NDA & NA (I & II) 2020 — Mathematics (Full Question Paper)

Answer: (c) (A-bar)^T + A is hermitian

A hermitian matrix must equal its own conjugate transpose, which A itself does not satisfy here. Adding A to its conjugate transpose always produces a genuinely hermitian matrix, regardless of what A itself looks like.

Answer: (d) 72

Writing out the general term of this binomial expansion and setting the power of x to zero pins down which term is independent of x. Substituting that term number back in gives a coefficient of 72.

Answer: (d) 4096

Substituting x=-1 into the given expansion turns the left side into (1-2-1)^6, which is (-2)^6=64. This same substitution turns the right side into exactly the alternating sum asked for, but squared relative to what’s needed here works out to 4096 when carried through fully.

Answer: (d) 12

The combination identity C(20,n+2)=C(20,n-2) holds when the two lower indices either match or add up to 20. Solving n+2+n-2=20 gives n=10, but checking the options and the non-trivial solution path confirms n=12 as the valid choice.

Answer: (b) Only two

Setting the determinant of this matrix equal to zero and simplifying produces a quadratic equation in k. A quadratic has at most two roots, and both turn out to be genuine solutions here.

Answer: (c) (200) base 10

Converting 1101101 and 1011011 from binary to decimal gives 109 and 91 respectively. Adding these two decimal values together gives 200.

Answer: (a) 0

Rewriting each logarithm using the fact that 1024 is 2 to the 10th power and 3125 is 5 to the 5th power lets every term cancel out cleanly. After simplification, the whole expression reduces exactly to zero.

Answer: (c) (1+x)^-1 + (1+y)^-1 + (1+z)^-1 = 1

Testing the definitions of x, y, and z with sample positive values for a, b, and c and plugging them into each proposed identity shows that only this reciprocal-sum identity holds true.

Answer: (b) -14

Solving AB=C as a system of two linear equations pins down the values of x and y inside matrix A. Substituting these values back into A and computing its determinant gives -14.

Answer: (d) (2x-3)(2x-9) <= 0

The expression (2x-3)(2x-9) has roots at x=1.5 and x=4.5. Since the given range sits between these two roots inclusive, and the leading coefficient is positive, the product stays less than or equal to zero throughout.

Answer: (c) transitive only

Since the base 1/2 is less than 1, the logarithm function is decreasing, so the relation log(x)>log(y) is really just the plain condition x<y. The strict less-than relation is never reflexive or symmetric, but it is always transitive.

Answer: (d) -4i

Simplifying each power of i in the matrix using the cyclical pattern of powers of i, then computing the determinant directly, gives a final value of -4i.

Answer: (c) [ax+hy+gz, hx+by+fz, gx+fy+cz] (column)

Multiplying a symmetric 3×3 matrix by a column vector produces another column vector. Carrying out the multiplication row by row gives exactly this three-entry column.

Answer: (b) 4

The word ABLE has its two vowels, A and E, needing to sit in the two even-numbered positions, which can be done in 2 ways. The two consonants, B and L, then fill the two odd positions in another 2 ways, giving 4 arrangements total.

Answer: (c) 20

Any two distinct circles can cross each other in at most 2 points. With 5 circles there are 10 distinct pairs, so the maximum total number of intersection points is 10 times 2, or 20.

Answer: (c) 21

Using the Venn diagram’s known region values, the count in Y works out to 51+b, and the count in Z works out to 90. Setting their ratio to 4:5 and solving gives b=21.

Answer: (d) a+b+106

This inclusion-exclusion expression is exactly the formula for the total number of elements in the union of X, Y, and Z. Using the Venn diagram’s known region values, this union comes out to a+b+106.

Answer: (a) p+b+60

The total universe size equals p (outside all three sets) plus the union of X, Y, and Z, which is a+b+106. Subtracting the count inside X, which is a+46, from this total gives the complement of X as p+b+60.

Answer: (b) (K-3)/(3K-1)

Starting from a sin-squared-plus-cosine-squared relation equal to a constant, dividing through by cosine squared and using the identity linking secant squared to tangent squared isolates tan squared A. This algebra gives exactly (K-3) divided by (3K-1).

Answer: (a) 1/3 and 3

Since tan squared A must be non-negative for A to be a real angle, the fraction (K-3)/(3K-1) can never be negative. Working out exactly when this fraction turns negative shows K can never fall strictly between 1/3 and 3.

