Q: Three students S_1, S_2 and S_1 are given a problem to solve. Consider the following events : U: At least one of S_1, S_2 and S_3 can solve the problem, V: S_1 can solve the problem, given that neithe

aligned & P ( U )=1- P ( ~S _1^ S _2^ S _3^ )= 1 2 \\ & P ( ~S _1^ S _2^ S _3^ )= 1 2 ; P ( ~S _1^ ) P ( ~S _2^ ) P ( ~S _3^ )= 1 2 \\ & (1- P ( ~S _1 ) ) (1- P ( ~S _2 ) ) (1- P ( ~S _3 ) )= 1 2 (1) aligned aligned & P ( V )= P ( S _1 ~S _2^ S _3^ ) P ( S _2^ S _3^ ) = 1 10 \\ & P ( S _1 ) P ( S _2^ ) P ( S _3^ )= 1 10 P ( S _2^ ) P ( S _3^ ) \\ & P ( S _1 )= 1 10 aligned aligned & P ( ~W )= P ( ~S _2 ~S _3^ )= 1 12 \\ & P ( ~S _2 ) P ( ~S _3^ )= 1 12 \\ & P ( ~S _2 ) (1- P ( ~S _3 ) )= 1 12 ..

Q: Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-ll contains 24 red, 9 blue and 15 green balls, Box-III contains 1 blue, 12 green and 3 yellow balls, Box-IV contains 10 green, 16

P Red from Box I = 8 16 = 1 2 P Blue from Box I = 3 16 P Green from Box I = 10 32 = 5 16 Now, P Green from Box II = 15 48 P Green from Box III = 12 16 = 3 4 P White from Box IV = 6 32 = 3 16 Event A : One of the chosen ball is white B : At least one of the chosen ball is green P A B = P A ∩ B P B = 5 16 × 3 16 1 2 × 15 48 + 3 16 × 3 4 + 5 16 × 1 = 5 52

Question 1

Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from

Accepted Answer

Total balls in Box I = 6 + 9 = 15 Total balls in Box II = 8 + 12 = 20 Total balls in the mixture = 15 + 20 = 35 Total red balls = 6 + 8 = 14 Total green balls = 9 + 12 = 21 The probabilities of the given events are: P(E_1) = 15 35 = 3 7 P(E_2) = 20 35 = 4 7 P(F_1) = 14 35 = 2 5 P(F_2) = 21 35 = 3 5 The joint probabilities are: P(E_1 F_1) = 6 35 P(E_2 F_2) = 12 35 For statement (A): P(E_1) P(F_1) = 3 7 2 5 = 6 35 = P(E_1 F_1) Thus, E_1 and F_1 are independent. Statement (A) is TRUE. For statement

Question 2

Three students S_1, S_2 and S_1 are given a problem to solve. Consider the following events : U: At least one of S_1, S_2 and S_3 can solve the problem, V: S_1 can solve the problem, given that neithe

Accepted Answer

aligned & P ( U )=1- P ( ~S _1^ S _2^ S _3^ )= 1 2 \ & P ( ~S _1^ S _2^ S _3^ )= 1 2 ; P ( ~S _1^ ) P ( ~S _2^ ) P ( ~S _3^ )= 1 2 \ & (1- P ( ~S _1 ) ) (1- P ( ~S _2 ) ) (1- P ( ~S _3 ) )= 1 2 (1) aligned aligned & P ( V )= P ( S _1 ~S _2^ S _3^ ) P ( S _2^ S _3^ ) = 1 10 \ & P ( S _1 ) P ( S _2^ ) P ( S _3^ )= 1 10 P ( S _2^ ) P ( S _3^ ) \ & P ( S _1 )= 1 10 aligned aligned & P ( ~W )= P ( ~S _2 ~S _3^ )= 1 12 \ & P ( ~S _2 ) P ( ~S _3^ )= 1 12 \ & P ( ~S _2 ) (1- P ( ~S _3 ) )= 1 12 ..

Question 3

A factory has a total of three manufacturing units, M_1, M_2, and M_3, which produce bulbs independent of each other. The units M_1, M_2, and M_3 produce bulbs in the proportions of 2:2:1, respectivel

Accepted Answer

Now given probability aligned & P ( Produced by M _2 defective )= 40 100 14- x 40 20 100 = 2 5 \ & 14- x 40 = 1 5 \ & 14- x =8 \ & x =6 aligned So, the required probability = 6 20 =0.3

Question 4

A bag contains N balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the

Accepted Answer

Given P ( W _1 G _2 B _3 )= 2 5 ~N and P ( B _3 W _1 G _2 )= 2 9 aligned & P ( B _3 W _1 G _2 ) P ( W _1 G _2 ) = 2 9 \ & 2 5 ~N N ( N -1) 3 6 = 2 9 \ & N =11 aligned

