5 mark questions X-STD MATHEMATICS

Prove That A – (B  C) = (A – B)  (A – C) A A A A A BBB B B C CC C C (1)B  C (2) A-(B  C) (3) A – B (4) A – C (5) (A – B)  (A – C) From the diagrams (2) and (5) A – (B  C) = (A – B)  (A – C)

Prove that A – (B  C) = (A – B)  (A – C) using Venn diagram A B C A A A A BB BB C C C C (1) B  C (2) A – ( B  C) (3) A – B (4) A – C (5) (A – B)  (A – C) From the diagram (2)and(5) A – (B  C) = (A – B)  (A – C)

Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. A B C A A A A B B B B CC C C B  C A  (B  C)A  B A  C (A  B)  (A  C) From the figures 2 and 5 A  (B  C)=(A  B)  (A  C)

Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. A B C A A A A B B B B CC C C B  C A  (B  C)A  B A  C (A  B)  (A  C) From the figures 2 and 5 A  (B  C)=(A  B)  (A  C)

Prove that (A  B)’ = A’  B’ using Venn diagram. ABB A A A AB B B UU U UU 1. A  B 2. (A  B)’ 3. A’ 4. B’ 5. A’  B’ From the diagrams 2 and 5 (A  B)’ = A’  B’

Prove that (A  B)’ = A’  B’ using Venn diagram. ABB A A A AB B B UU U UU 1. A  B 2. (A  B)’ 3. A’ 4. B’ 5. A’  B’ From the diagrams 2 and 5 (A  B)’ = A’  B’

Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that (i) A  (B  C) = (A  B)  C. (ii) Verify (i) using Venn diagram. Solution (i) B  C = {3, 4, 5, 6}  {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8} ` A  (B  C) = {1, 2, 3, 4, 5}  { 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}………… (1) A  B = {1, 2, 3, 4, 5}  {3, 4, 5, 6} = {1,2,3,4,5,6} `(A  B)  C = {1,2,3,4,5,6}  {5,6,7,8} = {1, 2, 3, 4, 5, 6, 7, 8}……………. (2) From (1) and (2), we have A  (B  C) = (A  B)  C.

Using Venn diagram, we have A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8} A B C A B C (1) B  C (2) A  (B  C)

Let A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. (i) Show that A  (B  C) = (A  B)  C. (ii) Verify (i) using Venn diagram. Solution Given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. B  C = {a,c,e}  {a,e} = {a,e} A  ( B  C) = {a,b,c,d}  {a,e} = {a}…… (1) A  B = {a,b,c,d.}  {a,c,e} = {a,c}. (A  B)  C = {a,c}  {a,e} = {a}…….. (2) From (1) and (2) A  (B  C) = (A  B)  C.

A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. A B C a e (1) B  C A B C a (2) A  (B  C)

A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. A B C a c (1) A  B A B C a (2) (A  B)  C from (2) and (4), it is verified that A  (B  C) =(A  B)  C

Let A = {0,1,2,3,4}, B = {1, - 2, 3,4,5,6} and C = {2,4,6,7}. (i) Show that A  (B  C) = (A  B)  (A  C). (ii) Verify using Venn diagram. Solution (i) B  C = {1, - 2, 3, 4, 5, 6}  {2, 4, 6, 7 } = {4, 6}; A  (B  C) = {0,1, 2, 3, 4}  {4, 6} = {0,1,2,3,4,6}…….(1) A  B = {0,1,2,3,4}  {1, - 2, 3,4,5,6} = {- 2, 0,1, 2, 3, 4, 5, 6}, A  C = {0,1,2,3,4}  {2,4,6,7} = {0,1, 2, 3, 4, 6, 7}. (A  B)  (A  C) = {- 2, 0,1, 2, 3, 4, 5, 6}  {0,1, 2, 3, 4, 6, 7} = {0,1, 2, 3, 4, 6}. …….(2) From (1) and (2),we get A  (B  C) = (A  B)  (A  C).

