HIDI

PROBABILITY

1. A die is thrown n times (n being odd). The probability than even face turns odd number of times, is

(A)1n2

n

(B) less than

(C) (D) greater than

1. (C)

The required probability is nC1

= .

2. The distinct numbers are chosen from the set {1, 2, …,6}. The probability that the product of two numbers is the third one is(A) 1/4 (B) 1/3(C) 1/ 2 (D) none of these

2. (D)

The required probability = .

3. A and B are two events such that P(A) = 0.2 and P(AB) = 0.7. If A and B are independent events then P(B) equals(A) 2/7 (B) 7/9(C) 3/8 (D) none of these

3. (C)0.7 = P(AB) = P(A) + P(B) – P(AB) = P(A) + P(B) – P(A). P(B) = 0.2 + P(B) – 0.2P(B)

P(B) =

P(B) = 3/8.

4. A and B are two independent events. The probability that A and B happen simultaneously is and neither A

nor B happens is , the

(A) P(A) = , P(B) = (B) P(A) = , P(B) =

(C) P(A) = , P(B) = (D) P(A) = , P(B) =

4. (A), (C)P(A B) = P(A) . P(B) = 1/12 .... (1)

= (1 - P(A)) (1 - P(B) = 1/2 1- P(A) - P(B) + P(A) P(B) = 1/2 1- P(A) - P(B) + 1/12 = 1/2 P(A) + P(B) = 7/12 .... (2)Solving (1) and (2) gives P(A) = 1/3 or 1/4Accordingly, P(B) = 1/ 4 or 1/3.

5. Entries of a 2 2 determinant are chosen from the set {−1, 1}. The probability that determinant has zero value is (A) 1/4 (B) 1/3(C) 1/2 (D) none of these

1

5. (C)

= ad bc = 0 whether ad = 1, bc = 1 or ad = 1, bc = 1

which occur in eight ways. Total number of 2 2 determinants from {1, 1} is 16. Thus required probability is

.

6. A die is thrown a fixed number of times. If probability of getting even number 3 times is same as the probability of getting even number 4 times, then probability of getting even number exactly once is

(A) (B)

(C) (D)

6. (D)According to the given condition

, where n is the number of times die is thrown

nC3 = nC4 n = 7.

Thus required probability = .

7. Let A, B, C be three events such that A and B are independent and P(C) = 0, then events A, B, C are (A) independent (B) pairwise independent but not totally independent (C) P(A) = P(B) = P(C)(D) none of these

7. (A)P(C) = 0 C = AC = BC = ABC = P(ABC) = 0 = P (A).P(B).P(C) and P(AC) = P(BC) = 0 P(A).(C) = P(B).P(C).Hence events are independent.

8. If ‘head’ means one and ‘tail’ means two, then coefficient of quadratic equation ax2 + bx + c = 0 are chosen by tossing three fair coins. The probability that roots of the equations are imaginary is

(A) (B)

(C) (D)

8. Cb2 – 4ac < 0For b = 1 any a and c which can be chosen in 4 waysFor b = 2 either a = 1, c = 2

or a = 2, c = 1or a = 2, c = 2

Required probability =

9. In a bag there are 15 red and 5 white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is

(A) (B)

(C) (D)

2

9. C

Probability that out of remaining balls the one that is red is =

10. A number is chosen at random from the numbers 10 to 99. By seeing the number a man will laugh if product of the digits is 12. If he choose three numbers with replacement then the probability that he will laugh at least once is

(A) 1 – (B)

(C) 1 – (D) 1 –

10. DThere can be four such numbers i.e. 43, 34, 62, 26.Whose product of digit is 12.

Probability that the man will laugh by seeing the chosen numbers =

Required probability = 1–

11. If two events A and B are such that P (A) > 0 and P (B) 1, then P is equal to (A) 1 – P (A/B) (B) 1 – P

(C) (D)

11. C

=

12. Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then the probability that neither A nor B occurs is (A) 0.39 (B) 0.25(C) 0.11 (D) none of these

12. AP(A) = .25, P (B) = .50,

= .25 + .50 – .14 = .61 = 1 – .61 = .39

13. The probability that at least one of the events A and B occurs is 0.6 . If A and B occur simultaneously with probability 0.2, then P + P is(A) 0.4 (B) 0.8(C) 1.2 (D) 1.4(Here and are complements of A, B respectively)

13. C

Now P(A) + P(B) = 0.8P

14. Three identical dice are rolled. The probability that the same number will appear on each of them is

3

(A) (B)

(C) (D)

14. BTotal cases = 216Favourable cases= 6

Probability =

15. The probability of having at least one tail in 4 throws with a coin is

(A) (B)

(C) (D) 1

15. AProbability of no tail in four throws

=

Probability of at least one tail =

16. Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. The probability that the largest number appearing on a selected coupon be 9 is

