TCS Sample Simple Probability
Questions
Dear Reader, Below are three simple questions based on probability. First two
are very simple while third employs a formula to arrive at the answer.
Question 1
12 leading companies were brought into an issue in deciding about the
allocation of land related properties by a Government agency. The companies were
numbered from 2 to 12. Among these the companies numbered 7,9 and 10 were three
leading business rivals. It was decided to use two dice and bet on the numbers
emerging out of such throws. It was seen that results range from 2 to 12. Out of
such throws which of the following: 7 or 9 or 10 is likely to appear more often
than the other numbers? (i.e will any of the companies will have advantage over
the others with the method followed.)
a) 10 b) 9 c) 7 d) all are equally possible.
Answer : c) 7
Solution :
Let (x,y) represent any throw of a dice where x represents number of I dice
and y represents number on II diceWhen two dice are thrown possible
occurences of 7 include : (1,6), (6,1), (2,5) (5,2) (3,4), (4,3)When two
dice are thrown possible occurences of 9 include : (3,6),(6,3) (4,5)
(5,4)When two dice are thrown possible occurences of 10 include : (5,5) ,
(5,5)Sum 7 can appear in at least 6 combinations as against 9 which occurs
in 4 combinations and 10 which occurs in 2 combinations. Hence company 7 should
have more chances of winning the bid over the others.
Question 2
Three school friends studying in a Higher Secondary School studying in
standard 10, 11 and 12 repectively met in a leading five star hotel in a tea
party. During discussions they had an argument about the chances of getting
specific numbers by the throw of 2 dice. Which two numbers from 9,10 and 11 will
have equal chances of appearing.
a) 9,10 b) 10,11 c) 9,11 d) none of the above
Answer : b) 10,11
Solution :
The solution to this question is very similar to that of the first
question.When two dice are thrown possible occurences of 9 include :
(3,6),(6,3), (4,5) (5,4)When two dice are thrown possible occurences of 10
include : (5,5), (5,5)When two dice are thrown possible occurences of 11
include : (5,6), (6,5)Therefore 9 can appear in 4 possible combinations, 10
can appear in 2 possible combinations and 11 can appear in 2 possible
combinations. Therefore numbers 10 and 11 have equal chances of appearing during
the throw of two dice.
Question 3
Mrs. Kalyani Chatterjee of Bhubaneshwar went abroad and settled in USA about
twenty years ago. She returned on an holiday trip to Bhubaneshwar and went to
Konarak Sun Temple for sightseeing. To her surprise she met ten of her class
mates at Konarak . What is the probability that at least one of the ten people
she met has the same birth day as that of Mrs.Kalyani Chatterjee. Her birth day
– 14th March.
a) 0.0028 b) 0.014 c) 0.128 d) none of these.
Answer : a) 0.0028.
Solution :
Probability of a friend having birthday on same date as that of Kalyani =
1/365
Probability of a friend not having birthday on same date as that of Kalyani =
1 - 1/365 = 364/365
Probability of 2 friends not having birthday on same date as that of Kalyani
= (364/365).(364/365) = (364/365)2
Similarly Probability of n friends not having birthday on same date as that
of Kalyani = (364/365)n
Now, Probability of n friends having birthday on same date as that of Kalyani
= 1 - Probability of n friends not having birthday on same date as that of
Kalyani= 1 - (364/365)n
Substituting n = 10 in above formula we get,
1 - (364/365)10 = 0.0028
Questions
Dear Reader, Below are three simple questions based on probability. First two
are very simple while third employs a formula to arrive at the answer.
Question 1
12 leading companies were brought into an issue in deciding about the
allocation of land related properties by a Government agency. The companies were
numbered from 2 to 12. Among these the companies numbered 7,9 and 10 were three
leading business rivals. It was decided to use two dice and bet on the numbers
emerging out of such throws. It was seen that results range from 2 to 12. Out of
such throws which of the following: 7 or 9 or 10 is likely to appear more often
than the other numbers? (i.e will any of the companies will have advantage over
the others with the method followed.)
a) 10 b) 9 c) 7 d) all are equally possible.
Answer : c) 7
Solution :
Let (x,y) represent any throw of a dice where x represents number of I dice
and y represents number on II diceWhen two dice are thrown possible
occurences of 7 include : (1,6), (6,1), (2,5) (5,2) (3,4), (4,3)When two
dice are thrown possible occurences of 9 include : (3,6),(6,3) (4,5)
(5,4)When two dice are thrown possible occurences of 10 include : (5,5) ,
(5,5)Sum 7 can appear in at least 6 combinations as against 9 which occurs
in 4 combinations and 10 which occurs in 2 combinations. Hence company 7 should
have more chances of winning the bid over the others.
Question 2
Three school friends studying in a Higher Secondary School studying in
standard 10, 11 and 12 repectively met in a leading five star hotel in a tea
party. During discussions they had an argument about the chances of getting
specific numbers by the throw of 2 dice. Which two numbers from 9,10 and 11 will
have equal chances of appearing.
a) 9,10 b) 10,11 c) 9,11 d) none of the above
Answer : b) 10,11
Solution :
The solution to this question is very similar to that of the first
question.When two dice are thrown possible occurences of 9 include :
(3,6),(6,3), (4,5) (5,4)When two dice are thrown possible occurences of 10
include : (5,5), (5,5)When two dice are thrown possible occurences of 11
include : (5,6), (6,5)Therefore 9 can appear in 4 possible combinations, 10
can appear in 2 possible combinations and 11 can appear in 2 possible
combinations. Therefore numbers 10 and 11 have equal chances of appearing during
the throw of two dice.
Question 3
Mrs. Kalyani Chatterjee of Bhubaneshwar went abroad and settled in USA about
twenty years ago. She returned on an holiday trip to Bhubaneshwar and went to
Konarak Sun Temple for sightseeing. To her surprise she met ten of her class
mates at Konarak . What is the probability that at least one of the ten people
she met has the same birth day as that of Mrs.Kalyani Chatterjee. Her birth day
– 14th March.
a) 0.0028 b) 0.014 c) 0.128 d) none of these.
Answer : a) 0.0028.
Solution :
Probability of a friend having birthday on same date as that of Kalyani =
1/365
Probability of a friend not having birthday on same date as that of Kalyani =
1 - 1/365 = 364/365
Probability of 2 friends not having birthday on same date as that of Kalyani
= (364/365).(364/365) = (364/365)2
Similarly Probability of n friends not having birthday on same date as that
of Kalyani = (364/365)n
Now, Probability of n friends having birthday on same date as that of Kalyani
= 1 - Probability of n friends not having birthday on same date as that of
Kalyani= 1 - (364/365)n
Substituting n = 10 in above formula we get,
1 - (364/365)10 = 0.0028