A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}. A B C 4 6 (1) B  C A B C 4 (2) A  (B  C)

A B C 4 3 (3) A  B A B C 0 (4) A  C from (2) and (4), it is verified that A  (B  C) =(A  B)  C A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}

A B C 4 (5) (A  B)  (A  C) from (2) and (5), it is verified that A  (B  C) = (A  B)  (A  C)

Given that U = {a,b,c,d,e, f,g,h}, A = {a, b, f, g}, and B = {a, b, c}, verify De Morgan’s laws of complementation. U = {a, b, c, d, e, f, g, h} A = {a, b, f, g} B = {a, b, c}. De Morgan’s laws (A  B)’ = A’  B’ (A  B)’ = A’  B’ VERIFICATION A  B = {a, b, c, f, g} (A  B)’ = {d, e, h}……..(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’  B’ = {d, e, h}……….(2) From (1) and (2) (A  B)’ = A’  B’ A  B = {a, b} (A  B)’ = {c, d, e, f, g, h}....(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’  B’ = {c, d, e, f, g, h}….(2) From (1) and (2) (A  B)’ = A’  B’

Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. De Morgan’s laws (1) A\(B  C) = (A\B)  (A\C)(2) A\(B  C) = A\B)  (A\C) VERIFICATION B  C = {1, 2, 3, 5, 7, 9, 10, 12, 13} A\(B  C) = {11, 15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)  (A\C) = {11, 15}……….(2) From (1) and (2) A\(B  C) = (A\B)  (A\C)

Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. B  C = { } A\(B  C) = {1, 3, 5, 7, 9,11,13,15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)  (A\C) = {1, 3, 5, 7, 9,11,13,15}……….(2) From (1) and (2) A\(B  C) = (A\B)  (A\C)

Let A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5,10,15, 20, 30} and C = {7, 8,15,20,35,45, 48}. Verify A\(B  C) = (A\B)  (A\C) A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5, 10, 15, 20, 30} C = {7, 8, 15, 20, 35, 45, 48} B  C = {15, 20} A\(B  C) = {10, 25, 30, 35, 40, 45, 50}……..(1) A\B = {25, 30, 35, 40, 45, 50} A\C = {10, 25, 30, 40, 50} (A\B)  (A\C) = {10, 25, 30, 35, 40, 45, 50}……….(2) From (1) and (2) A\(B  C) = (A\B)  (A\C)

13.Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A " B be defined by f(x) =. Represent f by (i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph AB ARROW DIAGRAM SET OF ORDERED PAIR f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}

X Y12456 Table Graph | | | | | | | | | – 7 – 6 – 5 – 4 – 3 – 2 – 1  (6, 1)  (9, 2)  (15, 4)  (18, 5)  (21, 6)

Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let f : A  B be a function given by f (x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph. Given A= {0, 1, 2, 3}, B = {1, 3, 5, 7, 9} f(x) = 2x + 1 f(0) = 2(0) + 1 = = 1 f(1) = 2(1) + 1 = = 3 f(2) = 2(2) + 1 = = 5 f(3) = 2(3) + 1 = = 7 (i)Set of ordered pairs {(0, 1), (1, 3), (2, 5), (3, 7)} (ii) Table x f(x)

A BARROW DIAGRAM Graph | | | | | | | | – 7 – 6 – 5 – 4 – 3 – 2 – 1  (4, 3)  (6, 4)  (8, 5)  (10, 6)

A function f: [1, 6)  R is defined as follows f(x) = Find the value of (i) f(5), (ii) f(3), (iii) f(1), (iv) f(2) – f(4) (v) 2f(5) – 3f(1) Since 5 lies between 4 and 6 f(x) = 3x 2 – 10 f(5) = 3(5) 2 – 10 = 75 – 10 = 65 Since 3 lies between 2 and 4 f(x) = 2x – 1 = 2(3) – 1 = 6 – 1 = 5 Since 1 lies in the interval 1  x < 2 f(x) = 1 + x f(1) = = 2 Since 2 lies in the interval 2  x < 4 f(x) = 2x – 1 = 2(2) – 1 = 4 – 1 = 3 Since 4 lies in the interval 4  x < 6 f(x) = 3x 2 – 10 f(4) = 3(4) 2 – 10 = 48 – 10 = 38

f(2) – f(4) = 3 – 38 = – 34 2f(5) – 3 f(1) = 2 (65) – 3(2) = 130 – 6 = 124

A function f: [-3, 7)  R is defined as follows Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (i) Since 5 lies between 4 and 6 f(x) = 3x – 2 f(5) = 3(5) – 2 = 15 – 2 = 13 Since 6 lies in the interval 4 < x  6 f(x) = 2x – 3 f(6) = 2(6) – 3 = 12 – 3 = 9 f(5) + f(6) = = 22

A function f: [-3, 7)  R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (ii) Since 1 lies in the interval -3  x < 2 f(x) = 4x 2 – 1 f(1) = 4(1) 2 – 1 = 4 – 1 = 3 Since –3 lies in the interval-3  x < 2 f(x) = 4x 2 – 1 = 4(-3) 2 – 1 = 36 – 1 = 35 f(1) – f(-3) = 3 – 35 = – 32