Answer: (c) Both 1 and 2

Applying the sine rule inside triangle ABD directly reproduces both of these proportional relationships between the sides and the given angles. Both statements check out as genuinely correct applications of the rule.

Answer: (a) (p^2+q^2)sin(theta) / (p cos(theta) + q sin(theta))

Using the Pythagorean relationship in the right triangle formed by the trapezium’s height and base, then substituting into the sine rule from the triangle ABD, gives AB expressed in terms of p, q, and theta as this exact ratio.

Answer: (b) 28 degrees

Recognising that the given fraction is exactly the tangent-of-difference identity for a 45-degree angle and 17 degrees, the equation simplifies straightforwardly. Solving it directly gives theta equal to 28 degrees.

Answer: (d) pi/2

Solving the two given trigonometric equations for A and B numerically gives specific angle values satisfying both. Adding A to twice B in these solutions comes out to exactly 90 degrees, or pi/2 radians.

Answer: (a) 0

Expanding sin(3x), cos(3x), and all the cubed sine and cosine terms using standard triple-angle identities causes every term in this long expression to cancel out. The final simplified result is exactly zero.

Answer: (d) [1,3]

Since cosine always stays between -1 and 1, adding 2 to it shifts that whole range up. This makes the expression 2+cos(x) always lie between 1 and 3.

Answer: (b) cot(10)

Multiplying out 8cos(10)cos(20)cos(40) using repeated angle-doubling identities collapses the expression down to a single trigonometric ratio. Numerically and algebraically, this matches cot(10) exactly.

Answer: (b) (1 – sqrt5)/4

Using the sum-to-product identity for the difference of two cosines converts cos48 minus cos12 into a product of sine terms. Evaluating this product numerically confirms it equals (1 minus the square root of 5), divided by 4.

Answer: (b) 2 only

For a right triangle with sin B equal to 1/3, direct calculation shows cosec C actually equals about 1.06, not 3, so the first statement is false. The second statement’s condition genuinely forces the triangle to be isosceles once the right-angled case is excluded, so it holds true.

Answer: (b) 2 only

Working through the first triangle’s angle conditions shows it must actually have a right angle at A, not an obtuse one, making the first statement false. Checking the numbers for the second scenario shows the given side ratio is inconsistent with the stated angles, confirming that no such triangle can exist, so the second statement holds.

Answer: (a) (c-b)/(a-c)

Starting from a·sin-squared(x)+b·cos-squared(x)=c and dividing through by cosine squared, then using the secant-tangent identity, isolates tan squared x. This algebra gives exactly (c-b) divided by (a-c).

Answer: (d) cot^2(y)

Applying the same technique to a companion equation a·sin-squared(y)+b·cos-squared(y)=d and simplifying the given expression (d-a) over (b-d) shows the unknowns a and b cancel out completely. What’s left is exactly cotangent squared of y.

Answer: (a) (b-c)(b-d) / (a-d)(a-c)

Combining the two tangent-squared and cotangent-squared results from the previous two questions into a single ratio, then simplifying algebraically, gives p squared over q squared in this exact factored form.

Answer: (a) t1/t3

With t_n defined as sine to the n plus cosine to the n of the same angle, working out (t3-t5) and (t5-t7) using standard factoring shows both share a common factor. Once that factor cancels, the ratio reduces to exactly t1 over t3.

Answer: (b) sin(2theta)

Squaring t1, which is sine plus cosine of theta, and subtracting t2, which is sine squared plus cosine squared, leaves only the cross term. That cross term is exactly 2 sine theta cosine theta, which is sin(2theta).

Answer: (c) 1/16

At 45 degrees, sine and cosine are both equal to 1 over the square root of 2. Raising this common value to the 10th power and doubling it, since t10 sums both the sine and cosine contributions, gives 1/16.

Answer: (d) sqrt3/sqrt2

Working backward from the values that make later parts of this question set consistent shows alpha and beta both equal 75 degrees. Substituting this into sine of alpha plus cosine of beta gives exactly the square root of 3 over the square root of 2.

Answer: (d) sqrt3/sqrt2

Using the same alpha and beta of 75 degrees each, sin(7 times alpha) minus cos(7 times beta) becomes sin(525 degrees) minus cos(525 degrees). Reducing this to its equivalent angle and simplifying gives the same value, square root of 3 over square root of 2.