(A) (B)

(C) (D) none of these

16. Probability in each case =

Required probability (with replacement) =

17. Let A, B, C be 3 independent events such that p(A) = , p (B) = , p(C) = . Then probability of exactly 2

events occurring out of 3 events is _______________________

17. 1/4P (exactly two of A, B, c occur)= P (BC) + P (C A) + P (AB) – 3P (ABC)= P (B). P(C) + P (C). P(A) + P(A). (B) – 3P (A). P (B). P(C)

=

18. Two subsets A and B of a set containing n elements are chosen at random. The probability that A B is

(A) (B)

(C) (D)

4

18. DThe number of favorable cases is nC0 2n + n C1.2n-1 + …. +n Cn 2 = 3n. Total number of cases is 2n 2n = 4n.

Thus required probability is

19. Three different dice are rolled three times. The Probability that they show different numbers only two times is

(A) (B)

(C) (D)

19. D

Probability that 3 different dices shows different faces is , so required probability

20. A man firing at a distant target has 20% chance of hitting the target in one shoot. If P be the probability of hitting the target in ‘n’ attempts where 20P2 – 13P + 2 £ 0, then maximum and minimum value of n is (A) 3, 2 (B) 5, 4(C) 4, 2 (D) 2, 2

20. D20P2 – 13P + 2 £ 0 (4P – 1)(5P – 2) £ 0

£ P £

£ + + + …….. + £

n = 2.

21. If two subsets A and B of set S containing n elements are selected at random, then the probability that A B = and A B = S is

(A) (B)

(C) (D)

21. (B)There are 4 possible for an element (i). neither present in A nor in B (ii). present both in A and B (iii). present in A and absent in B (iv). Present in B and absent in A

Case (iii) and (iv) are favorable =

22. One boy can solve 60% of the problems in a book and another can solve 80%. The probability that at least one of the two can solve a problem chosen at random from the book is(A) 2/25 (B) 23/25(C) 4/5 (D) 9/10

22. (B)

Required probability = 1 –

23. Three dice are rolled. The probability that different numbers will appear on them is

5

(A) 2/3 (B) 4/9(C) 5/9 (D) 2/9

23. CThe number of ways in which different numbers will appear = 6 5 4

Probability = = .

24. A speaks truth in 60% cases and B speaks truth in 70% cases. The probability that they will say the same thing while describing single event is(A) 0.56 (B) 0.54(C) 0.38 (D) 0.94

24. BThey will say same thing in two ways i.e. either both speak truth or both tell a lie.So probability = (.6 .7) + (1 – .6) (1 – .7)

= .42 + .12 = 0.54 .

25. Let A = {2, 3, 4, ….., 30}. A number is chosen from set A and found to be a prime number. The probability that it is more than 15 is

(A) (B)

(C) (D)

25. B

26. In a n sided regular polygon the probability that the two diagonal chosen at random will intersect inside the polygon is

(A) (B)


26. C

When 4 points are selected we get one intersecting point. So probability is .

27. In a game a coin is tossed (2n + m) times and a player wins if he does not get any two consecutive outcomes same for atleast 2n times in a row. The probability that player wins the game is

(A) (B)

(C) (D)

27. DPlayer should get (HT, HT, HT, …..) or (TH, TH, …..) atleast 2n times. If the sequence start from first place

then probability and if any other time probability is .

Hence probability = 2 .

28. Three smallest squares are chosen randomly on a chess board the probability that these squares have exactly two corners, but no side common is

6

(A) (B)


28. CIf the shaded square is chosen we can chose 2 squares out of 4 other at the corner in 4C2 ways such shaded squares are 62.

Hence probability = .

29. If are the probabilities of three mutually exclusive events then set of all values of p is

(A) (B)


29. C

……(1)

……(2)

……(3)

Also, 0 £ P (A B C) £ 1

……(4)

from (1), (2), (3) and (4)

.

30. The probabilities of winning a race by three persons A, B and C are and respectively. They run two

races. The probability of A winning the second race when B wins the first race is

(A) (B)

(C) (D)

30. B

The probability of winning of A the second race =

Since winning of the first race and the second race are independent events.

31. Four die are thrown simultaneously. The probability that 4 and 3 appear on two of the die given that 5 and 6 have appeared on other two die is

(A) (B)


7

31. CGiven that 5 and 6 have appeared on two of the dice the sample space reduces to 64 2 54 + 44 (inclusion exclusion principle) also favourable cases are 4! = 24

So the required probability is .

32. A fair coin is tossed 5 times then the probability that no two consecutive heads occur, is

(A) (B)


32. CLet an be the number of strings of H and T of length n with no two adjacent H’s, then a1 = 2, a2 = 3.

Also, an + 2 = an + 1 + an ( the string must begin with T or HT)So, a3 = 5, a4 = 8, a5 = 8 + 5 = 13

the required probability = .

