b is not even, then a and b are not both odd." Yes, that makes sense. So, I'm confident now that Option B is the correct contrapositive.

answer：Okay, so I have this piecewise function f(x) defined as follows:f(x) = e^x - kx, when x ≥ 0,andf(x) = kx² - x + 1, when x < 0.There are four conclusions given, and I need to figure out which ones are correct. Let me go through each conclusion one by one.**Conclusion ①: When k = 1, the function f(x) has no zero points.**Alright, let's set k = 1 and analyze both parts of the function.For x ≥ 0: f(x) = e^x - x.I know that e^x grows exponentially, and x grows linearly. At x = 0, f(0) = e^0 - 0 = 1, which is positive. As x increases, e^x will dominate, so f(x) will always be positive. Also, the derivative f’(x) = e^x - 1. For x > 0, e^x > 1, so the function is increasing. Since it starts at 1 and increases, it never crosses zero.For x < 0: f(x) = x² - x + 1.This is a quadratic equation. Let me check its discriminant to see if it has real roots.Discriminant D = b² - 4ac = (-1)² - 4*1*1 = 1 - 4 = -3.Since D is negative, there are no real roots. So, for x < 0, f(x) doesn't cross zero either.Therefore, when k = 1, f(x) has no zero points. So, conclusion ① is correct.**Conclusion ②: When k < 0, the function f(x) has exactly one zero point.**Let me consider k < 0.First, for x ≥ 0: f(x) = e^x - kx.Since k is negative, -kx becomes positive. So, f(x) = e^x + |k|x. Both e^x and |k|x are positive and increasing functions. At x = 0, f(0) = 1. As x increases, f(x) increases without bound. So, f(x) is always positive for x ≥ 0 when k < 0.Now, for x < 0: f(x) = kx² - x + 1.Since k is negative, this is a downward-opening parabola. Let's analyze its behavior.First, let's find its derivative to see if it has any critical points.f’(x) = 2kx - 1.Set f’(x) = 0: 2kx - 1 = 0 ⇒ x = 1/(2k).But since k is negative, x = 1/(2k) is negative. So, the vertex of the parabola is at x = 1/(2k), which is in the domain x < 0.Now, let's evaluate f(x) at x = 1/(2k):f(1/(2k)) = k*(1/(2k))² - (1/(2k)) + 1= k*(1/(4k²)) - 1/(2k) + 1= 1/(4k) - 1/(2k) + 1= (1 - 2)/(4k) + 1= (-1)/(4k) + 1.Since k is negative, -1/(4k) is positive. So, f(1/(2k)) = positive + 1, which is positive.Now, let's check the limits:As x approaches negative infinity, f(x) = kx² - x + 1. Since k is negative, the x² term dominates and goes to negative infinity. So, f(x) approaches negative infinity.At x = 0, f(0) = 1.So, the function f(x) for x < 0 starts at negative infinity, reaches a maximum at x = 1/(2k) (which is positive), and then goes back to f(0) = 1.Since it goes from negative infinity to positive, it must cross the x-axis exactly once in the interval x < 0.Therefore, when k < 0, f(x) has exactly one zero point. So, conclusion ② is correct.**Conclusion ③: There exists a real number k such that the function f(x) has two zero points.