A function f: [-3, 7)  R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iii) Since -2 lies in the interval -3  x < 2 f(x) = 4x 2 – 1 f(1) = 4(-2) 2 – 1 = 16 – 1 = 15 Since 4 lies in the interval 2  x  4 f(x) = 3x – 2 = 3(4) – 2 = 12 – 2 = 10 f(-2) – f(4) = 15 – 10 = 5

A function f: [-3, 7)  R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iv) Since 3 lies in the interval 2  x  4 f(x) = 3x – 2 f(3) = 3(3) – 2 = 9 – 2 = 7 Since -1 lies in the interval -3  x < 2 f(x) = 4x 2 – 1 = 4(-1) 2 – 1 = 4 – 1 = 3 f(6) = 9,f(1) = 3

A function f: [-7, 6)  R is defined as follows Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (i) Since -4 lies in the interval -5  x  2 f(x) = x + 5 f(-4) = – = 1 Since 2 lies in the interval -5  x  2 f(x) = x + 5 f(2) = = 7 2f(-4) + 3f(2) = 2(1) + 3(7) = = 23

A function f: [-7, 6)  R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (ii) Since -7 lies in the interval -7  x < -5 f(x) = x 2 + 2x + 1 f(-7) = (–7) 2 + 2(–7) + 1 = 49 – = 36 Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(2) = = 2 f(-7) – f(–3) = 36 – 2 = 34

A function f: [-7, 6)  R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(-3) = = 2 Since -6 lies in the interval -7  x < -5 f(x) = x 2 + 2x + 1 f(-6) = (–6) 2 + 2(–6) + 1 = 36 – = 25 4 lies in the interval 2 < x < 6 f(x) = x – 1 f(4) = 4 – 1 = 3 Since 1 lies in the interval -5  x  2 f(x) = x + 5 f(1) = = 6

A function f: [-7, 6)  R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) f(-3) = 2 f(-6) = 25 f(4) = 3 f(1) = 6

If verify that (AB) T = B T A T. LHS = RHS  (AB) T = B T A T.

If verify that (AB) T = B T A T.

LHS = RHS  (AB) T = B T A T.

If then show that A 2 – 4A + 5I 2 = O A 2 – 4A + 5I 2 = O

7. Find X and Y if 2X + 3Y =

11. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets. TVDVDVIDEOGAMESCD players WEEK I Store I Store II Store III WEEK II Store I Store II Store III Find the sum of the items sold out in two weeks using matrix addition.

If verify (AB)C = A(BC). LHS = RHS  (AB)C =A(BC)

Ifthen prove that (A + B) 2  A 2 + 2AB + B 2.

LHS  RHS (A + B) 2  A 2 + 2AB + B 2.

If find (A + B)C and AC + BC. Is (A + B)C = AC + BC ?

If find (A + B)C and AC + BC. Is (A + B)C = AC + BC ?  (A+B)C = AC + BC

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term. Given t 10 = 41 t 18 = 73 t n = a + (n – 1)d t 10 = 41 a + (10 – 1)d = 41 a + 9d = 41…………(1) t 18 = a + (18 – 1)d = 73 a + 17d = 73 …….(2) (2)  a + 17d = 73 (1)  a + 9d = 41 (2) – (1)  8d = 32 d = 32/8 d = 4 Sub d = 4 in (1) a + 9(4) = 41 a + 36 = 41 a = 41 – 36 a = 5 t 27 = a + (27 – 1)d = 5 + (26)4 = = 109 (-)

The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers. Let a – d, a, a + d be three consecutive terms in an A.P. Sum = 6 a – d + a + a + d = 6 3a = 6 a = 6/3 = 2 Product = – 120 (a – d)  a  (a + d) = – 120 (2 – d)  2  (2 + d) = – 120 (2 2 – d 2 )  2 = – – d 2 = – 120/2 = – 60 – d 2 = – 60 – 4 – d 2 = – 64 d 2 = 64 d =  8 The given numbers are 2 – 8, 2, – 6, 2, 10 (or) 10, 2, –6