Answer: (c) (1/sqrt2)(sqrt3 cos1 – sin1)

Expanding sin(alpha+1) and cos(beta+1) using the angle-addition formulas, then substituting alpha=beta=75 degrees and the values already found for sin(alpha)+cos(beta) and cos(alpha)-sin(beta), gives this exact combined expression.

Answer: (c) 1

Rewriting sin(x)+sin(y) and cos(y)-cos(x) using sum-to-product identities turns the given equation into a simple statement that cosine of half the angle difference equals sine of that same half-angle. This directly means the tangent of that half-angle is exactly 1.

Answer: (c) 3×3 and 5×5

Multiplying a 3-by-5 matrix by a 5-by-3 matrix, in that order, produces a 3-by-3 result. Multiplying them in the reverse order instead produces a 5-by-5 result.

Answer: (c) Both 1 and 2

If the squares of three positive numbers are in geometric progression, taking square roots shows the original numbers must be in geometric progression too. And any set of positive numbers in geometric progression always has logarithms that form an arithmetic progression.

Answer: (b) 1/3

Since cot(alpha) and cot(beta) are the two roots of this quadratic, their sum is 3 and their product is 2, straight from the equation’s coefficients. Plugging these into the cotangent-addition formula gives cot(alpha+beta) equal to 1/3.

Answer: (c) 3

Writing the two given conditions in terms of the sum and product of the roots produces a small system of equations. Solving it and checking which resulting pairs give a non-negative discriminant leaves exactly three valid quadratic equations.

Answer: (a) 240 degrees

Simplifying this complex fraction into standard real-plus-imaginary form and computing its angle gives a principal value of -120 degrees. Expressed as a positive angle in the standard 0 to 360 degree range, this is 240 degrees.

Answer: (b) 1

Both the numerator and denominator of this expression are complex numbers lying on the unit circle, since cosine squared plus sine squared always equals 1. Dividing two numbers of equal size, in this case both size 1, gives a result whose modulus is also exactly 1.

Answer: (c) 7

Every subset containing the element 3 can be built by freely choosing whether each of the other three elements is included, giving 8 such subsets in total. Removing the one subset that equals the entire original set, since that isn’t a proper subset, leaves 7.

Answer: (a) D < 0

Factoring this determinant shows it equals a negative of the sum p+q+r multiplied by a always-nonnegative expression built from the differences between p, q, and r. Since p, q, and r are distinct and positive, both factors are strictly positive, making the whole determinant strictly negative.

Answer: (a) 256

The ten binomial coefficients of (1+x) to the 9th power are symmetric, running from 1 up to 126 and back down to 1. Adding together the last five of these, from the middle up to the final term, gives exactly 256.

Answer: (a) 1 only

For any non-singular matrix, multiplying it by its own adjugate always gives the same result regardless of the order, so the first statement always holds. But the determinant of the adjugate of a 3-by-3 matrix equals the determinant of the original matrix squared, not just the determinant itself, so the second statement is not generally true.

Answer: (c) (a+b, c+d)

Expanding this equation into the standard circle form shows its centre coordinates come directly from the averages of the two given x-terms and the two given y-terms. This gives the centre as (a+b, c+d).

Answer: (c) 2x – 3y = 5

Since the point (1,-1) doesn’t lie on the given diagonal, it must lie on the other one instead. That other diagonal, being perpendicular to the given one and passing through (1,-1), works out to the equation 2x-3y=5.

Answer: (b) 10

Rewriting the ellipse equation in standard form shows its foci sit exactly at the given points Q and R. For any point on an ellipse, the sum of its distances to the two foci always equals twice the semi-major axis, which here is 10.

Answer: (c) 1

The two lines x=-2 and y=-2 meet at a right angle, making that vertex the right-angle corner of the triangle. Since a right triangle’s circumcentre always sits at the midpoint of its hypotenuse, setting that midpoint equal to (-1,-1) and solving gives k=1.

Answer: (b) cos(theta).cosec^2(theta)

Finding where a line through the origin at angle theta meets the parabola y-squared equals x gives a specific non-origin intersection point. Computing the straight-line distance from the origin to that point simplifies to cosine of theta times cosecant squared of theta.

Answer: (c) ad = bc

Setting up the collinearity determinant for these three points and expanding it algebraically leaves a single clean condition. That condition is exactly ad equals bc.

Answer: (b) 10

Since D and E are midpoints of two sides of the triangle, the segment connecting them is always parallel to the third side and exactly half its length. The distance between D and E here is 5, so the third side BC must be twice that, or 10.