**Hmm, so I need to find a k where f(x) has two zeros. Let's think about how this can happen.Looking at the function, for x ≥ 0, f(x) = e^x - kx. For x < 0, f(x) = kx² - x + 1.We already saw that for k = 1, f(x) has no zeros. For k < 0, f(x) has exactly one zero.What about for k > 0?Let me consider k > 0.For x ≥ 0: f(x) = e^x - kx.At x = 0, f(0) = 1. The derivative f’(x) = e^x - k.If k > e^x at some point, the function will have a minimum. Let's find where f’(x) = 0:e^x - k = 0 ⇒ x = ln(k).So, if k > 1, then ln(k) is positive, and f(x) has a critical point at x = ln(k).Let's evaluate f(x) at x = ln(k):f(ln(k)) = e^{ln(k)} - k*ln(k) = k - k ln(k).For f(ln(k)) to be negative, we need k - k ln(k) < 0 ⇒ 1 - ln(k) < 0 ⇒ ln(k) > 1 ⇒ k > e.So, if k > e, then f(x) will have a minimum below zero, meaning f(x) crosses the x-axis twice for x ≥ 0.But wait, for x < 0, f(x) = kx² - x + 1.Since k > 0, this is an upward-opening parabola. Let's check its discriminant:D = (-1)^2 - 4*k*1 = 1 - 4k.If D < 0, then f(x) has no real roots for x < 0.So, if 1 - 4k < 0 ⇒ k > 1/4.So, if k > e (which is greater than 1/4), then for x < 0, f(x) has no real roots, and for x ≥ 0, f(x) has two real roots. Therefore, total zeros would be two.Wait, but conclusion ③ says "there exists a real number k such that f(x) has two zero points." So, if I choose k > e, say k = 2e, then f(x) would have two zeros for x ≥ 0 and none for x < 0, totaling two zeros.Alternatively, maybe for some k, f(x) has one zero for x < 0 and one for x ≥ 0, totaling two.Wait, let me check for k between 0 and e.Suppose k = 2, which is less than e (~2.718).For x ≥ 0: f(x) = e^x - 2x.f(0) = 1.f’(x) = e^x - 2.Set to zero: e^x = 2 ⇒ x = ln(2) ≈ 0.693.f(ln(2)) = 2 - 2*ln(2) ≈ 2 - 1.386 ≈ 0.614 > 0.So, the minimum is above zero, meaning f(x) is always positive for x ≥ 0.For x < 0: f(x) = 2x² - x + 1.Discriminant D = (-1)^2 - 4*2*1 = 1 - 8 = -7 < 0.So, no real roots for x < 0.Thus, for k = 2, f(x) has no zeros.Wait, so when k is between 0 and e, f(x) might not have any zeros?Wait, let me try k = 3, which is greater than e.For x ≥ 0: f(x) = e^x - 3x.f(0) = 1.f’(x) = e^x - 3.Set to zero: e^x = 3 ⇒ x = ln(3) ≈ 1.0986.f(ln(3)) = 3 - 3*ln(3) ≈ 3 - 3*1.0986 ≈ 3 - 3.2958 ≈ -0.2958 < 0.So, f(x) has a minimum below zero, meaning it crosses the x-axis twice for x ≥ 0.For x < 0: f(x) = 3x² - x + 1.Discriminant D = (-1)^2 - 4*3*1 = 1 - 12 = -11 < 0.So, no real roots for x < 0.Thus, total zeros are two.Therefore, for k > e, f(x) has two zeros.So, conclusion ③ is correct.**Conclusion ④: There exists a real number k such that the function f(x) has three zero points.**Hmm, three zeros. Let's see.From the previous analysis, for k > e, f(x) has two zeros for x ≥ 0 and none for x < 0.For k < 0, f(x) has one zero for x < 0 and none for x ≥ 0.For k = e, f(x) has one zero for x ≥ 0 (since the minimum is zero) and none for x < 0.For 0 < k < e, f(x) has no zeros for x ≥ 0 and none for x < 0.Wait, is there a k where f(x) has two zeros for x < 0 and one for x ≥ 0, totaling three?Let me check.For x < 0: f(x) = kx² - x + 1.This is a quadratic. For it to have two real roots, the discriminant must be positive.D = (-1)^2 - 4*k*1 = 1 - 4k > 0 ⇒ k < 1/4.So, if k < 1/4, f(x) for x < 0 has two real roots.But wait, for x < 0, the quadratic is kx² - x + 1.If k < 0, it's a downward-opening parabola, which we already saw crosses the x-axis once.If 0 < k < 1/4, it's an upward-opening parabola with two real roots.But wait, for x < 0, the function is defined for x < 0, so even if the quadratic has two roots, both roots must be less than zero.Let me check.Suppose k = 1/5, which