Question 5

A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining quest

Accepted Answer

C Correct G Guess K Knows array ll P ( C G )= 1 2 , P ( C K )=1 \ P ( G C )= 1 6 array aligned & Let required probability = x \ & P ( G C )= (1- x ) P ( C G ) (1- x ) P ( C G )+ x P ( C K ) \ & 1 6 = (1- x ) ( 1 2 ) (1- x ) ( 1 2 )+( x )(1) \ & x = 5 7 Option (3) is correct. aligned

Question 6

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is 1 3 , then the probability that the

Accepted Answer

Given, P ( H ) = 1 3 So,  P ( T ) = 1 - 1 3 = 2 3 Now probability of the event is given by, P ( E ) = P ( H H ) + P ( T H H ) + P ( H T H H ) + P ( T H T H H ) + P ( H T H T H H ) + P ( T H T H T H H ) + &#8230; &#8658; P ( E ) = 1 3 &#215; 1 3 + 2 3 &#215; 1 3 &#215; 1 3 + 1 3 &#215; 2 3 &#215; 1 3 &#215; 1 3 + 4 3 5 + 4 3 6 + 8 3 7 + 8 3 8 + &#8230; &#8658; P ( E ) = 1 3 2 + 2 3 3 + 2 3 4 + 4 3 5 + 4 3 6 + 8 3 7 + 8 3 8 + &#8230; &#8658; P ( E ) = 1 3 2 + 2 3 4 + 4 3 6 + &#8230; . + 2 3 3

Question 7

Consider the 6 &#215; 6  square in the figure. Let A 1 , A 2 , &#8230; , A 49 be the points of intersections (dots in the picture) in some order. We say that A i and A j are friends if they are a

Accepted Answer

Plotting the diagram of the given data we get, Now from above diagram we get, Number of points having 0 friend = 0 Number of points having 1 friend = 0 Number of points having 2 friends = 4 Number of points having 3 friends = 5 &#215; 4 = 20 Number of points having 4 friends  = 49 - 24 = 25 So, probability distribution table will be, Now from above table expected value will be given by, 7 &#183; E X = 7 0 + 0 + 4 49 &#215; 2 + 20 49 &#215; 3 + 25 49 &#215; 4 &#8658; 7 &#183; E X = 7 100 + 6

Question 8

Let X = x , &#160; y &#8712; &#8484; &#215; &#8484; &#160; : &#160; x 2 8 + y 2 20 &#60; 1 &#160; and &#160; y 2 &#60; 5 x . Three distinct point P , &#160; Q and R are randomly chosen from X . Then t

Accepted Answer

Given, Equation of ellipse  x 2 8 + y 2 20 = 1 And equation of parabola  &#160; y 2 = 5 x Now intersection point of ellipse and parabola will be  2 , 10 &#160; &#38; &#160; 2 , - 10 Now plotting the region  x 2 8 + y 2 20 &#60; 1 &#160; and &#160; y 2 &#60; 5 x  we get, Now randomly chosen points will be between origin and the intersecting point, So, the integer points inside region are  2 , &#160; 1 , &#160; 2 , &#160; - 1 , &#160; 2 , &#160; 2 , &#160; 2 , &#160;

Question 9

Suppose that Box-I contains 8  red, 3 blue and 5 green balls, Box-ll contains 24 red, 9 blue and 15 green balls, Box-III contains 1 blue, 12 green and 3 yellow balls, Box-IV contains 10 green, 16

Accepted Answer

P Red&#160;from&#160;Box &#160; I = 8 16 = 1 2 P Blue&#160;from&#160;Box &#160; I = 3 16 P Green&#160;from&#160;Box &#160; I = 10 32 = 5 16 Now, P Green&#160;from&#160;Box &#160; II = 15 48 P Green&#160;from&#160;Box &#160; III = 12 16 = 3 4 P White&#160;from&#160;Box &#160; IV = 6 32 = 3 16 Event A : One of the chosen ball is white B : At least one of the chosen ball is green P A B = P A &#8745; B P B = 5 16 &#215; 3 16 1 2 &#215; 15 48 + 3 16 &#215; 3 4 + 5 16 &#215; 1 = 5 52

Question 10

Two players, P 1  and P 2 , play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let x and y de

Accepted Answer

Given, P (draw in 1 round) = 6 36 = 1 6 P (win in 1 round) = 1 2 1 - 1 6 = 5 12 P (loss in 1 round) = 5 12 Now finding the probability of all we get, P X 2 &#62; Y 2 = P 10 , 0 + P 7 , 2 = 5 12 &#215; 5 12 + 5 12 &#215; 1 6 &#215; 2 = 45 144 = 5 16 &#8594; R P X 2 = Y 2 = P 5 , 5 + P 4 , 4 = 5 12 &#215; 5 12 &#215; 2 + 1 6 &#215; 1 6 = 25 + 2 72 = 3 8 P X 2 &#8805; Y 2 = 5 16 + 3 8 = 11 16 &#8594; Q P X 3 = Y 3 = P 6 , 6 + P 7 , 7 = 1 6 &#215; 6 &#215; 6 + 5 12 &#215; 1 6 &#215; 5 12 &#215; 6 =

