KENDRIYA VIDYALAYA SAMPLE PAPER 12 (2022-23) MATHS XII PDF

KENDRIYA VIDYALAYA GACHIBOWLI, GPRA CAMPUS, HYD–32 SAMPLE PAPER TEST 12 FOR BOARD EXAM (2022-23) (ANSWERS) SUBJECT: MATHEMATICS (041) MAX. MARKS : 80 CLASS : XII...

KENDRIYA VIDYALAYA GACHIBOWLI, GPRA CAMPUS, HYD–32 SAMPLE PAPER TEST 12 FOR BOARD EXAM (2022-23) (ANSWERS) SUBJECT: MATHEMATICS (041) MAX. MARKS : 80 CLASS : XII DURATION: 3 HRS General Instructions: 1. This Question paper contains - five sections A, B, C, D and E. Each section is compulsory. However, there are internal choices in some questions. 2. Section A has 18 MCQ’s and 02 Assertion-Reason based questions of 1 mark each. 3. Section B has 5 Very Short Answer (VSA)-type questions of 2 marks each. 4. Section C has 6 Short Answer (SA)-type questions of 3 marks each. 5. Section D has 4 Long Answer (LA)-type questions of 5 marks each. 6. Section E has 3 source based/case based/passage based/integrated units of assessment (4 marks each) with sub parts. SECTION – A Questions 1 to 20 carry 1 mark each. 1. The integrated factor of the differential equation: (1+x2) + = is (a) (b) 2 (c) 3 (d) Ans: (d) 2. For any square matrix A, AAT is a (a) unit matrix (b) symmetric matrix (c) skew-symmetric matrix (d) diagonal matrix Ans: (b) symmetric matrix  2   3 3. If A   0 2 5  then A-1 exist if    1 1 3  (a) λ = 2 (b) λ = 0 (c) λ ≠ 2 (d) λ ≠ 0 Ans: (c) λ ≠ 2 4. If A is square matrix such that A2 = I , then (A - I)3 + (A + I)2 – 7A is equal to (a) -A (b) I – A (c) I + A (d) 3A Ans: (a) -A 5 x 5. If A=   and A = A’, then  y 0 (a) x = 0, y = 5 (b) x + y = 5 (c) x = y (d) none of these Ans: (c) x = y 6. The differential coefficient of sec(tan-1 x ) w.r.t. x is (a) √ (b) (c) x √1 + (d) √ Ans: (a) √ 7. If A is a square matrix of order ,then | ( )| = (a) | | (b) | | (c) | | (d) | | Ans: (b) | | Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1- 8. P is a point on the line joining the points (0,5, -2) and (3, -1,2). If the x-coordinate of P is 6, then its z-coordinate is (a) 10 (b) 6 (c) -6 (d) -10 Ans: (b) 6 √ 9. The value of ∫ (a) (b) (c) (d) Ans: (d) 10. The sum of the order and degree of the following differential equation = 0, is (a) 5 (b) 4 (c) 3 (d) 2 Ans: (c) 3 11. The value of such that the vectors ⃗ = 2 ̂ + ̂ + and ⃗= ̂ + 2 ̂ + 3 are orthogonal is (a) 0 (b) 1 (c) (d) - Ans: (d) - 12. The corner points of the shaded bounded feasible region of an LPP are (0,0),(30,0),(20,30) and (0,50) as shown in the figure. The maximum value of the objective function Z = 4x+y is (a) 120 (b) 130 (c) 140 (d) 150 Ans: (a) 120 13. The projection of the vector 2i  3 j  2k on the vector i  2 j  k is √ √ (a) ( ) ( ) ( ) √ Ans: (a) 14. If ⃗ + ⃗ = 60, ⃗ − ⃗ = 40 and | ⃗|= 22 then ⃗ = (a) 36 (b) 22/60 (c) 46 (d) None of these Ans: (c) 46 15. If the function f(x) defined by f(x) = , ≠3 , = 3 is continuous at x = 3, then the value of k is (a) 6 (b) 3 (c) -6 (d) 3 Ans: (a) 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 2- 16. Let A and B be two events. If P(A)=0.2,P(B)=0.4 ,P(AUB)=0.6 then P is equal to (a) 1 (b) 0 (c) 0.2 (d) 0.4 Ans: (b) 0 17. The corner points of the feasible region determined by the following system of linear inequalities: 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0,0), (5,0), (3,4), (0,5). Let Z= px + qy, where p,q > 0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is (a) p = q (b) p = 2q (c) p = 3q (d) q = 3p Ans: (d) q = 3p 18. The value of ∫ dx is (a) 2 (b) 3/4 (c) 0 (d) – 2 Ans: (c) 0 ASSERTION-REASON BASED QUESTIONS In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. 