is less than 1/4.f(x) = (1/5)x² - x + 1.Discriminant D = 1 - 4*(1/5)*1 = 1 - 4/5 = 1/5 > 0.So, two real roots.Let me find the roots:x = [1 ± sqrt(1 - 4*(1/5))]/(2*(1/5)) = [1 ± sqrt(1 - 4/5)]/(2/5) = [1 ± sqrt(1/5)]/(2/5).sqrt(1/5) ≈ 0.447.So,x1 = [1 + 0.447]/(2/5) ≈ 1.447 / 0.4 ≈ 3.6175,x2 = [1 - 0.447]/(2/5) ≈ 0.553 / 0.4 ≈ 1.3825.But wait, these roots are positive, but we are considering x < 0.So, actually, for k = 1/5, the quadratic f(x) = (1/5)x² - x + 1 has roots at x ≈ 3.6175 and x ≈ 1.3825, both positive. Therefore, for x < 0, f(x) doesn't cross zero.Wait, that's confusing. Let me double-check.Wait, the quadratic is f(x) = kx² - x + 1.For x < 0, we need to see if the quadratic has roots in x < 0.Given that the quadratic is kx² - x + 1, with k > 0.The roots are x = [1 ± sqrt(1 - 4k)]/(2k).For the roots to be negative, we need both roots to be negative.But since the quadratic is kx² - x + 1, the sum of roots is 1/k, which is positive, and the product is 1/k, which is positive.So, both roots are positive or both are negative.But since the sum is positive and the product is positive, both roots are positive.Therefore, for 0 < k < 1/4, the quadratic has two positive roots, so for x < 0, f(x) doesn't cross zero.Therefore, for x < 0, f(x) doesn't have any zeros when 0 < k < 1/4.Wait, but if k is negative, we already saw that f(x) has one zero for x < 0.So, is there a k where f(x) has two zeros for x < 0 and one for x ≥ 0?But for x < 0, when k is negative, f(x) is a downward-opening parabola with one zero.When k is positive, for x < 0, f(x) has no zeros.Therefore, it's impossible for f(x) to have two zeros for x < 0.Thus, the maximum number of zeros f(x) can have is two, when k > e, with two zeros for x ≥ 0 and none for x < 0.Therefore, conclusion ④ is incorrect.So, the correct conclusions are ①, ②, and ③.

answer：First, I need to determine the number of lollipops Joann ate on the first day. Let's denote this number as ( a ).Since Joann increases the number of lollipops she eats by 3 each day, the number of lollipops she eats each day forms an arithmetic sequence:- Day 1: ( a )- Day 2: ( a + 3 )- Day 3: ( a + 6 )- Day 4: ( a + 9 )- Day 5: ( a + 12 )- Day 6: ( a + 15 )- Day 7: ( a + 18 )The total number of lollipops eaten over the seven days is the sum of this arithmetic sequence:[text{Total} = a + (a + 3) + (a + 6) + (a + 9) + (a + 12) + (a + 15) + (a + 18) = 7a + 63]Given that the total is 150 lollipops:[7a + 63 = 150]Solving for ( a ):[7a = 150 - 63 = 87 a = frac{87}{7}]Now, to find the number of lollipops eaten on the fourth day:[text{Day 4} = a + 9 = frac{87}{7} + 9 = frac{87}{7} + frac{63}{7} = frac{150}{7}]

answer：Alright, so I've got this inequality: ( ax^2 + 2ax + 1 > 0 ), and I need to find the range of real numbers ( a ) for which this inequality holds true for all real numbers ( x ). Hmm, okay. Let me think about how to approach this.First off, I remember that quadratic inequalities can sometimes be tricky, especially when the coefficient of ( x^2 ) isn't fixed—it's ( a ) in this case. So, the behavior of the quadratic depends on the value of ( a ). I think I need to consider different cases based on the value of ( a ).Let me start by recalling that a quadratic equation ( ax^2 + bx + c ) will always be positive if the parabola opens upwards (i.e., ( a > 0 )) and doesn't intersect the x-axis, meaning it has no real roots. For that to happen, the discriminant ( D = b^2 - 4ac ) must be less than zero. On the other hand, if the parabola opens downwards (( a < 0 )), it can't always be positive because it will eventually go to negative infinity as ( x ) increases or decreases without bound. So, maybe ( a ) has to be positive?But wait, what if ( a = 0 )? Then the inequality simplifies to ( 1 > 0 ), which is always true. So, ( a = 0 ) is definitely part of the solution. Okay, so ( a ) can be zero or positive, but not negative. Let me write that down:- If ( a = 0 ), the inequality is always true.- If ( a > 0 ), we need to check if the quadratic is always positive.- If ( a < 0 ), the quadratic will eventually be negative, so it's not always positive.So, focusing on ( a > 0 ), I need to ensure that the quadratic ( ax^2 + 2ax + 1 ) is always positive. As I thought earlier, this requires the discriminant to be negative. Let's calculate the discriminant:( D = (2a)^2 - 4(a)(1) = 4a^2 - 4a )For the quadratic to have no real roots, we need ( D < 0 ):( 4a^2 - 4a < 0 )Let me simplify this inequality:( 4a^2 - 4a < 0 )Factor out 4a:( 4a(a - 1) < 0 )Divide both sides by 4 (since 4 is positive, the inequality sign doesn't change):( a(a - 1) < 0 )Now, this is a quadratic inequality in terms of ( a ). To solve ( a(a - 1) < 0 ), I can consider the critical points where the expression equals zero: ( a = 0 ) and ( a = 1 ). These points divide the number line into intervals. I'll test each interval:1. When ( a < 0 ): Let's pick ( a = -1 ). Then, ( (-1)(-1 - 1) = (-1)(-2) = 2 ), which is positive. So, this interval doesn't satisfy the inequality.2. When ( 0 < a < 1 ): Let's pick ( a = 0.5 ). Then, ( 0.5(0.5 - 1) = 0.5(-0.5) = -0.25 ), which is negative. So, this interval satisfies the inequality.3. When ( a > 1 ): Let's pick ( a = 2 ). Then, ( 2(2 - 1) = 2(1) = 2 ), which is positive. So, this interval doesn't satisfy the inequality.Therefore, the solution to ( a(a - 1) < 0 ) is ( 0 < a < 1 ).But wait, earlier I considered ( a = 0 ) separately and found that it also satisfies the original inequality. So, combining both results, the range of ( a ) is ( 0 leq a < 1 ).Let me double-check this. If ( a = 0 ), the inequality becomes ( 1 > 0 ), which is always true. If ( a = 0.5 ), then the quadratic is ( 0.5x^2 + x + 1 ). The discriminant is ( 1^2 - 4*0.5*1 = 1 - 2 = -1 ), which is negative, so the quadratic is always positive. If ( a = 1 ), the quadratic becomes ( x^2 + 2x + 1 = (x + 1)^2 ), which is always non-negative but equals zero when ( x = -1 ). Since the inequality is strict (( > 0 )), ( a = 1 ) is not included. If ( a > 1 ), say ( a = 2 ), the quadratic is ( 2x^2 + 4x + 1 ). The discriminant is ( 16 - 8 = 8 ), which is positive, so the quadratic has real roots and thus is not always positive. If ( a < 0 ), as I thought earlier, the parabola opens downward, so it will eventually be negative.Therefore, the range of ( a ) is all real numbers from 0 up to but not including 1. So, in interval notation, that's ( [0, 1) ).**Final Answer**The range of real number ( a ) is boxed{[0,1)}.