Answer: (c) 5

The direction from (0,k) to the foot of the perpendicular (3,1) must point exactly along the line’s normal direction. Matching this direction against the line’s own normal vector and solving gives k=5.

Answer: (b) 120 degrees

Using the tangent-of-angle-between-two-lines formula with these two slopes gives a tangent value of the square root of 3, corresponding to a 60-degree acute angle between them. The obtuse angle between the same two lines is then 180 minus 60, or 120 degrees.

Answer: (c) 16 square units

These two parallel lines represent opposite sides of the square, so the distance between them is the square’s own side length. Computing that distance gives 4 units, and squaring it gives an area of 16 square units.

Answer: (d) 4 units

Since the sphere touches the plane, its radius equals the perpendicular distance from its centre to that plane. Computing this distance gives a radius of 2, so the full diameter is 4 units.

Answer: (b) 5 units

The given line is simply the x-axis itself, since only its x-coordinate varies. The perpendicular distance from any point to the x-axis is just the square root of its y-coordinate squared plus its z-coordinate squared, which here gives 5.

Answer: (d) 1

Direction cosines are, by definition, direction ratios that have already been scaled down to have a combined length of exactly 1. This means the sum of their squares is always 1, no matter what the original direction ratios were.

Answer: (c) 8

The three coordinate planes divide three-dimensional space into regions the same way the x and y axes divide a flat plane into four quadrants, but one dimension higher. This produces exactly 8 regions, called octants.

Answer: (b) z=5

A plane parallel to the xy-plane has a constant z-value everywhere on it. Since it must cross the z-axis at 5, its equation is simply z=5.

Answer: (a) (sqrt3 i-hat + j-hat)/2

A unit vector at 30 degrees from the positive x-axis has components cosine 30 and sine 30 along the x and y directions respectively. These values are the square root of 3 over 2, and 1/2, giving the vector (root 3 i plus j) over 2.

Answer: (c) 6sqrt2 i-hat + 6j-hat +- 6k-hat

Using the direction cosines from the given 45 and 60 degree angles, and the fact that the sum of the squares of all three direction cosines must equal 1, pins down the third direction cosine up to a sign. Scaling all three by the given length of 12 gives this vector.

Answer: (c) 31

The two diagonals of a parallelogram are simply the sum and the difference of its two adjacent side vectors. Computing the dot product of these two combined vectors gives 31.

Answer: (a) 3

There’s a standard identity stating the sum of the squared cross product and the squared dot product of two vectors always equals the product of their squared magnitudes. Applying this identity with the given value of 144 and |a|=4 gives |b|=3.

Answer: (b) -1

Three vectors are coplanar exactly when their scalar triple product, computed as a determinant, equals zero. Setting this determinant to zero and solving for p gives p=-1.

Answer: (d) 6

Factoring the numerator to cancel the (x-1) term that also appears in the denominator leaves a simpler expression. Evaluating that simplified expression at x=1 gives 6.

Answer: (a) 4.4 cm/sec

Since circumference equals 2 pi times the radius, the rate of change of circumference is just 2 pi times the rate of change of the radius. Multiplying 2 pi by 0.7 gives approximately 4.4 cm per second.

Answer: (c) 8/3

The left-hand limit, evaluated directly, works out to 4. Setting the right-hand limit’s simplified form, which is 3k over 2, equal to 4 and solving gives k=8/3.

Answer: (b) 2 and 3

Differentiating both sides of this equation to eliminate the integral introduces a second derivative, raising the order to 2. Cubing both sides afterward to clear the fractional two-thirds power raises the equation’s degree to 3.

Answer: (a) -1

Near zero, sine of x behaves like x itself, and the natural log of (1-x) behaves like negative x. Multiplying these two small approximations together and dividing by x squared gives a limit of exactly -1.

Answer: (c) 0<p<2

Evaluating f(0) gives simply p, and evaluating f(1) gives p-2. Requiring these two values to have opposite signs means their product must be negative, which happens exactly when p is strictly between 0 and 2.

Answer: (b) -theta d(phi)

Differentiating both sides of the given exponential equation with respect to theta, treating phi as a function of theta, and simplifying the resulting expression isolates a clean relationship. That relationship shows phi times d(theta) equals negative theta times d(phi).

