#### 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 C*

*ombinatorics & 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 | 1**≤

**x**≤

**10 : 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.