19. Assertion(A): The value of + − = Reason(R) : (− ) = − Ans: (b) Both A and R are true but R is not the correct explanation of A. 20. Assertion(A): = = and = = are coplaner. Reason (R) : Let line passes through the point ( , , ) and parallel to the vector whose direction ratios are , , , : Let line passes through the point ( , , ) and parallel to the vector whose direction ratios are , , ,. Then both lines are coplaner if and only if | − − − | =0 Ans: (a) Both A and R are true and R is the correct explanation of A. SECTION – B Questions 21 to 25 carry 2 marks each. 21. Find the value of sin-1[cos( )] Ans: = 6 + 3 3 = = − = − = − sin 5 2 5 10 10 =− =− , 10 10 ∵ ∈ − , 10 2 2 OR Let A = R- {3} and B = R- {1}. Consider the function f : A → B defined by f(x) =. Is f is one-one and onto? Justify your Answer. Ans: A = R- {3} and B = R- {1} function f : A → B defined by f(x) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3-  x1  2 x2  2 Now, f ( x1 )  f ( x2 )   x1  3 x2  3  ( x2  3)( x1  2)  ( x2  2)( x1  3)  x1 x2  3 x1  2 x2  6  x1 x2  3x2  2 x1  6  3x1  2 x2  3x2  2 x1   x1   x2  x1  x2 Hence, f(x) is one-one function. x2 x2 Let f ( x)   y  y( x  3)  x  2  yx  3 y  x  2 x3 x 3 3y  2  x( y  1)  3 y  2  x  ( y  1) 3y  2 3 y  2  2( y  1) 2 x2 y 1 y 1  f ( x)    x3 3y  2 3 y  2  3( y  1) 3 y 1 y 1 3y  2  2 y  2 3y  2 y   y 3 y  2  3 y  3 2  3 f(x) is onto. So f(x) is bijective and invertible 22. If √1 − + 1− = ( − ), show that =. Ans: Let x  sin A; y  sin B  (1) f ( x)  1  sin 2 A  1  sin 2 B  a(sin A  sin B)  cos A  cos B  a(sin A  sin B) We know that [1  sin 2 A  cos2 A] AB cos   AB  A B  2  a  2 cos    a  2sin     2   2  AB sin    2  AB  cos    cot    a   cot    2   sin   AB   cot 1 a  sin 1 x  sin 1 y  2 cot 1 a 2 1 1 dy   0 1  x2 1  y 2 dx dy 1  y2   dx 1  x2 23. The volume of a cube is increasing at the rate of 9 cubic cm per sec. How fast is the surface area increasing when the length of an edge is 10 cm. Ans: Let V and S be the volume and surface area of a cube of side x cm respectively. dV Given,  9 cm3 / sec dt  ds  We require    dt  x 10 cm Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4-  dv dx dx Now, V  x 3   3x 2   9  3x 2  dt dt dt dx 9 3    dt 3x 2 x 2 Again, S  6 x 2  By formula for surface area of a cube ds dx 3 36  ds  36   12 x   12 x  2      3.6 cm 2 / sec dt dt x x dt   x 10 cm 10 24. Find the vector of magnitude 6, which is perpendicular to both the vectors 2i  j  2k 4i  j  3k Ans: Any vector perpendicular to ⃗ = 2 ̂ − ̂ + 2 ⃗ = 4 ̂ − ̂ + 3 is ⃗× ⃗= ̂ ̂ 2 −124 −13 =- ̂+2 ̂+2 = ⃗ A vector of magnitude 6 in the direction of ⃗ ⃗ =| ⃗ |. 6 =-2 ̂ + 4 ̂ + 4 OR Find the equation of a line in vector and cartesian form which passes through the point (1,2,3) and is parallel to the vector 3i  2 j  3k. Ans: 25. If ⃗+ ⃗+ ⃗ =0⃗ and | ⃗| =3, ⃗ =5 and | ⃗| =7 then what is the angle between ⃗ and ⃗. Ans: ⃗+ ⃗+ ⃗ =0⃗ ( ⃗ + ⃗) = (− ⃗) , ⇒ ⃗+ ⃗. ⃗+ ⃗ = ⃗. ⃗, | ⃗| + ⃗ + 2 ⃗. ⃗ = | ⃗| ⇒ 2 ⃗. ⃗=15, ⇒ 2 | ⃗| ⃗ cos = 15, = 60° SECTION – C Questions 26 to 31 carry 3 marks each. 26. Find: ∫ √ Ans: 5  4 x  2 x 2  (2 x 2  4 x  5)  5  5  2  5   7   2  x 2  2 x    2  x 2  2 x  1  1    2  ( x  1)2    2  ( x  1)2    2  2  2   2  7  7  2  2 2  2   ( x  1)   2    ( x  1)  2   2     Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5-  dx dx 1 dx     5  4 x  2 x2  72  2  72  2  ( x  1) 2    ( x  1) 2   2   2  dx x Using   sin 1  c , we get a2  x2 a 1 x 1 1 2  sin 1 c  sin 1 ( x  1)  c 2 7 2 7 2 OR 3 x  x 1 Evaluate:  x2 1 dx Ans: 27. Two cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the mean of the number of aces drawn. Ans: Let X denote the number of King in a draw of two card 48 C 48  47 188  P( X  0)  P( no ace)  52 2   C2 52  51 221 4 C1 48 C1 4  48 32  P( X  1)  P ( one ace and one non ace )  52   C2 52  51 221 4 C2 4 3 1  P( X  2)  P( two ace)  52   C2 52  51 221 n 188 32 2 34  Mean of X  xi p ( xi )  0   1   2 1 221 221 221 221 OR Three friends go for coffee. They decide who will pay the bill, by each tossing a coin and then letting the “odd person” pay. There is no odd person if all three tosses produce the same result. If there is no odd person in the first round, they make a second round of tosses and they continue to do so until there is an odd person. What is the probability that exactly three rounds of tosses are made? Ans: P(not obtaining an odd person in a single round) 1 1 1 1 = P(All three of them throw tails or All three of them throw heads) =    2  2 2 2 4 P(obtaining an odd person in a single round) 3 = 1 – P(not obtaining an odd person in a single round) = 4 ∴ Required probability Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6-  = P(‘In first round there is no odd person’ and ‘In second round there is no odd person’ and ‘In 1 1 3 3 third round there is an odd person’) =    4 4 4 64  /4 28. Evaluate:  log(1  tan x)dx 0 Ans: 29. Solve the following Linear Programming Problem graphically: Maximize: Z = 100 + 120 Subject to : 5 + 8 ≤ 200, 5 + 4 ≤ 120, , ≥ 0 Ans: Plotting the constraints in graph Corner points(0,0),(24,0),(8,20),and (0,25) At(0,0) Z=0 At(0,25) Z=3000 At(24,0) Z=2400 At(8,20) Z=3200 (Maximum) Maximum value of Z is Rs.3200 at point (8,20) d2 y 30. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find dx2 Ans. Given that x = a (cos t + t sin t) and y = a (sin t – t cos t) Differentiating both sides w.r.t. t, we get Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7-  dx d  d d   a  cos t   t sin t  sin t (t )    a[ sin t  (t sin t  sin t )]  at cos t dt  dt  dt dt   dy d  d d  and  a  sin t   t cos t  cos t (t )    a[cos t  (t sin t  cos t )]  at sin t dt  dt  dt dt   dy dy dt at sin t Now,    tan t dx dx at cos t dt Again, Differentiating both sides w.r.t. x, we get d2 y d 2 dt sec2 t 1 2  tan t  sec t    sec3 t dx dx dx at cos t at 31. Find the particular solution of the differential equation (1 + e2x)dy + (1 + y2) ex dx = 0 given that y = 1 when x = 0. Ans: (1  e 2 x )dy  (1  y 2 )e x dx  0 dy e x dx 1 e x dx    tan y   (1  y 2 ) (1  e 2 x ) (1  e 2 x ) Let t  e x  dt  e x dx dt  tan 1 y    tan 1 y   tan 1 t  c (1  t 2 )  tan 1 y   tan 1 (e x )  c When x = 0 and y = 1, tan 1 1   tan 1 (e 0 )  c       c  c  4 4 2   Hence, tan 1 y   tan 1 (e x )   tan 1 y + tan 1 (e x )  2 2 OR dy  Solve the following differential equation:  2 y tan x  sin x , given that, when x  , y = 0 dx 3 dy Ans: The given differential equation is of the form  Py  Q dx Comparing the two equations, P  2 y tan x & Q  sin x Integrating Factor (IF) = e  e e  Pdx 2tan xdx 2 tan xdx 2  e2[logsec x ]  elog(sec x )  sec2 x ∴ Solution to the differential equation  y( IF )  (Q  IF )dx  C  ysec2 x   sin xsec 2 xdx  C sin x sin x 1  y sec 2 x   2 dx  C    dx  C cos x cos x cos x  y sec2 x   tan x sec xdx  C  y sec2 x  sec x  C 1 C y  2  cos x  C (cos 2 x) sec x sec x  When x  , y  0 3    1 1  y  0  cos  C cos 2     C  0  C  2 3 3 2 4 2  y  cos x  2cos x Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8-  