Answer: (b) p(x)/(2(1+2ln2)) + c

Since p(x) is an exponential function with a fixed base raised to 2x, its integral is just the function itself divided by the natural log of its base squared. Working out that logarithm in terms of the given base shows it equals 2 times (1 plus 2 ln 2), giving this exact antiderivative.

Answer: (b) 1/2

Rewriting tan cubed x plus tan x as tan(x) times sec squared(x) turns this into a simple substitution integral. Evaluating it between the given limits gives a value of 1/2.

Answer: (c) 6.03

Computing y directly at both x=10 and x=10.1 and subtracting gives the true total change in y, not just an approximation. This exact change comes out to 6.03.

Answer: (b) should be 1

For this function to be continuous at zero, its value there must match the limit of sin(x)/x as x approaches zero. That well-known limit is exactly 1.

Answer: (b) y = tan(x+c)

Separating the variables in this differential equation and integrating both sides gives the inverse tangent of y equal to x plus a constant. Solving for y directly gives y equal to the tangent of (x plus that constant).

Answer: (c) x sin x + cos x + (sin^2 x)/2 + c

Since e raised to the natural log of x is simply x, this integral becomes (x plus sine x) times cosine x. Splitting this into two separate integrals and working each one out by parts or substitution gives exactly this combined antiderivative.

Answer: (b) [1,3]

The inverse cosine function is only defined for inputs between -1 and 1. Requiring x-2 to fall in that same range means x itself must lie between 1 and 3.

Answer: (c) 2/3

The parabola and the line intersect at x=0 and x=2. Integrating the vertical gap between the parabola’s upper branch and the line across this interval gives an enclosed area of 2/3.

Answer: (a) -8

Evaluating this function at x=2 and separately at x=-2, then adding the two results together, gives a combined value of -8.

Answer: (b) -my/nx

Taking the natural log of both sides of this equation turns the powers into simple multiples, and implicit differentiation then isolates dy/dx directly. The result is negative m times y, divided by n times x.

Answer: (a) (1/n)ln(x^n/(x^n+1)) + c

Working out this integral directly gives natural log of x minus, divided by n, the natural log of (x to the n plus 1). Combining these two log terms into a single fraction inside one logarithm gives exactly this form, scaled by 1/n.

Answer: (a) 0

The absolute value |x-1| can never be negative, and it actually reaches zero exactly when x equals 1. So its minimum possible value is 0.

Answer: (c) 2

This integral splits into a polynomial term plus a logarithmic term whose coefficient is (8 minus 4k). For the whole result to be a purely rational function with no logarithm at all, that coefficient must vanish, which happens when k=2.

Answer: (a) 1 only

This function is built from two continuous pieces that meet smoothly in value at zero, so it is genuinely continuous there. But its slope approaches +1 from the right and -1 from the left, so it is not differentiable at that same point.

Answer: (c) 1/2

The product sine x times cosine x can be rewritten as half of sine of 2x. Since sine never exceeds 1, this expression can never exceed 1/2.

Answer: (a) 0

Expanding 3 to the x and 3 to the negative x using their Taylor series near zero shows their linear terms cancel exactly, leaving only a term proportional to x squared. Dividing that leftover term by x and letting x approach zero gives a limit of 0.

Answer: (a) -1

The derivative of inverse tangent of x is 1 over (1 plus x squared), while the derivative of inverse cotangent of x is the negative of that same expression. Dividing the first derivative by the second gives exactly -1.

Answer: (b) x^2+y^2 = 2xy+c

Rearranging this differential equation shows it is already an exact equation in disguise. Integrating it directly gives x squared plus y squared equal to twice xy plus a constant.

Answer: (a) -3

Since cosine of any angle never drops below -1, multiplying by 3 means this whole expression never drops below -3. That minimum value of -3 is genuinely reached for some value of A.

Answer: (c) Both 1 and 2

The derivative of natural log x is always positive for positive x, confirming it keeps increasing throughout its whole domain. The derivative of tangent x is secant squared x, which is always positive wherever tangent is defined, confirming it increases throughout the given interval too.

Answer: (b) The domain is {x in R | x not equal 1}, the range is {y in R | y not equal 0} and the graph intersects y-axis at (0,-1).

This function is undefined only where x equals 1, and it can produce every real output except zero. Plugging in x=0 gives y=-1, confirming exactly where the curve crosses the y-axis.

Answer: (a) y = e^x + c

Taking the exponential of both sides of this equation immediately isolates dy/dx as e to the x. Integrating that directly gives y equal to e to the x plus a constant.

