Question #61624

25 students of your school participated in a tournament of three games namely : cricket, Football and
basket ball. 15 students received medals in cricket, 12 in Football, 11 in basketball, 5 in cricket and
basket ball, 9 in cricket and football, 4 in football and basketball and 3 in all the three games. How many
students received medals in (i) none of the games (ii) cricket only.
1

Expert's answer

2016-09-01T09:36:03-0400

Answer on Question #61624 – Math – Statistics and Probability

Question

25 students of your school participated in a tournament of three games namely: cricket, Football and basketball.

15 students received medals in cricket,

12 in Football,

11 in basketball,

5 in cricket and basketball,

9 in cricket and football,

4 in football and basketball

and 3 in all the three games.

How many students received medals in:

(i) none of the games;

(ii) cricket only.

Solution

Venn diagram



It is given that

S=25|S| = 25 students participated in a tournament of three games namely: cricket, football, and basketball.

C=15|C| = 15 students received medals in cricket,

B=11|B| = 11 students received medals in basketball,

F=12|F| = 12 students received medals in football,

CB=5|C \cap B| = 5 students received medals in cricket and basketball,

CF=9|C \cap F| = 9 students received medals in cricket and football,

BF=4|B \cap F| = 4 students received medals in basketball and football, CBF=3|C \cap B \cap F| = 3 students received medals in cricket, basketball, and football.

(i) Using inclusion-exclusion principle

CBF=C+B+FCBCFBF+CBF=15+11+12594+3=23|C \cup B \cup F| = |C| + |B| + |F| - |C \cap B| - |C \cap F| - |B \cap F| + |C \cap B \cap F| = 15 + 11 + 12 - 5 - 9 - 4 + 3 = 23 students received medals in either of the games or in all three games.

Then using a complement set obtain

CBF=SCBF=2523=2|\overline{C \cup B \cup F}| = |S| - |C \cup B \cup F| = 25 - 23 = 2 students received medals in none of the games.

(ii) Using the union of disjoint sets

CBˉF=CFCBF=93=6|C \cap \bar{B} \cap F| = |C \cap F| - |C \cap B \cap F| = 9 - 3 = 6 students received medals in cricket and football, but did not receive ones in basketball;

CBFˉ=CBCBF=53=2|C \cap B \cap \bar{F}| = |C \cap B| - |C \cap B \cap F| = 5 - 3 = 2 students received medals in cricket and basketball, but did not receive ones in football;

CBˉFˉ=CCBˉFCBFˉCBF=15623=1511=4|C \cap \bar{B} \cap \bar{F}| = |C| - |C \cap \bar{B} \cap F| - |C \cap B \cap \bar{F}| - |C \cap B \cap F| = 15 - 6 - 2 - 3 = 15 - 11 = 4 students received medals in cricket only.

Answer: (i) 2; (ii) 4.

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