Q 11 :    

An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, and 7 is                    [2024]

  • 825

     

  • 950

     

  • 1425

     

  • 2150

     

(4)

Let A, B, C be the events represents the numbers divisible by 4, 6 and 7 respectively.

   n(A)=12,  n(B)=8,  n(C)=7

       n(AB)=4,  n(BC)=1,  n(AC)=1

and n(ABC)=0

So, n(ABC)=12+8+7-4-1-1=21

   Required probability=2150

 



Q 12 :    

Two integers x and y are chosen with replacement from the set {0, 1, 2, 3,…,10}. Then the probability that |x-y|>5, is                  [2024]

  • 60121

     

  • 31121

     

  • 62121

     

  • 30121

     

(4)

Let E={(x,y):|x-y|>5} and x,y{0,1,2,,10}

E={(0,6),(0,7),(0,8),(0,9),(0,10),(1,7),(1,8),(1,9),(1,10),(2,8),(2,9),(2,10),(3,9),(3,10),(4,10),(6,0),(7,0),(8,0),(9,0),(10,0),(7,1),(8,1),(9,1),(10,1),(8,2),(9,2),(10,2),(9,3),(10,3),(10,4)}

   n(E)=30

Total number of possible outcomes =11×11=121

    Required probability=30121

 



Q 13 :    

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is                  [2024]

  • 310

     

  • 14

     

  • 19

     

  • 13

     

(4)

Let E1 be the event that bag A is selected, E2 be the event that bag B is selected and E be the event that white ball is drawn.

  P(E1)=P(E2)=12,  P(EE1)=310,  P(EE2)=35

Required probability,

P(E1E)=P(E1)·P(EE1)P(E1)·P(EE1)+P(E2)·P(EE2)

=12×31012×310+12×35=13

 



Q 14 :    

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability that the first drawn marble is red and the second drawn marble is white, is                           [2024]

  • 225

     

  • 475

     

  • 23

     

  • 425

     

(2)

Total marbles = 10 + 30 + 20 + 15 = 75

Let E be the event of drawing first drawn marble is red and second drawn marble is white.

Now, probability of drawing first red marble and 2nd white marble is

         P(E)=1075×3075=475

 



Q 15 :    

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is               [2024]

  • 227

     

  • 29

     

  • 127

     

  • 19

     

(2)

Let x be the probability of getting a tail.

 P(T)=x;  P(H)=2x

 x+2x=1x=13

 P(T)=13 and P(H)=23

Now, required probability=C23×13×13×23

=3×227=29



Q 16 :    

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 13 and 23 respectively. Let x be the number of matches that the team wins, and y be the number of matches that the team loses. If the probability P(xy2)  is p, then 39p equals _______ .                [2024]



(8288)

|x-y|10 and 0x10 and 0y10

 p=P(|x-y|2)

=P(x=4,y=6)+P(x=5,y=5)+P(x=6,y=4)

=C410(13)4(23)6+C510(13)5(23)5+C610(13)6(23)4

=210(26310)+252(25310)+210(24310)

=25310[420+252+105]p=25310×777

Hence, 39p=39×25310×777=253×777=8288



Q 17 :    

Let the mean and the standard deviation of the probability distribution 

X α 1 0 -3
P(X) 13 K 16 14

 

be μ and σ, respectively. If σ-μ=2, then σ+μ is equal to _____________.              [2024]



(5)

We have, P(X)=1

13+K+16+14=18+4+624+K=1

K=1-1824=624=14

Now, μ=α3+14-34=α3-12

and σ=(α23+14+94)-(α3-12)2 =(2α29+α3+94)

Given, σ-μ=2

σ=μ+2

σ2=(μ+2)2 2α29+α3+94=α29+94+α

α29-2α3=0

α=0,6                    [α0  X=0 is already given]

α=6μ=32 and σ=72

So μ+σ=5

 



Q 18 :    

Let a,b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that ax2+bx+c=0 has all real roots is mn,gcd(m,n)=1, then m+n is equal to _______ .                  [2024]



(19)

We have, ax2+bx+c=0

For real roots, b2-4ac0                    ...(i)

a,b,c{1,2,3,4}                                    ...(ii)

Ordered triplet (a,b,c) satisfying (i) and (ii) are

 (1,2,1),(1,3,1),(2,3,1),(1,3,2),(1,4,1),(1,4,2),(2,4,1),(2,4,2),(4,4,1),(1,4,4),(3,4,1),(1,4,3)

i.e. total 12 favourable outcomes.

Total number of outcomes = 4×4×4=64

 Required probability=1264=316=mn        (Given)

Here, m+n=3+16=19