Q 1 :

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(AB) is             [2008]

  • 35

     

  • 0

     

  • 1

     

  • 25

     

(3)

A (number is greater than 3)={4,5,6}

 P(A)=36=12

B (number is less than 5)={1,2,3,4}P(B)=46=23

 AB={4}P(AB)=16

 P(AB)=P(A)+P(B)-P(AB)

                        =12+23-16=3+4-16=1



Q 2 :

If P(B)=34, P(ABC¯)=13 and P(A¯BC¯)=13, then P(BC) is            [2003]

  • 112

     

  • 16

     

  • 115

     

  • 19

     

(1)

Given : P(B)=34, P(ABC¯)=13

P(A¯BC¯)=13

[IMAGE 372]

From above venn diagram, we see

BC=B-(ABC¯)-(A¯BC¯)

P(BC)=P(B)-P(ABC¯)-P(A¯BC¯)

P(BC)=34-13-13=9-4-412=112



Q 3 :

A number is chosen at random from the set {1, 2, 3, ..., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is ________.                   [2021]



(214)

E= set of numbers divisible by 3

E={3,6,9,12,,1998}

  n(E)=666

F= set of numbers divisible by 7

F={7,14,21,,1995}

  n(F)=285

EF= set of numbers divisible by 3 and 7={21,42,,1995}

  n(EF)=666+285-95=856

Required probability=8562000=P

So,  500P=8562000×500=214



Q 4 :

Let E,F and G be three events having probabilities P(E)=18, P(F)=16, P(G)=14, and let P(EFG)=110.

For any event H, if Hc denotes its complement, then which of the following statements is(are) TRUE?            [2021]

  • P(EFGc)140

     

  • P(EcFG)115

     

  • P(EFG)1324

     

  • P(EcFcGc)512

     

Select one or more options

(1, 2, 3)

 Given that

P(E)=18, P(F)=16, P(G)=14, P(EFG)=110

(3)  P(EFG)= P(E)+P(F)+P(G)-P(EF)-P(FG)-P(GE)+P(EFG)

      =18+16+14-P(EF)+110

       =1324+110-P(EF)

        P(EFG)1324   [Option (3) is correct]

(4)   Now, P(EcFcGc)

        =1-P(EFG)1-1324

       P(EcFcGc)1124  [Option (4) is incorrect]

(1)    P(E)P(EFG)+P(EFGc)

        18P(EFGc)+110

         18-110P(EFGc)

          140P(EFGc)   [Option (1) is correct]

(2)    P(F)P(EcFG)+P(EFG)

        16P(EcFG)+110

         16-110P(EcFG)

          460P(EcFG)

           115P(EcFG)  [Option (2) is correct]



Q 5 :

Two players, P1 and P2, 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 denote the readings on the die rolled by P1 and P2, respectively. If x>y, then P1 scores 5 points and P2 scores 0 point. If x=y, then each player scores 2 points. If x<y, then P1 scores 0 point and P2 scores 5 points. Let Xi and Yi be the total scores of P1 and P2, respectively, after playing the ith round.       [2022]

  List - I   List - II
(I) Probability of (X2Y2) is (P) 38
(II) Probability of (X2>Y2) is  (Q) 1116
(III) Probability of (X3=Y3) is (R) 516
(IV) Probability of (X3>Y3) (S) 355864
    (T) 77432

 

The correct option is:

  • (I) → (Q); (II) → (R); (III) → (T); (IV) → (S)

     

  • (I) → (Q); (II) → (R); (III) → (T); (IV) → (T)

     

  • (I) → (P); (II) → (R); (III) → (Q); (IV) → (S)

     

  • (I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

     

(1)

P(Xi>Yi)+P(Xi<Yi)+P(Xi=Yi)=1

and  P(Xi>Yi)=P(Xi<Yi)=p

For i=2

P(X2=Y2)=P(5,5)+P(4,4)

=512×512×2+16×16=2572+136=2772=38

P(X2>Y2)=P(10,0)=512×512+512×16×2=516

P(X2Y2)=516+38=1116

IQ, IIR

For i=3

P(X3=Y3)=P(6,6)+P(7,7)

=16×6+512×16×512×6=77432

P(X3>Y3)=12(1-77432)=355864

IIIT, IVS