answer：Okay, so I have this problem with two parts, both involving the function ( f(x) = |x + m| + |2x - 1| ) where ( m ) is a real number. Let me try to tackle each part step by step.Starting with part (I): When ( m = -1 ), I need to find the solution set of the inequality ( f(x) leq 2 ). So, substituting ( m = -1 ) into the function, it becomes ( f(x) = |x - 1| + |2x - 1| ). The inequality is ( |x - 1| + |2x - 1| leq 2 ).Hmm, dealing with absolute value inequalities often requires breaking them down into different cases based on the critical points where the expressions inside the absolute values change their sign. For ( |x - 1| ), the critical point is at ( x = 1 ), and for ( |2x - 1| ), the critical point is at ( x = frac{1}{2} ). So, these critical points divide the real line into three intervals:1. ( x < frac{1}{2} )2. ( frac{1}{2} leq x < 1 )3. ( x geq 1 )I'll analyze each interval separately.**Case 1: ( x < frac{1}{2} )**In this interval, ( x - 1 ) is negative, so ( |x - 1| = -(x - 1) = 1 - x ). Similarly, ( 2x - 1 ) is also negative because ( 2x < 1 ), so ( |2x - 1| = -(2x - 1) = 1 - 2x ). Therefore, the inequality becomes:( (1 - x) + (1 - 2x) leq 2 )Simplifying:( 1 - x + 1 - 2x leq 2 )( 2 - 3x leq 2 )Subtract 2 from both sides:( -3x leq 0 )Divide by -3 (remembering to reverse the inequality sign when dividing by a negative):( x geq 0 )But in this case, we're considering ( x < frac{1}{2} ). So, combining these, the solution in this interval is ( 0 leq x < frac{1}{2} ).**Case 2: ( frac{1}{2} leq x < 1 )**Here, ( x - 1 ) is still negative, so ( |x - 1| = 1 - x ). However, ( 2x - 1 ) is non-negative because ( x geq frac{1}{2} ), so ( |2x - 1| = 2x - 1 ). The inequality becomes:( (1 - x) + (2x - 1) leq 2 )Simplifying:( 1 - x + 2x - 1 leq 2 )( x leq 2 )But in this interval, ( x ) is between ( frac{1}{2} ) and 1. So, the solution here is ( frac{1}{2} leq x < 1 ).**Case 3: ( x geq 1 )**In this interval, both ( x - 1 ) and ( 2x - 1 ) are non-negative. So, ( |x - 1| = x - 1 ) and ( |2x - 1| = 2x - 1 ). The inequality becomes:( (x - 1) + (2x - 1) leq 2 )Simplifying:( x - 1 + 2x - 1 leq 2 )( 3x - 2 leq 2 )Add 2 to both sides:( 3x leq 4 )Divide by 3:( x leq frac{4}{3} )But in this case, ( x geq 1 ). So, the solution here is ( 1 leq x leq frac{4}{3} ).Now, combining all three cases:- From Case 1: ( 0 leq x < frac{1}{2} )- From Case 2: ( frac{1}{2} leq x < 1 )- From Case 3: ( 1 leq x leq frac{4}{3} )Putting it all together, the solution set is ( 0 leq x leq frac{4}{3} ).So, for part (I), the solution set is ( [0, frac{4}{3}] ).Moving on to part (II): Let ( A ) be the solution set of the inequality ( f(x) leq |2x + 1| ) regarding ( x ), and it's given that ( [1, 2] subseteq A ). I need to find the range of the real number ( m ).So, ( A ) is the set of all ( x ) such that ( |x + m| + |2x - 1| leq |2x + 1| ). Since ( [1, 2] subseteq A ), this means that for all ( x ) in ( [1, 2] ), the inequality ( |x + m| + |2x - 1| leq |2x + 1| ) must hold.Let me write that inequality again:( |x + m| + |2x - 1| leq |2x + 1| )I need this to be true for all ( x ) in ( [1, 2] ). Let me analyze the terms involved.First, let's note that for ( x ) in ( [1, 2] ), ( 2x - 1 ) is positive because ( 2x geq 2 ) when ( x geq 1 ), so ( 2x - 1 geq 1 ). Similarly, ( 2x + 1 ) is also positive because ( 2x geq 2 ), so ( 2x + 1 geq 3 ). Therefore, both ( |2x - 1| ) and ( |2x + 1| ) can be simplified without the absolute