1.Thirty cars are assembled in a factory. The options available are a transistor, an air conditioner and power windows. It is known that 15 of the cars have transistor, 8 of them have conditioners and 6 of them have power windows. Moreover, 3 of them have all three options. Determine at least how many cars do not have any options at all. (5m) Jan 2014
Solution: Given data:
|U| = 30 Total cars |A| = 15 Transistors |B| =8 Air conditioners
|C|= 6 Power windows |A ∩ B ∩ C| = 3 All options
i) |A U B U C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C|
Here A ∩ B, B ∩ C, C ∩ A is a subset of A ∩ B ∩ C So
|A U B U C|
(A U B U C) is the set of cars that do not have any option.
|A U B U C|- |A U B U C|
≥ 30 – 23 = 7
Therefore a minimum of 7 cars has none of the options.
2.A survey on a sample of 25 new cars showed that the cars had the following
15 cars had air conditioners
12 cars had radios
11 cars had power windows
5 cars had air conditioners and power windows
9 cars had air conditioners and radios
4 cars had radios and power windows
3 cars had all the three options
Find the number of cars that had
i)only power windows
ii)at least one option (7m) Jan 2014
Solution: Given data:
|U| = 25 Total cars
|A| = 15 Air conditioners |R| = 12 Radios
|W|= 11 Power windows
|A ∩ R| = 9 Air conditioners and Radios
|A ∩ W| = 5 Air conditioners and Power windows
|R ∩ W| = 4 Radios and Power windows
|A ∩ R ∩ W| = 3 All options i) only power windows:
|W| = |W – A – R |
= |W| - |A ∩ W| - |R ∩ W| + |A ∩ R ∩ W|
= 11 – 5 – 4 + 3
= 5
5 CARS HAVE ONLY POWER WINDOWS ii) At least one option:
|W U A U R| = |W| + |A| + |R| - |A ∩ R| - |A ∩ W| - |R ∩ W|
+ |A ∩ R ∩ W|
= 11 + 15 + 12 – 5 – 9 – 4 + 3
= 23
3.A survey of 500 television viewers of sports channel produced the following information: 285 watch cricket, 195 watch hockey, 115 watch foot ball, 45 watch cricket and foot ball, 70 watch cricket and hockey, 50 watch hockey and foot ball and 50 do not watch any of the three kinds of games
i)How many viewers in survey watch all three kinds of games?
ii)How many viewers watch exactly one sport? (8m) Jan 2014
Solution: At least one:
|CU H U F| = |U| - |C U F U H|1
=500 – 50 = 450 do watch any of the games
|C U H U F| = |C| + |F| +|F| -|C∩F|-|C∩H|-|H∩F|+|C∩F∩H|
But |C∩F∩H| means those viewers who watch all the 3 games.
So, |C∩F∩H|= |C U H U F| -|C| - |H| -|F| +|C∩F|+|C∩H|+|H∩F|
= 450 – 285 – 195 – 115 +45 +70 +50
= 20
Therefore 20 viewers watch all the 3 games.
2)Only cricket viewers:
|C1| = |C – H - F|
= |C|-|C∩F|-|C∩H|+|C∩F∩H|
= 285 -70 -45 +20
=190
3)only hockey viewers:
|H1| = | H - C - F|
= |H|-|H∩F|-|C∩H|+|C∩F∩H|
= 195 -50 -70 +20
= 95
4)only football viewers:
|F1| = | F- C - H|
= |F|-|H∩F|-|C∩F|+|C∩F∩H|
=115 – 45 -50 +20
= 40
Therefore number of viewers those who watch exactly 1 game is
|C1| + |H1| + |F1|
= 190 + 95 + 40
= 325.
4.The freshman class of a private engineering college has 300 students. It is known that 180 can program in PASCAL, 120 in FORTRAN, 30 in c++, 12 in PASCAL and c++, 18 in FORTRAN and c++, 12 in PASCAL and FORTRAN, and 6 in all three languages If two students are selected at random, what is the probability that they can
i)Both program in PASCAL?
ii)Both program only in PASCAL? (6m) Jan 2014
300
5.In a survey of 120 passengers, an airline found that 48 enjoyed wine with their meals, 78 enjoyed mixed drinks, 66 enjoyed iced tea. In addition, 36 enjoyed any given pair of these beverages and 24 enjoyed them all. If two passengers are selected at random from thee survey sample of 120, what is the probability that they both want only iced tea with their meals? (7m) Jan 2014 Solution: from the information provided, we construct the Venn diagram .The sample space constants of the pairs of passengers we can select from the sample of 120.
120
So |S| = c =7140
2
A)Only iced tea | T – W – C |
= |T| -|T∩W| - |T∩C| - |T∩W∩C|
= 66 – 36 – 36 + 24 = 18
The Venn diagram indicates that there are 18 passengers who drink only ice tea.
So |A| 153
Therefore p (A) = |A| / |S|
= 153 / 7140 = 51 / 2380.
B)Exactly 2 of the 3 type
Only T and C = |T∩C| - |T∩W∩C| ………………….i
36 – 24 = 12
Only C and W = |W∩C| - |T∩W∩C| ………………..ii
36 – 24 = 12
Only W and T =|W∩T| - |T∩W∩C| ………………iii
36 – 24 = 12
Adding i, ii, iii we get = 12 + 12 + 12 = 36 = |B|