SECTION – D Questions 32 to 35 carry 5 marks each. 32. Find the equation of a line passing through the pointg P(2, −1, 3) and perpendicular to the lines      r  i  j  k   (2i  2 j  k ) and r  2i  j  3k   (i  2 j  2 k ) x  2 y 1 z  3 Ans: Let the equation of line passing through the point (2, -1, 3) be   …(1) a b c  Given lines are r  i  j  k   (2i  2 j  k ) ……. (2)  and r  2i  j  3k   (i  2 j  2 k ) …… (3) Since (1), (2) and (3) are perpendicular to each other  2 a  2b  c  0 and a  2b  2c  0 a b c a b c       l 4  2 1  4 4  2 6 3 6  a  6l ,b  3l , c  6l x  2 y 1 z  3 Putting it in (1) we get required equation of line as   6l 3l 6l x2 z 3   y 1  2 2 OR Find the coordinates of the foot of perpendicular drawn from the point A(–1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1). Hence find the image of the point A in the line BC. Ans: Let P be the foot of the perpendicular drawn from point A on the line joining points B and C. Equation of the line joining the points B(0,–1,3) and C(2,–3,–1) is x  x1 y  y1 z  z1   x2  x1 y2  y1 z2  z1 x  0 y 1 z 3 x  0 y 1 z  3       2  0 3  1 1  3 2 2 4 x  0 y 1 z  3 Let    2 2 4 General coordinates of P is (2λ,−2λ−1,−4λ+3) Direction ratios of AP(2λ+1,−2λ−9,−4λ−1) ∵ Both the lines AP and BC are perpendicular to each other. ∴ 2(2λ+1)−2(−2λ−9)−4(−4λ−1)=0 ⇒ 24λ+24=0 ⇒ λ=−1 ∴ P(−2, 1, 7) So, the coordinates of the foot of perpendicular drawn from the point A to BC is P(−2, 1, 7). Let Q(h, p, s) be the image of A in the line BC, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9-  So P must be the mid-point of AQ.  h 1 p  8 s  4  P , ,   P(2,1, 7)  2 2 2  On comparing the coordinates, we get h = -3, p=-6, s= 10, Hence the image is Q(-3, -6, 10). 33. Make a rough sketch of the region {( , ): 0 ≤ ≤ + 1,0 ≤ ≤ + 1,0 ≤ ≤ 2} and find the area of the region using integration. Ans: Required Area=∫ ( + 1) + ∫ ( + 1) = 34. Let = {1,2,3, … ,9} and be the relation in × defined by (a, b) R (c, d) if a+ d = b+ c, for ( , ), ( , ) in ×. Prove that is an equivalence relation and also obtain the equivalence class [(2,5)]. Ans: Ans: A = {1,2,3...9} R in A × A (a,b) R (c,d) if (a,b)(c,d) ∈ A∈A a+b=b+c Consider (a,b) R (a,b) (a,b) ∈ A × A a+b=b+a Hence, R is reflexive. Consider (a, b) R (c, d) given by (a, b) (c, d) ∈ A × A a+d=b+c⇒c+b=d+a ⇒(c, d) R (a, b) Hence R is symmetric. Let (a, b) R (c, d) and (c, d) R (e, f) (a, b),(c, d),(e, f) ∈ A × A a + b = b + c and c + f = d + e a+b=b+c ⇒ a − c = b − d.....(1) c + f = d + e......(2) Adding (1) and (2) a−c+c+f=b−d+d+e ⇒a+f=b+e ⇒ (a, b) R (e, f) R is transitive. R is an equivalence relation. We select from set A={1,2,3,....9} a and b such that 2 + b = 5 + a so b = a + 3 Consider (1,4) (2, 5) R (1, 4) ⇒ 2 + 4 = 5 + 1 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 10-  [(2,5)=(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)] is the equivalent class under relation R. OR x Show that the function f : R→ {x  R : −1 < x 0 But, 2xy is negative. Then, 2xy ≠ x − y. Thus, the case of x being positive and y being negative can be ruled out. Under a similar argument, x being negative and y being positive can also be ruled out. Therefore, x and y have to be either positive or negative. x y When x and y are both positive, we have f ( x)  f ( y)    x  xy  y  xy  x  y 1 x 1 y x y When x and y are both negative, we have f ( x)  f ( y)    x  xy  y  xy  x  y 1 x 1  y Therefore, f is one-one. Now, let y  R such that −1 < y < 1. y If y is negative, then there exists x   R