Engineering Data Analysis Exam (18 Problem-Solving Questions with 4 Additional
Questions)
1. In how many ways can you invite one or more of your five friends to a party?
Solution:
First of all, calculate the total number of ways in which the friends can be invited i.e without
imposing any condition on the number of friends that can come.
Each friend has 2 ways i.e the friend will be invited or the friend will not be invited and also
that whether the friend will be invited or not does depend on anybody else so invitation of
each friend is an independent event.
So 1 friend leads to 2 ways then, 5 friends lead to 2×2×2×2×2=2^5 ways.
Now the condition is one or more friend has to be invited so if we subtract the case where 0
friends were invited from a total number of ways we will get the desired answer.
No. of ways in which no friends are invited =1
So our answer is =2^5 −1= 31 ways
Answer: 31 ways
2. An office building has 9 doors. In how many ways can a person enter and leave by
different doors?
Solution:
By any door
no. of ways = 9*9
no. of ways = 81 ways
By different door
no. of ways = 9*8
no. of ways = 72 ways
Answer: 81 and 72 OR 72 only
3. In how many ways can the letters in the word "MATHEMATICS" can be arranged?
Solution:
M=2
A=2
T=2
"MATHEMATICS"=11 letters
11! / 2! 2! 2! = 4989600 ways
Answer: 4989600 ways
4. In a class of 40 students. 27 like statistics and 25 like probability. How many like
both statistics and probability?
Solution:
Venn Diagram
Statistics = 27-x
center of venn diagram = x
probability = 25-x
(27-x)+x+(25-x)=40
x = 12
Answer: 12
, 5. In how many ways can 10 trees be planted in a circular lot?
Solution:
There are a total of 10! (=3628800) ways to plant ten different trees.
However, the trees are in a circle, in which each "way" is counted ten times, due to rotations.
This leaves us 10!/10 = 9! = 362880 ways to plant six different trees in a circle.
Answer: 362880
6. If you roll a pair of dice one time, what is the probability of getting a sum of 9?
Solution:
The simplest approach to the question is
Sample space-{(1,1),(1,2),…..,(1,6), (2,1),(2,2),…..,(2,6),………………..,(6,5),(6,6)}
which constitutes a total of 6*6 elements i.e 36 in total
But we need the elements which produce a sum of 9 together
So our elements are (3,6),(4,5),(5,4),(6,3)
Total number of likely results = 4 nos
So our required answer is 4/36 i.e 0.1111
Answer: 11.11%
7. A die is constructed so that a 1 or a 2 occurs twice as a 3, 4 or a 6 and a 5 occurs
thrice as a 1 or a 2. If the die is tossed once. Find the probability that an even number
will occur?
Solution:
The probability of an even number is the probability of a 2, 4, 6:
6/18 +1/18 +1/18 = 8/18 = 4/9
Answer: 4/9
8. In a basketball game, the free throw average is 0.65. Find the probability that a
player misses one shot of the three free throws?
Solution:
0.65*.0.65 = 0.4225 = 0.422
Answer: 0.422
9. Find the probability of getting a prime number thrice by tossing a fair die 5 times.
Solution:
On a die, 3 numbers are prime (2, 3, 5) and 2 numbers are not prime (1, 4).
2 outcomes each roll (Prime or Not Prime), with 10 rolls, gives a total number of possible
outcomes of 2^10 = 1024.
In 5 rolls, 3 Primes can occur in 5c3 ways.
Therefore, the probability is 5c3/(2^5).
Answer: 0.3125
Questions)
1. In how many ways can you invite one or more of your five friends to a party?
Solution:
First of all, calculate the total number of ways in which the friends can be invited i.e without
imposing any condition on the number of friends that can come.
Each friend has 2 ways i.e the friend will be invited or the friend will not be invited and also
that whether the friend will be invited or not does depend on anybody else so invitation of
each friend is an independent event.
So 1 friend leads to 2 ways then, 5 friends lead to 2×2×2×2×2=2^5 ways.
Now the condition is one or more friend has to be invited so if we subtract the case where 0
friends were invited from a total number of ways we will get the desired answer.
No. of ways in which no friends are invited =1
So our answer is =2^5 −1= 31 ways
Answer: 31 ways
2. An office building has 9 doors. In how many ways can a person enter and leave by
different doors?
Solution:
By any door
no. of ways = 9*9
no. of ways = 81 ways
By different door
no. of ways = 9*8
no. of ways = 72 ways
Answer: 81 and 72 OR 72 only
3. In how many ways can the letters in the word "MATHEMATICS" can be arranged?
Solution:
M=2
A=2
T=2
"MATHEMATICS"=11 letters
11! / 2! 2! 2! = 4989600 ways
Answer: 4989600 ways
4. In a class of 40 students. 27 like statistics and 25 like probability. How many like
both statistics and probability?
Solution:
Venn Diagram
Statistics = 27-x
center of venn diagram = x
probability = 25-x
(27-x)+x+(25-x)=40
x = 12
Answer: 12
, 5. In how many ways can 10 trees be planted in a circular lot?
Solution:
There are a total of 10! (=3628800) ways to plant ten different trees.
However, the trees are in a circle, in which each "way" is counted ten times, due to rotations.
This leaves us 10!/10 = 9! = 362880 ways to plant six different trees in a circle.
Answer: 362880
6. If you roll a pair of dice one time, what is the probability of getting a sum of 9?
Solution:
The simplest approach to the question is
Sample space-{(1,1),(1,2),…..,(1,6), (2,1),(2,2),…..,(2,6),………………..,(6,5),(6,6)}
which constitutes a total of 6*6 elements i.e 36 in total
But we need the elements which produce a sum of 9 together
So our elements are (3,6),(4,5),(5,4),(6,3)
Total number of likely results = 4 nos
So our required answer is 4/36 i.e 0.1111
Answer: 11.11%
7. A die is constructed so that a 1 or a 2 occurs twice as a 3, 4 or a 6 and a 5 occurs
thrice as a 1 or a 2. If the die is tossed once. Find the probability that an even number
will occur?
Solution:
The probability of an even number is the probability of a 2, 4, 6:
6/18 +1/18 +1/18 = 8/18 = 4/9
Answer: 4/9
8. In a basketball game, the free throw average is 0.65. Find the probability that a
player misses one shot of the three free throws?
Solution:
0.65*.0.65 = 0.4225 = 0.422
Answer: 0.422
9. Find the probability of getting a prime number thrice by tossing a fair die 5 times.
Solution:
On a die, 3 numbers are prime (2, 3, 5) and 2 numbers are not prime (1, 4).
2 outcomes each roll (Prime or Not Prime), with 10 rolls, gives a total number of possible
outcomes of 2^10 = 1024.
In 5 rolls, 3 Primes can occur in 5c3 ways.
Therefore, the probability is 5c3/(2^5).
Answer: 0.3125