A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year. Given t 7 = 1000 t 10 = 1450 tn = a + (n – 1)d t 7 = 1000 a + (7 – 1)d = 1000 a + 6d = 1000…………(1) t 10 = a + (10 – 1)d = 1450 a + 9d = 1450 …….(2) (2)  a + 9d = 1450 (1)  a + 6d = 1000 (2) – (1)  3d = 450 d = 450/3 d = 150 Sub d = 150 in (1) a + 6(150) = 1000 a = 1000 a = 1000 – 900 a = 100 t 15 = a + (15 – 1)d = (15)150 = = 2350

Find the sum of the first 40 terms of the series 1 2 – – …… – – …….. = ….. to 40 terms = (1 – 4) + (9 – 16) + (25 – 36) + …… to 20 terms. = (-3) + (-7) + (-11) +………20 terms It is an AP with a = –3, d = –4, n = = 10(-82) = – 820

Find the sum of all 3 digit natural numbers, which are divisible by 9. First number = (9 – 1) = 108 Last number = 999 The AP is ……+ 999 a = 108, l = 999, d = 9 n = 110 = 55  1107 = The three digit numbers are from 100 to 999

Find the sum of all natural numbers between 300 and 500 which are divisible by 11. First number = (11 – 3) = 308 Last number = 500 – 5 = 495 The AP is ……+ 495 a = 308, l = 495, d = 11 n = 18 = 9  803 =

A sum of Rs is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P.? If so, find the total interest at the end of 30 years. Every year deposit = Rs Rate of interest = 8% = 0.08 First year interest = 1000  0.08 = 80 Second year interest = 2000  0.08 = 160 Third year interest = 3000  0.08 = 240 The interest 80, 160, 240,…. Forms an AP. Total interest = ……30terms = 15  2480 = Rs.37200

Find the sum of all numbers between 100 and 200 which are not divisible by 5. The required sum = ( ……+ 200) – ( … + 200) ………… = ( ………….+ 200) – ( ………+ 99) = 100  201 – 99  50 = – 4950 = …… Here a = 100, d = 5, l =

Find the sum of all numbers between 100 and 200 which are not divisible by 5. = 21  150 = The required sum = – 3150 = 12000

Calculate the standard deviation of the following data. 10, 20, 15, 8, 3, , xd = x –  xd 2 10 – 10 = 0 20 – 10 = – 10 = 5 8 – 10 = -2 3 – 10 = -7 4 – 10 =

Calculate the standard deviation of the following data. 38, 40, 34,31, 28, 26, , xd = x –  xd 2 38 – 33 = 5 40 – 33 = 7 34 – 33 = 1 31 – 33 = – 33 = – 33 = – 33 =

Calculate the standard deviation of the following data xfd = x – 3 d 2 fdfd 2 3 – 3 = 0 8 – 3 = 5 13 – 3 = – 3 = – 3 = X f

Standard deviation

11. A group of 45 house owners contributed money towards green environment of their street. The amount of money collected is shown in the table below. Calculate the variance and standard deviation. 0 – – – – – 100 Time f mid x d = (x – 30)/20 d 2 fd fd 2 Amount No of house owners

Standard deviation Variance = (S.D) 2 = (19) 2 = 361

Calculate the coefficient of variation of the following data: 20, 18, 32, 24, , xd = x –  xd 2 20 – 24 = – 24 = – 24 = 8 24 – 24 = 0 26 – 24 =

The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below. Calculate the variance and standard deviation of the data. 5 – – – – – 30 Time f mid x d = x –17.5 d 2 fd fd 2 Time (sec) No of people

Variance

Show that the points A(2, 3), B(4, 0) and C(6, -3) are collinear. Area of the  ABC = ½ {(x 1 y 2 + x 2 y 3 + x 3 y 1 ) – (x 1 y 3 + x 3 y 2 + x 2 y 1 )}sq. units = ½{(2)(0) + (4)(–3) + (6)(3)} – {(2)(–3) + (6)(0) + (4)(3)} = ½ {(0 – ) – (– )} = ½ {(18 – 12) – (12 – 6)} = ½ (6 – 6) = 0 The given points are collinear.

find the value of k for which the given points are collinear. (2, - 5),(3, - 4) and (9, k) If the points are collinear, Area of the  ABC = 0 {(x 1 y 2 + x 2 y 3 + x 3 y 1 ) – (x 1 y 3 + x 3 y 2 + x 2 y 1 )} = k-5 {(2)(-4) + (3)(k) + (9)(-5)} – {(2)(k) + (9)(-4) + (3)(-5)} = 0 {(–8 + 3k – 45) – (2k – 36 – 15)} = 0 {(3k – 53) – (2k – 51)} = 0 3k – 2k – = 0 k – 2 = 0 k = 2

