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ISI MMA PAPER SOLVED

25 SOLVED PROBLEMS FROM MMA PAPER(ISI) ON COMBINATORICS & BASIC PROBABILITY

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,
n(ABC)=[1000/105]=9
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
=4/20=1/5.

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
P(E)=n(E)/n(S)=1/40.

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)
=31.5
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
is

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.