such that 1 y  y  y    y   1 y  1 y y f ( x)  f     y  1  y  1 y 1    y  1 y  y  1 y  1 y  y If y is positive, then there exists x   R such that 1 y  y  y    y   1 y  1 y y f ( x)  f     y  1 y  1 y  y  1 y  y 1   1 y  1 y  Therefore, f is onto. Hence, f is one-one and onto.  1 1 0   2 2 4 35. Given A =  2 3 4  and B =  4 2 4 , verify that BA = 6I, how can we use the result    0 1 2   2 1 5  to find the values of x, y, z from given equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7  1 1 0   2 2 4 Ans: We have A =  2 3 4  and B =  4 2 4    0 1 2   2 1 5  Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 11-   1 1 0   2 2 4   2  4  0 2  2  0 4  4  0  Now, AB   2 3 4   4 2 4    4  12  8 4  6  4 8  12  20   0 1 2   2 1 5   0  4  4 0  2  2 0  4  10  6 0 0 1 0 0    0 6 0   6 0 1 0     0 0 6  0 0 1  1  AB  61  A 1  B 6  2 2 4  1  A   4 2 4  1 6  2 1 5  The given system of linear equations can be written in matrix form as AX = B, where  1 1 0  x  3 A   2 3 4  , X   y  , B  17       0 1 2   z   7   X  A 1 B  2 2 4   3   2  3  2  17  4  7  1 1  X   4 2 4  17   X   4  3  2  17  4  7     6 6  2 1 5   7   2  3  1 17  5  7  x   2    y    1  x  2, y  1, z  4  z   4  SECTION – E(Case Study Based Questions) Questions 36 to 38 carry 4 marks each. 36. Case-Study 1: Read the following passage and answer the questions given below. Some young entrepreneurs started an industry “Young achievers” for casting metal into various shapes. They put up an advertisement online stating the same and expecting order to cast method for toys, sculptures, decorative pieces and more. A group of friends wanted to make innovative toys and hence contacted the “Young achievers” to order them to cast metal into solid half cylinders with a rectangular base and semi circular ends. (i) If r, h and V are radius, length and volume respectively casted half cylinder, then find the total surface area function S of the casted half cylinder in terms of V and r. (ii) For the given volume V, Find the condition for the total surface area S to be minimum. (iii) Use second derivative test to prove that Surface area is minimum for given volume. OR (iii) Find the ratio h : 2r for S to be minimum. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 12-  1 2V Ans: (i) The volume of half - cylinder is, V   r 2 h  h  2 ….. (i) 2 r Total surface area = area of rectangular base + area of 2 semicircular ends + area of curved surface 1 1  S  h  2r  2   r 2   2 rh 2 2 2 S  2rh   r   rh From equation (i) let's substitute the value of h,  2V   2V  4V 2V By substituting we get, S  2r   2    r 2   r  2    r2  r  r  r r (ii) In order to find the extrema, we need to find the first differentiation and equate it to zero. dS dS 4V 2V Therefore, 0  2  2 r  2  0 dr dr  r r 2V 4V V 2V  2 r  2  2  r 3   2 r r   d 2S (iii) To find whether it’s maxima or minima we need to differentiate again. If  0 then it’s a dr 2 d 2S minima and if  0 it’s a maxima. dr 2 d 2S 1  4V   2  2  3   2V  dr r    d 2S As all the numbers are positive, 0 dr 2 Therefore, its minima. OR 3 1 2  r (iii) Now, r 3  V   2   V     1 2    2    Substituting in equation (i) we get, r3 1    r 2h 1 2  2   2    2r  1 2  2r 2 2r   2    2   1   h    h  h  h    2r   2 37. Case-Study 2: In a street two lamp posts are 600 feet apart. The light intensity at a distance d from the first 1000 (stronger) lamp post is , the light intensity at distance d from the second (weaker) lamp post d2 125 is 2 (in both cases the light intensity is inversely proportional to the square of the distance to d the light source). The combined light intensity is the sum of the two light intensities coming from both lamp posts. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 13-  Based on the above information, answer the following questions. (i) If you are in between the lamp posts, at distance x feet from the stronger light, then find the formula for the combined light intensity coming from both lamp posts as function of x (ii) If I(x) denote the combined light intensity, then find the value of x for which I(x) is minimum. OR (ii) Find the darkest spot between the two lights Ans: (i) Since, the distance is x feet from the stronger light, therefore the distance from the weaker light will be 600 – x. 1000 125 So, the combined light intensity from both lamp posts is given by 2  x (600  x) 2 1000 125 (ii) We have, I ( x)  2  x (600  x) 2 2000 250 6000 750  I '( x)  3  3  I ''( x)  4  x (600  x) x (600  x) 4 2000 250 For maxima/minima, I′(x) = 0 ⇒  3  3  8(600  x)3  x3 x (600  x) Taking cube root on both sides, we get, 2(600 – x) = x ⇒ 1200 = 3x ⇒ x = 400 Thus, I(x) is minimum when you are at 400 feet from the strong intensity lamp post. OR (ii) Since, I(x) is minimum when x = 400 feet, therefore the darkest spot between the two light is at a distance of 400 feet from stronger lamp post, i.e., at a distance of 600 – 400 = 200 feet from the weaker lamp post. 38. Case-Study 3: The reliability of a COVID 19 test is specified as follows: Of people having COVID 19, 90% of the test detect the disease but 10% go undetected. Of people free of COVID 19, 99% of the test are judged COVID 19 negative but 1% are diagnosed as showing COVID 19 positive. From a large population of which only 0.1% have COVID 19, one person is selected at random, given the COVID19 test, and the pathologist reports him/her as COVID 19 positive. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 14-  (i) What is the probability of the ‘person to be tested as COVID 19 positive’ given that ‘he is actually having COVID 19’ and What is the probability of the ‘person to be tested as COVID 19 positive’ given that ‘he is actually not having COVID 19’? (ii) What is the probability that the ‘person is actually not having COVID 19’? (iii) What is the probability that the ‘person is actually having COVID 19’ given that ‘he is tested as COVID 19 positive’? OR (iii) What is the probability that the ‘person selected will be diagnosed as COVID 19 positive’? Ans: (i) Let E: the person selected is actually having COVID 19, E : the person selected is not having COVID 19, A: person’s COVID19 test is diagnosed as positive. Clearly, probability of the ‘person to be tested as COVID 19 positive’ given that ‘he is actually having COVID 19’, P (A | E) =90% =0.9. And probability of the ‘person to be tested as COVID 19 positive’ given that ‘he is not actually having COVID 19’, P(A | E )= 1% = 0.01 (ii) Refer to (i). P(E) = 0.1% = 0.001 So, P( E ) = 1 – P(E) = 1 – 0.001 = 0.999 (iii) Refer to (i). We have P(E) = 0.1% = 0.001, P( E ) = 1 – P(E) = 1 – 0.001 = 0.999 P(A|E) = 90% = 0.9 , P(A| E ) = 1% = 0.01 P ( A | E ) P( E ) By Bayes’ Theorem, P( E | A)  P( A | E ) P( E )  P( A | E ) P( E ) OR (iii) Refer to (i). We have P(E) = 0.1% = 0.001, P( E )=1 – P(E) = 1 – 0.001 = 0.999 P(A|E) = 90% = 0.9 , P(A| E ) = 1% = 0.01 We have P ( A)  P ( A | E ) P ( E )  P ( A | E ) P ( E )  P( A)  (0.9  0.001)  (0.01 0.999)  0.01089 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 15-