Find the area of the quadrilateral whose vertices are (6, 9), (7, 4), (4,2) and (3,7) Area of the quadrilateral ABCD = ½ {(x 1 y 2 + x 2 y 3 + x 3 y 4 + x 4 y 1 ) – (x 2 y 1 + x 3 y 2 + x 4 y 3 + x 1 y 4 )} = ½ {(4)(4) + (7)(9) + (6)(7) + (3)(2)} – {(2)(7) + (4)(6) + (9)(3) + (7)(4)} = ½ {( ) – ( )} A(4,2) D(3,7) B(7,4) C(6,9) = ½ {127 – 93} = ½ (34) = 17 sq. units

Find the area of the quadrilateral whose vertices are (-3, 4), (-5,- 6), (4,- 1) and (1, 2) Area of the quadrilateral ABCD = = ½ {(x 1 y 2 + x 2 y 3 + x 3 y 4 + x 4 y 1 ) – (x 2 y 1 + x 3 y 2 + x 4 y 3 + x 1 y 4 )} = ½ {(-5)(-1) + (4)(2) + (1)(4) + (-3)(-6)} – {(-5)(4) + (-3)(2) + (1)(-1) + (4)(-6)} = ½ {( ) – (-20 – 6 – 1 – 24)} A(-5,-6) D(-3,4) B(4,-1) C(1,2) = ½ {35 – (–51)} = ½ ( ) = ½ (86) = 43 sq. units

Find the area of the quadrilateral whose vertices are (-4, 5), (0, 7), (5,- 5) and (-4,- 2) Area of the quadrilateral ABCD = = ½ {(x 1 y 2 + x 2 y 3 + x 3 y 4 + x 4 y 1 ) – (x 2 y 1 + x 3 y 2 + x 4 y 3 + x 1 y 4 )} = ½ {(-4)(-5) + (5)(7) + (0)(5) + (-4)(-2)} – {(-4)(5) + (-4)(7) + (0)(-5) + (5)(-2)} = ½ {( ) – (-20 – 28 – 0 – 10)} = ½ {63 – (–58)} = ½ ( ) = ½ (121) = 60.5 sq. units A(-4,- 2) D(-4,5) B(5,-5) C(0,7)

A coin is tossed three times. Find the probability of getting (i) head and tail alternatively (ii) at least two heads (iii) exactly two heads (iv) no heads Tossing three coins once is as same as a coin is tossed three times Sample space S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} n(S) = 8 (i) Let the event of getting head and tail alternatively be A A = {HTH, THT} n(A) = 2 The probability of getting head and tail alternatively is (ii) Let the event of getting at least two heads be B B = {HHH, HHT, HTH, THH} n(B) = 4 The probability of getting at least two heads is

(iii) Let the event of getting exactly two heads be C C = {HHT, HTH, THH} n(C) = 3 The probability of getting exactly two heads is (iv) Let the event of getting no head be D D = {TTT}  n(D) = 3 The probability of getting no head is

A card is drawn from a pack of 52 cards. Find the probability that it is either red card or king card. Total number of cards = 52 n(S) = 52 Let A be the event drawing a red card n(A) = 26 The probability of drawing a red card is Let B be the event drawing a king card n(B) = 4 The probability of drawing a king card is

n(A  B) = 2 P(A  B) = P(A) + P(B) – P(A  B)

Two dice are rolled once. Find the probability of getting an even number on the second die or the total of face numbers 10. When two dice are rolled the sample space S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}  n(S) = 36 Let the event of getting even in the second die be A A = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6) (3, 2), (3, 4), (3, 6), (4, 2), (4, 4), (4, 6) (5, 2), (5, 4), (5, 6), (6, 2), (6, 4), (6, 6)}  n(A) = 18

Let the event of getting the total of the face numbers is 10 be B B = {(4, 6), (5, 5), (6, 4)}  n(B) = 3 A  B = {(4, 6), (6, 4)}  n(A  B) = 2 P(A  B) = P(A) + P(B) – P(A  B)

16.Two persons X and Y appeared in an interview for two vacancies in an office. The chance for X’s selection is 1/5 and the chance for Y’s selection is 1/7. Find the chance that (i) both of them are selected (ii) only one of them is selected, (iii) none of them is selected. The probability of X’S selection = P(X) = 1/5 The probability of Y’S selection = P(Y) = 1/7 (i) The probability of both of them to be selected (ii) The probability of only one of them to be selected

(iii) The probability of none of them to be selected