in a class of 60 students 23 play hockey, 15 play basket-ball and 20 p

1.	in a class of 60 students 23 play hockey, 15 play basket-ball and 20 play cricket.7 play hockey and basketball,5play cricket and basketball,4 play hockey and cricket,15 do not play any of these games. find how many play.1. all three games2. hockey and basketball but not cricket3. hockey and cricket but not basketbal4. exactly one game.5. atleast 2 games6. almost 2 games
Answer» Let B, C and H be the sets of students that plays Basketball, Cricket and Hockey n(H) = 23, n(B) = 15, n(C) = 20 n(H intersection B) = 7, n(C intersection B) = 5, n(H intersection C) = 4 n(H union B union C) = 60 – 15 = 45 1. Hence, students that plays all games, n(H intersection B intersection C) = n(H union B union C) – n(H) – n(B) – n(C) + n(H intersection B) + n(B intersection C) + n(C intersection H) = 45 – 23 – 15 – 20 + 7 + 5 + 4 = 3 2. Who plays hockey not cricket n(H) – n(H intersection C) = 23 – 4 = 19 3. Plays cricket and hockey not basketball n(H intersection C) – n(H intersection C intersection B) 4 -3 = 1

in a class of 60 students 23 play hockey, 15 play basket-ball and 20 play cricket.7 play hockey and basketball,5play cricket and basketball,4 play hockey and cricket,15 do not play any of these games. find how many play.1. all three games2. hockey and basketball but not cricket3. hockey and cricket but not basketbal4. exactly one game.5. atleast 2 games6. almost 2 games

Answer»

Let B, C and H be the sets of students that plays Basketball, Cricket and Hockey

n(H) = 23, n(B) = 15, n(C) = 20

n(H intersection B) = 7, n(C intersection B) = 5, n(H intersection C) = 4

n(H union B union C) = 60 – 15 = 45

1. Hence, students that plays all games,

n(H intersection B intersection C) = n(H union B union C) – n(H) – n(B) – n(C) + n(H intersection B) + n(B intersection C) + n(C intersection H)

= 45 – 23 – 15 – 20 + 7 + 5 + 4 = 3

2. Who plays hockey not cricket

n(H) – n(H intersection C) = 23 – 4 = 19

3. Plays cricket and hockey not basketball

n(H intersection C) – n(H intersection C intersection B)

4 -3 = 1