Question 11

A number is chosen at random from the set  1 , 2 , 3 , &#8230; , 2000 . Let  p  be the probability that the chosen number is a multiple of 3 or  7 . Then the value of  500 p &

Accepted Answer

Multiple of 3 = 3 , 6 , 9 &#8230; . 1998  are  666 Multiple of 7 = 7 , 14 &#8230; . 1995  are  285 Multiple of 21 = 21 , 42 &#8230; 1995  are  95 p = 666 + 285 - 95 2000 = 856 2000 = 214 500 = p 500 p = 214

Question 12

Consider three sets E 1 = 1 , 2 , 3 , F 1 = 1 , 3 , 4 and G 1 = 2 , 3 , 4 , 5 . Two elements are chosen at random, without replacement, from the set E 1 , and let S 1 denote the set of these chosen el

Accepted Answer

We need to find p , the conditional probability of the event S 1 = 1 , 2 , given  E 1 = E 3 . So,  p = P S 1 &#8745; E 1 = E 3 P E 1 = E 3 = P B 1 , 2 P ( B ) P ( B ) = P B 1 , 2 + P B 1 , 3 + P B 2 , 3 P B 1 , 2 &#8594;  If  1 , 2  chosen in the beginning  P B 1 , 3 &#8594; If 1 , 3  chosen in the beginning P B 2 , 3 &#8594; If  2 , 3  chosen in the beginning (i) If 1 , 2  is chosen in the begining, then F 2 = 1 , 2 , 3 , 4 and S 2 must contain

Question 13

Three numbers are chosen at random, one after another with replacement, from the set S=\ 1,2,3, , 100\ . Let p_ 1 be the probability that the maximum of chosen numbers is at least 81 and p_ 2 be the p

Accepted Answer

Maximum of the chosen numbers is at least 81 It means we have to choose at least one number from 81 to  100 Total number of possible selections  = 100 &#215; 100 &#215; 100 = 100 3 Favourable cases = Total - unfavourable cases Unfavourable cases are those in which we have selected all the three numbers form 1 to 80 &#160;   = 80 &#215; 80 &#215; 80 = 80 3 Total number of favourable cases  = 100 3 - 80 3 So,  p 1 = 100 3 - 80 3 100 3 = 20 3 5 3 -

Question 14

Three numbers are chosen at random, one after another with replacement, from the set S=\ 1,2,3, , 100\ . Let p_ 1 be the probability that the maximum of chosen numbers is at least 81 and p_ 2 be the p

Accepted Answer

Minimum of the chosen numbers is at most 40 It means we have to choose all the numbers from 1 to 40 only from 41 to 100 Total number of possible selections  = 100 &#215; 100 &#215; 100 = 100 3 Favourable cases = Total - Unfavourable cases Unfavourable cases are those in which we have selected all the three numbers form 41 to 100 &#160; = 60 &#215; 60 &#215; 60 = 60 3 Total number of favourable cases  = 100 3 - 60 3 So,  p 2 = 100 3 - 60 3 100 3 = 20 3 5 3

Question 15

Let E , &#160; F and G be three events having probabilities  P ( E ) = 1 8 , P ( F ) = 1 6  and  P ( G ) = 1 4 ,  and let  P ( E &#8745; F &#8745; G ) = 1 10 For any event H ,

Accepted Answer

(i) P E &#8745; F &#8745; G c = P E &#8745; F - P E &#8745; F &#8745; G &#8804; P E - 1 10 = 1 8 - 1 10 = 1 40 (ii) P E c &#8745; F &#8745; G = P ( F &#8745; G ) - P ( E &#8745; F &#8745; G ) &#8804; P ( F ) - 1 10 = 1 6 - 1 10 = 1 15 (iii) P E &#8746; F &#8746; G &#8804; P E + P F + P G = 1 8 + 1 6 + 1 4 = 3 + 4 + 6 24 = 13 24 (iv) P E c &#8745; F c &#8745; G c max = 1 - P ( E &#8745; F &#8745; G ) min &#8757; P ( E &#8745; F &#8745; G ) min = 1 4 Thus P E c &#8745; F c &#8745; G c &#8804; 3 4

Question 16

The probability that a missile hits a target successfully is 0 . 75 . In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that h

