Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian

Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.



In MMA paper you will get 30 questions among which 5-8 will be from Combinatorics & Probability.So, I am giving you some basic problems with solutions from MMA paper.

Q1. Let A be a 2*2 matrix to be written down using the numbers 1,-1 as
elements. The probability that the matrix is non-singular is
A. 1/2     B.3/8    C.5/8     D.none.

Ans: (A) A 2*2 matrix has 4 elements each of which can be chosen in 2
ways. So, total number of 2*2 matrices with elements 1,-1 is 2^4=16. Among
which the number of non-singular matrices is=16 - 8 = 8.
Required probability is =8/16=1/2.

Q2.Let (1-3p)/2,(1+4p)/3,(1+p)/6 are the probabilities of 3 mutually
exclusive & exhaustive events. Then the set of all values of p is
A.[-1/4,1/3]     B.(0,1)     C.(-1,1)       D.none

Ans :(A) since the givens are probabilities so the given 3 terms are
greater than or equal to 0. since they are exhaustive so we have
(1-3p)/2 + (1+4p)/3 + (1+p)/6 =1 implying -1/4<=p<=1/3.

Q3. The number of ways to give 16 different things to 3 persons according
as A<B<C so that B gets 1 more than A and C gets 2 more than B is
A. 4!5!6!     B. 4!5!6!/16!     C. 16!/(4!5!6!)    D.None

Ans : (C) here x+y+z=16, x=y+1, y=z+2
Solving we get x=4, y=5, z=7
The required no. of ways = 16C4*12C5*7C7 = 16!/(4!5!6!)

Q4. The number of different seven-digit numbers can be written using only
three digits 1,2,3 under the condition that the digit 2 occurs twice in
each number is
A. 512     B. 31     C. 32       D. none

Ans: (C) we have to put 2 twice in each numbers. So, any 2 out of the 7
places can be chosen in 7C2 ways. The remaining 5 places can be filled with
the other two numbers in 2^5 ways. The required no. of numbers are =
7C2*2^5 =672.

Q5. The balls are distributed to 3 boxes at random. Find the number of
ways in which we set at most 1 box empty?

Ans : zero box empty+one box empty = 3 balls in 3 boxes+ {3C1 * 3 balls
in 2 boxes} = 3! + 3*3P2 = 24.

Q6. A pair of fair dice is rolled together till a sum of either 5 or 7 is
obtained. The probability that 5 comes before 7 is
A. 1/3      B. 2/3     C.2/5      D.none

Ans : (C) Define 3 events:
A : The event that a sum of 5 occurs
B: The event that the sum of 7 occurs
C: Neither a sum of 5 nor a sum of 7 occurs
Thus P(A occurs before B) = P(A)+P(C)P(A)+P(P(C)P(C)P(A)+ . . . . . =P(A)/(1-P(C)) =2/5.
where, P(A)= 4/36,P(B)=6/36,P(C)=26/36.

Q7. There are four machines & it is known that exactly two are faulty when
they tested one by one in a random order till both the faulty machines are
identified then the probability of that only two tests are needed is
A. 1/4       B.1/5      C.1/6      D.None

Ans: (C) The probability that only two tests are needed = (Probability
that the first machine is tested is faulty)*(Probability that the second
machine tested is faulty) =2/4 * 1/3 =1/6.

Q8. The no of non-empty subset of a set containing 6 elements are
A. 63      B.65       C.64        D.None

Ans. (A) 2^6 -1=63.

Q9. The no. of squares that can be formed in a chessboard is
A. 204       B.224      C.230       D.None

Ans : (A)
A chessboard has 9 equi-spaced horizontal & vertical lines to make a 1*1
square which is done in 8*8=8^2 ways. Similarly 2*2 squares can be done in
7^2 ways.
Total no. of ways= 8^2 + 7^2 + . . . . + 2^2 +1^1 =204.

Q10. The no.of non-negative integer solutions of x+y+z=10 is
A. 10C2       B.10C3        C.12C2       D.none

Ans : (C) n=10,r=3 required answer is =(n+r-1)C(r-1)=12C2.

Q11. Let S be the set of positive integers not exceeding 1000 then find
number of sets of such integers which are not divisible by 3,5,7 ?

Ans : A,B,C denote integers divisible by 3,5,7 respectively.
Required answer is =(AUBUC)'= 1000-(AUBUC)
n(A)=[1000/3]=333,     n(B)=[1000/5]=200,       n(C)=[1000/7]=142
n(AB)=[1000/15]=66,      n(BC)=[1000/35]=28,       n(AC)=[1000/21]=47,
Now Use inclusion-exclusion principle.