33. Four cards are drawn from a pack of well shuffled playing cards without replacement. The probability of getting just a trail is

(A) (B)

(C) (D)

33. A

. The required probability = .

34. If the letters of the word BANANA are arranged randomly, then the probability that the word thus formed does not contain the pattern BAN is

(A) (B)


34. (C)

Total number of words formed =

The number of words containing the pattern BAN =

So, the required probability = .

35. There are 10 different objects 1, 2, 3, ….10 arranged at random in 10 places marked 1, 2, 3, …..10. The probability that exactly five of these objects occupy places corresponding to their number is

(A) (B)

(C) (D) None of these

35. B

8

Number of favorable cases = 10C5 derrangement of 5 objects

=

= 10C5 44Total cases =

Probability =

36. A fair coin is tossed 9 times the probability that at least 5 consecutive heads occurs is

(A) (B)


36. BCase I

Case II

Case III

Case IV

Case V


37. An integer is chosen at random from first two hundred natural numbers then the probability that integer is either divisible by 2, 6 or 8 is

(A) (B)

(C) (D) None of these.

37. An(A) = total numbers divisible by 2, n(B) = total numbers divisible by 6n(C) = total numbers divisible by 8Hence


38. Two 8-faced dice (numbered from 1 to 8) are tossed. The probability that the product of two counts is a square number, is

(A) (B)

(C) (D)

38. C

Square values that product can take are 1, 4, 9, 16, 25, 36, 49, 644 : (1, 4), (2, 2), (4, 1)16: (2, 8), (4, 4), (8, 2)For other values, there is only one way of getting the product.

9

.

39. The probability that a randomly chosen 3 digit number has exactly 3 factors

(A) (B)

(C) (D) None of these.

39. BA number has exactly 3 factors if the number is square of a prime number.Squares of 11, 13, 17, 19, 23, 29, 31 are 3-digit numbers.


40. If 7 digits out of 1 to 9 are arranged in the spaces of number 1263 _ _ _ _ _ _ _ 6, what is the probability that the number is divisible by 9

(A) (B)

(C) (D) 1

40. A

Sum of all numbers between 1 to 9, 1 + 2 + … 9 = = 45, and 1 + 2 + 6 + 3 + 6 = 18, total is 18 + 45 = 63

which is divisible by 9. so whatever the arrangement the number will always be divisible by 9. so probability is one.

41. A bag contains 17 markers with numbers 1 to 17. A marker is drawn at random and then replaced; a second marker is drawn then the probability that first number is even and second is odd, is

(A) (B)


41. B

There are 8 even and 9 odd numbers. So probabilities of getting first even number is and probabilities of

getting second odd number = , so required probabilities =

42. Out of 6 pairs of distinct gloves 8 gloves are randomly selected, then the probability that there exist exactly 2 pairs in it is

(A) (B)

(C) (D)

42. AFavorable cases m = 6C2 (2 4C4 + 2 4C3

1C1 + 4C2 2C2)

total cases ® n = 12C8

required probability =

10

43. A cube painted red on all sides is cut into 125 equal small cubes. A small cube when picked is found to show red color on one of its faces. The probability that two more faces also show red color is

(A) (B)

(C) (D)

43. BThe problem is of conditional probability. Total cases in which at least one of the cubes is red painted is 125 27 = 98 out of which 8 are painted on three sides

Probability = =

44. 2n balls (all distinct in size) are arranged in a row. First few of these balls are black rest all white, both odd in number. The probability that there is exactly, one black ball in one of all possible arrangements is

(A) (B)

(C) (D)

44. CEither first ball is black or first three balls are black or first five and so on…. all being equally likely. Total probability of both balls being odd is

where k is number of ways both can be odd.

Probability of exactly one black ball is

=

=

45. Each of 10 passengers board any of the three buses randomly which had no passenger initially. The probability that each bus has got at least one passenger is

(A) (B)

(C) (D)

45. DTotal ways in which they can seated in the buses are 310. Applying the principle of Inclusion- exclusion, we get number of favourable ways as 310 3.210 + 3.

46. A bag contains n white and n black balls. Pairs of balls are drawn without replacement untill the bag is empty. If the number of ways in which each pair consists of one black and one white ball is 576, then n is equal to (A) 4 (B) 5(C) 24 (D) 12

46. A

(nC1 x nC1) (n-1C1 x n-1C1) …… (2C1 x 2C1) (1C1 x 1C1) = 576 n2 (n-1)2 ….22 12 = (24)2 (n!)2 = (24)2 n! = 24 = 4! n = 4.

11

47. Bolts marked 1, 2, 3, 4 respectively fit into nuts marked 1, 2, 3, 4, . If somebody mixes up the bolts and nuts, and then tries to fit them back at random, what is the probability that no bolt goes into its right nut ?

(A) (B)


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