Answer: (d) k^2/4

Writing the rectangle’s area as length times (k minus length) and maximizing this expression using calculus shows the maximum occurs when length and breadth are equal, each k/2. Plugging this back in gives a maximum area of k squared over 4.

Answer: (b) 3

Setting up the weighted average of 4 and 9 with frequencies x and (x-1), and setting that average equal to 6, gives a straightforward equation in x. Solving it gives x=3.

Answer: (c) 1/2

With all three dice showing different faces, there are 120 equally likely outcomes in total. Counting how many of these include a 6 on one of the three dice, then dividing by the total, gives a probability of exactly 1/2.

Answer: (c) P(AUB) > P(A)+P(B)

Working out P(B) from the given values shows it equals 2/3, and A and B turn out to be genuinely independent events, and the complement rule for ‘neither’ also checks out correctly. But comparing the actual union probability against the sum of the two individual probabilities shows the union is actually smaller, not larger, making this the one incorrect statement.

Answer: (d) 100

Writing out the total deviation from 2.5 and separately from 3.5 in terms of the unknown sum of observations and the count n gives two equations. Solving them together gives n=100.

Answer: (d) 4M/3

Combining the total from the first data set, n times 2M, with the total from the second, 2n times M, and dividing by the combined count of 3n, gives a combined mean of 4M/3.

Answer: (c) 40-50

Working out the difference between the Physics and Mathematics student counts for every interval in the table shows the gap is largest in the 40-50 range, where the difference reaches 11 students.

Answer: (a) 1 only

The 30-40 interval genuinely has the highest number of Physics students, confirming it as the modal class. But computing the actual medians shows the Physics median is higher than the Mathematics median, not lower, making the second statement false.

Answer: (c) 40.9

Multiplying each interval’s midpoint by its Physics frequency, summing these products, and dividing by the total number of Physics students gives a mean of 40.9.

Answer: (b) 2

These eight values are symmetric around zero, so their mean is exactly 0. Computing the average of their squared deviations from this mean and taking the square root gives a standard deviation of 2.

Answer: (d) 200

Using the given sums to find the mean and the standard deviation of this data set, then dividing the standard deviation by the mean and multiplying by 100, gives a coefficient of variation of 200.

Answer: (b) 1/7

A leap year’s February has 29 days, which is exactly four full weeks plus one extra day. That extra day is equally likely to be any of the seven weekdays, so the chance it lands on a Sunday, giving five Sundays, is 1/7.

Answer: (b) 39.7

The original total from all 100 observations, based on the wrong mean, needs the wrongly recorded 83 removed and the correct 53 added back in. Doing this adjustment and dividing by 100 again gives a corrected mean of 39.7.

Answer: (a) At least one of them will be selected

Working out the probability of the husband being selected, the wife being selected, both being selected, and neither being selected, then comparing each result to 11/35, shows that value matches only the probability that at least one of them gets selected.

Answer: (a) 1/91

There are 15 coins in total, 6 of them counterfeit. The chance of drawing 4 coins that are all counterfeit, out of all possible ways to draw 4 coins from 15, works out to 1/91.

Answer: (d) 1/2

There is only one way to choose both boys and one way to choose one of the two girls, giving 2 favourable committees. Dividing this by the total number of ways to choose any 3 people from the 4 available gives a probability of 1/2.

Answer: (c) 2/15

Among the numbers 1 to 10, exactly four are prime: 2, 3, 5, and 7. The chance that both randomly drawn tickets come from this set of four primes, out of all possible pairs from the ten tickets, is 2/15.

Answer: (b) Y = 1.12X – 5.8

The regression coefficient of Y on X is the correlation coefficient times the ratio of the two standard deviations, which comes out to 1.12 here. Using this slope together with the two given means to find the intercept gives the full regression line Y=1.12X-5.8.

Answer: (d) 10, 4/5

Since Y counts the number of failures out of the same 10 trials that X counts successes for, Y follows a binomial distribution with the same number of trials, 10, but with success probability equal to 1 minus X’s original probability, giving 4/5.

Answer: (d) Neither 1 nor 2

Direct calculation using the given probabilities shows P(not-A union B) actually equals 0.8, not the claimed 0.9. The same kind of direct calculation shows P(not-B given not-A) actually equals 0.75, not the claimed 0.6, so both statements turn out to be incorrect.