value signs in this interval.So, ( |2x - 1| = 2x - 1 ) and ( |2x + 1| = 2x + 1 ). Therefore, the inequality becomes:( |x + m| + (2x - 1) leq (2x + 1) )Simplify this:( |x + m| + 2x - 1 leq 2x + 1 )Subtract ( 2x ) from both sides:( |x + m| - 1 leq 1 )Add 1 to both sides:( |x + m| leq 2 )So, the inequality reduces to ( |x + m| leq 2 ) for all ( x ) in ( [1, 2] ).This means that for every ( x ) in ( [1, 2] ), the expression ( x + m ) must lie between ( -2 ) and ( 2 ). In other words:( -2 leq x + m leq 2 )Which can be rewritten as:( -2 - x leq m leq 2 - x )So, for each ( x ) in ( [1, 2] ), ( m ) must satisfy ( -2 - x leq m leq 2 - x ).But since this must hold for all ( x ) in ( [1, 2] ), ( m ) must be simultaneously greater than or equal to ( -2 - x ) for all ( x ) in ( [1, 2] ) and less than or equal to ( 2 - x ) for all ( x ) in ( [1, 2] ).To find the range of ( m ), we need to find the maximum of the lower bounds and the minimum of the upper bounds.First, let's consider the lower bound: ( -2 - x ). Since ( x ) is in ( [1, 2] ), ( -2 - x ) will be minimized when ( x ) is maximized. So, the minimum value of ( -2 - x ) is ( -2 - 2 = -4 ), and the maximum value is ( -2 - 1 = -3 ). Therefore, the maximum of the lower bounds is ( -3 ).Next, consider the upper bound: ( 2 - x ). Since ( x ) is in ( [1, 2] ), ( 2 - x ) will be maximized when ( x ) is minimized. So, the maximum value of ( 2 - x ) is ( 2 - 1 = 1 ), and the minimum value is ( 2 - 2 = 0 ). Therefore, the minimum of the upper bounds is ( 0 ).Therefore, combining these, ( m ) must satisfy:( -3 leq m leq 0 )So, the range of the real number ( m ) is ( [-3, 0] ).Let me just double-check my reasoning. For ( |x + m| leq 2 ) to hold for all ( x ) in ( [1, 2] ), ( m ) must adjust such that ( x + m ) doesn't exceed 2 or go below -2. Since ( x ) is at least 1, ( m ) has to compensate by being at most 1 (to keep ( x + m leq 2 )) and at least -3 (to keep ( x + m geq -2 )). That seems consistent with my earlier conclusion.Yeah, I think that's solid.**Final Answer**(I) The solution set is (boxed{left[0, dfrac{4}{3}right]}).(II) The range of ( m ) is (boxed{[-3, 0]}).

answer：To determine the probability of selecting two shirts, one pair of shorts, and one pair of socks when randomly removing four articles of clothing from a drawer containing 6 shirts, 7 pairs of shorts, and 8 pairs of socks, follow these steps:1. **Calculate the total number of ways to choose 4 articles of clothing from the drawer:** - Total articles = 6 shirts + 7 shorts + 8 socks = 21 articles. - The number of ways to choose 4 articles from 21 is given by the combination formula ( binom{21}{4} ). - ( binom{21}{4} = frac{21!}{4!(21-4)!} = frac{21 times 20 times 19 times 18}{4 times 3 times 2 times 1} = 5985 ).2. **Calculate the number of favorable outcomes (selecting 2 shirts, 1 short, and 1 sock):** - Ways to choose 2 shirts from 6: ( binom{6}{2} = 15 ). - Ways to choose 1 short from 7: ( binom{7}{1} = 7 ). - Ways to choose 1 sock from 8: ( binom{8}{1} = 8 ). - Total favorable outcomes = ( 15 times 7 times 8 = 840 ).3. **Determine the probability:** - Probability = ( frac{text{Favorable outcomes}}{text{Total outcomes}} = frac{840}{5985} ). - Simplify the fraction by dividing numerator and denominator by 15: ( frac{840 div 15}{5985 div 15} = frac{56}{399} ).Therefore, the probability of selecting two shirts, one pair of shorts, and one pair of socks is (boxed{frac{56}{399}}).