Q12.Let A={2,3,4, . . . . ,20}. A number is chosen at random from A &
found that it is a prime. What is the probability that it is more than 10 ?

Ans : No of primes between 10 to 20 is=4, So,the required probability is

Q13. A point is chosen at random from the intersection of a circle. What
is the probability that the point is closed to the centre than the boundary
of the circle?

Ans : n(S)= the area of the circle with radius r
n(E)= the area of the circle of radius r/2.
P(E)=n(E)/n(S) ={ pie(r/2)^2 }/{pie(r)^2}=1/4.

Q14. A straight line of unit length,say (0,1) is divided into 3 intervals
by chosing 2 points at random. What is the probability that the three line
segment form a triangle?

Ans : The answer is =1/4. Use geometric probability.

Q15.Three numbers are chosen at random without replacement from the set
A={ x | 1x10 : x€N } . The probability that the minimum of the chosen
number is 3 & the maximum is 7 is
A. 1/3     B. 1/40       C.1/10      D.none

Ans : (D) n(S)=3 , we have to choose from 4,5,6. n(E)= 3C1

Q16. If S={1,2,3,4,5} . Then the total number of possible functions of S,
where f(f(s)) does not equals to s is
A. 32        B.40        C.44         D.none

Ans:(C) Total no. of required functions = Number of dearrangement of 5
objects=5!(1/2! - 1/3! + 1/4! - 1/5!)=44.

Q17. A bag contains unlimited contains unlimited number of white,red,black
& blue balls .The number of ways of selecting 10 balls so that there is at
least one ball of each colour is
A. 180        B.270        C.192        D.none

Ans : (D) No of ways = coefficient of x^10 in x^4 (1-x)^-4 = 84.

Q18. The number of ways of selecting r balls with replacement out of n
balls numbered 1,2,3, . . . . ,100 such that the largest numbered selected
is 10 is
A. 3        B. 4      C. 5        D. none

Ans : (A) From the given condition we can write 10^r - 9^r = 271
Applying trial & error method,putting r=1,2,3, . . . . We see that r=3
satisfies the given conditions.

Q19. n men & n woman sit along a line alternatively in x ways & along a
circle in y ways such that x=10y , then the number of ways in which n men
can sit at a round table so that all shall not have same neighbours is
A. 6       B. 12     C. 36       D. None

Ans :(B) x/y = 2*n!*n! / [(n-1)!*n!] =2n
We have x=10y, so we get n=5.
The required number is = 1/2 * 4! =12.

Q20. Let 1 to 20 are placed in anywhere around a circle. Then the sum of
some 3 consecutive numbers must be at least
A. 30        B.31        C.32        D.None

Ans : (C) Suppose x1,x2, . . . . ,x20 be the numbers placed around the
circle. Now the mean of the 20 sums of 3 consecutive numbers
 (x1+x2+x3), . . . ,(x20+x1+x2) is = 1/20*{3(x1+x2+x3+ . . . . +x20)} =(3*20*21)/(2*20)
Thus from pigeon hole principle that at least one of the sums must be
greater than equals to 32.

Q21. A mapping is selected at random from the set of all the mappings of
the set A={1,2, . . . ,n} into itself. What is the probability that the mapping is an injection?

Ans : Total no of mapping from A to A=n^n
Out of the functions n! are injections.
Hence the required probability is = (n-1)/{n^(n-1)} .

Q22. The squares are chosen at random from a chessboard. Probability that
the selected squares are in a diagonal line is
A. 7/144       B.7/744     C.1/744       D.None

Ans : (B) Total no of ways = 64C3 = 41664
Favourable cases are =2(4C3+4C3+5C3+6C3+7C3+8C3+7C3+6C3+5C3+4C3+3C3)=392.
Required probability is = 392/41664 =7/744

Q23. A & B toss a fair coin each simultaneously 50 times.What is the
probability that both of them will not get tail in the same toss ?

Ans : {HH,HT,TH,TT}
Required probability is (3/4)^50.

Q24. The probabilities of solving a problem by three students A,B,C are
1/2,1/3,1/4 respectively. The probability that the problem will be solved

A.1/2       B.1/4     C.3/4       D.none

Ans : (C) 1- P(A')P(B')P(C') =3/4.

Q25. Dialling a telephone number a boy forgets the last two digits
remembering only that these are different dialled at random. The
probability that the number is dialled correctly is
A. 1/45          B. 1/90       C. 1/100        D.none

Ans : (A) There are 10 digits,now last two digits can be dialled in 10P2
ways out of which one is favourable. Hence the required probability = 1/90.