Accepted Answer

Let the number of missiles be  n , P (of destroying the target) &#8805; 0 . 95 C 3 n &#183; 3 4 3 &#183; 1 4 n - 3 + C 4 n 3 4 4 1 4 n - 4 + &#8230; . . + C n n 3 4 n &#8805; 0 . 95 1 - C 0 n 3 4 0 &#183; 1 4 n + C 1 n 3 4 1 &#183; 1 4 n - 1 + C 2 n 3 4 2 1 4 n - 2 &#8805; 0 . 95 1 - 95 100 &#8805; 1 4 n + 3 n 4 n + n n - 1 2 &#183; 9 4 n 4 n 20 &#8805; 2 + 6 n + 9 n 2 - 9 n 2 2 2 n - 1 5 &#8805; 9 n 2 - 3 n + 2 2 2 n - 1 &#8805; 5 9 n 2 - 3 n + 2 n = 3 &#8658; 32 &#8805; 5 &#215; 74

Question 17

Two fair dice, each with faces numbered  1 , 2 , 3 , 4 , 5 and 6 , are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime

Accepted Answer

Perfect Square can be  4 or  9 , which can be obtained by  1 , 3 , 3 , 1 , 2 , 2 , 6 , 3 3 , 6 , 5 , 4 , 4 , 5 . Prime Number can be  2 or 3 or 5 or 7 or 11 , which can be obtained by  1 , 1 , 1 , 2 , 2 , 1 , 1 , 4 , 4 , 1 , 2 , 3 , 3 , 2 , 1 , 6 , 6 , 1 , 5 , 2 , 2 , 5 , 3 , 4 , 4 , 3 , 6 , 5 , 5 , 6 P (square or prime) = 22 36 P &#160;Perfect square&#160;is&#160;odd&#160; &#160;perfect square&#160;before prime&#160; = 4 36 + 14 36 . 4 36 + 14 36 . 14 36 . 4 36 + .

Question 18

Let C 1 and C 2 be two biased coins such that the probabilities of getting head in a single toss are 2 3 and 1 3 , respectively. Suppose &#945; is the number of heads that appear when C 1 is toss

Accepted Answer

For tossing two coins independently, maximum and minimum heads that can appear will be  2 &#160; &#38; &#160; 0  respectively. Roots of equation x 2 &#8722; &#945; x + &#946; = 0 are real and equal, then discriminant D = 0 &#8658; &#945; 2 &#8722; 4 &#946; = 0 &#8658; &#945; 2 = 4 &#946; &#8658; &#945; = 0 , &#160; &#946; = 0 or &#945; = 2 , &#160; &#946; = 1 Thus, the required probability = C 0 2 2 3 0 1 3 2 . C 0 2 1 3 0 2 3 2 + C 2 2 2 3 2 1 3 0 . C 1 2 1 3 1 2 3 1 = 1 9 &#215; 4 9

Question 19

Let X denote the number of elements in set X . Let S = 1 , &#160; 2 , &#160; 3 , &#160; 4 , &#160; 5 , &#160; 6 be a sample space, where each element is equally likely to occur. If A and B are indepen

Accepted Answer

Since A and B are independent events P A &#8745; B = P A P B A &#8745; B S = A S &#183; B S S &#160; A &#8745; B = A &#183; B &#8756; x = n u m b e r &#160; o f &#160; e l e m e n t &#160; i n &#160; s e t S = 6 6 A &#8745; B = A B 6 n A &#8745; B = n A n B s e t &#160; x = n x Case I : n A &#8745; B = 1 &#8658; n A n B = 6 a &#160; n A = 6 , &#160; n B = 1 N u m b e r &#160; o f &#160; w a y s = C 6 &#160; 6 &#215; C 1 &#160; 6 = 6 b &#160; n A = 3 , &#160; n B = 1 &#160; &#160; Number &#160; o

Question 20

There are three bags B 1 , &#8201; B 2 and B 3 . The bag B 1 contains 5 red and 5 green balls, B 2 contains 3 red and 5 green balls, and B 3 contains 5 red and 3 green balls, Bags B 1 , &#8201; B 2 an

Accepted Answer

Let us define following events B 1 : Event of selection of bag 1 ( 5 Red, 5 Green) B 2 &#160; : Event of selection of bag 2 ( 3 Red, 5 Green) B 3 : Event of selection of bag 3 ( 5 Red, 3 Green) R = Event of selection of Red ball G = Event of selection of Green ball Option ( C ) : P G B 3 = 3 8 (as bag 3 has been selected now total balls in bag 3 = 8 , green ball = 3 ) Option ( B ) : P G = P B 1 P G B 1 + P B 2 P G B 2 + P B 3 P G B 3 = 3 10 &#215; 5 10 + 3 10 &#215; 5 8 + 4 10 &#215; 3 8 = 39 80

Probability — Previous Year Questions & Solutions