Objective Mathematics - Mathematics for IIT-JEE - Home

Chapter No -6 ( Probability )

4. For real roots of x2  px  2  0 , D  0

1.

 p2  8  0  p  2 2  farourable values of p  3 , 4 , 5 , 6 required probability 



 A B  P A B C  P    P (C )  C  



 P A  P ( B ) or P B  P  A 

2. n( S )  total no. of ways of distribution of three identical dice among six different persons. = number of non-negative integral solutions of x1  x2  x3  x4  x5  x6  3

 n( S )  3 61C61  8C5  56 

P( E ) 

6 3  56 28

3.

1 n 5. P(r  1)  1  P(0)  1  C0   2

n

n

1  1     0.95 n

s c i t   a m e h t a a m r M a E e h JE iv .S T K t . II c L .  je Er b O 

P (C )  P ( A  C )  P ( B  C )  1  P ( A)  P ( B ) P (C )

 

4 2  6 3



(2) n  20



min. value of n  5

6. P( A  B)  P( A) P( B)

 A and B are not independent.



P( A  B)  P( A)  P( B)  P( A  B)

 0.70  0.60  0.50  0.80



P( A  B)  B  P ( A  B)  B Now , P    A( A  B ) P( A  B)  A B 



 P ( A  B)  P( A  B )  P( A)





P( A)  P( A  B) P( A  B )



0.70  0.50 0.80

0.2  0.25 0.80

7. Probability that minimum no. is 3 1



C1  7 C2 10

C3



7 40

Probability that maximum no. is 7 from venn diagram , x 

1 1 1 1    12 12 6 2

1  x 6

1



10

1 P AC  B  6



C3



1 8

Probability that minimum is 3 and maximum is 7 1



C1  6 C2



Mathematics for IIT-JEE (Hints/Solutions)



C1  1C1  3C1 10

C3



1 40

[45]

Required probability  

7 1 1   40 8 40 11 40



 3  2  1  14.      6  6  6 

8. ax  by  2 ; cx  dy  4 for unique solution ,   0 

a b c d



n(S )  16

  Indian men seated with wife 15. P    and American men seated adjacent to his wife 

Now , favorable cases exist if ad  bc



 n( E )  6 

P( E ) 

6 3  16 8



 1  1  1   1  1  1   1  1  1                 2  2  2   2  2  4   2  4  2 

3

4



4!(32) 5!(16)

2 5

1 1 and P A  B  12 2



Let P ( A)  x and P( B)  y  xy 

1 1 and (1  x )(1  y )  12 2

 1  ( x  y) 

1 1    8 2 P( E ) 

(6  1)! (2!)

16. P ( A  B) 

1 1 1 1    8 16 16 4 Case (ii) (Student passing in three exams) 

(5  1)! (2!)5

 s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

9. Case (i) (Student passing in two exams)



1 36

 0  ( ad  bc )  0

Sample space is having (2) 4 cases 

1 3

 1 x 

1 1 3   4 8 8



1 1  12 2

1 5   12 x 2  7 x  1  0 12 x 12

1 1 1 1 or  y  or . 4 3 3 4

x

10. P6 can never reach to final round  Required probability = 0.

1 1 7 Now , 3 x  4 y  3    4     4   3  12

11. 0.80  p  c  m   0.35  3(0.20)   0.20

or

1  1  3   4   0  3  4 

 p  c  m  0.80  0.95  0.20  1.55 or

12.

100

17.

31 20

50

C5  p  (1  p ) 50  100 C51 ( p )51 (1  p ) 49

1 p   p

100

C51

100

C50

50  51

 51  51 p  50 p  p 

51 101

13. Only two tests will be needed if both defective or both non-defeetive bulls are identified in first two tasts 

 2C  1   2C  1  P( E )   4 1      4 1     C  3   C  3   1  1

Mathematics for IIT-JEE (Hints/Solutions)

I

II

W

B

B  (2 / 4)(2 / 3)(3/ 4)  1/ 4

B

W

B  (2 / 4)(2 / 5)(3 / 4)  3/ 20

W

W

B  (2 / 4)(1/ 3)(2 / 2)  1/ 6

B

B

B  (2 / 4)(3 / 5)(4 / 6)  1/ 5

P( E )  

III

1 3 1 1    4 20 6 5

23 20

18. P ( A  B )  P( A)  P ( B )  P ( A  B )

[46]

4

 5   5   5  1   5   1  19. P(k  3)               ....   6   6   6  6   6   6 

22. x2  9 x  18  0  ( x  3)( x  6)  0  x  (3 , 6)



3 2   5   1   5  5       1      ....    6   6   6  6  

Let x1 , x2 , x3 be number on dice

  3 3  5   1  1   5          6   6 1 5   6     6 5



x1  x2  x3  4 or 5



Coeff of x 4 in ( x  x 2  ....  x6 )3  coeff of x 5



in x  x 2  .... x 6



3



6



4 3 6  Coeff of x in x 1  x

5 1 5 1 P(k  6)            ....  6 6 6 6

Coeff of x 5 in

  5 5  5  1 1   5           6   6  1  5   6   6



k  4 , 5



P( E ) 

63 (6)

3



3

 1  x 

3



3

 x (1  x ) (1  x)   6 3

6 3

3

1 24

s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

Required probability =

(5 / 6)5 (5 / 6)

3



25 36

23. P ( I )  Probability that matrix have exactly 5 entries as 1 P (S )  Probability that matrix is symmetric

H 1 H 2 20. P    and P    F 2 B 3

 S  P( S  I ) Now , P    P( I ) I

8  1  1     9 15  2  2   F  P   8 1 1    7  2  1  16  HT         15  2  2  15  3  3 

 9!    5! 4! 126 P( I )   9   9 2 2 Number of symmetric matrix with exactly 5 entries as one = 3 + 9 = 12

21. Let C , S , B , T be the events of the person going by car , seooter , bus or train respectively. Given that P (C )  P (T ) 

1 3 , P(S )  , P(B)  2 , 7 7 7



2  S  12 P    I 126 21  

24.

1 7

Let L be the event of the person reaching the office in tiem. L 7 L 8 L 5 L 8  P    , P    , P    , P    C  9 S 9  B 9 T  9 L P   .P (C ) C C  P      L P L

 

1 7  1 7 9   1 7 3 8 2 5 8 1 7        7 9 7 9 7 9 9 7

Mathematics for IIT-JEE (Hints/Solutions)

1  1    2 4 4  8  P( E )    1 1 16       1        4  4   P( E ) 

 8   16

[47]

225   25.  N    30 N    N  15 is only applicable because it is given that

28. P  B0   P( B1 )  ....  P( B15 )  1

225   N    30 N   

P( E ) 

3  1 1  11     4  4 24  24



Let P  Bn   kn2

1  0.01 100



k  4k  9k  ....225k  1

 k

 A 1 B 1 26. P ( A)  P    and P     B 4  A 2 

P( B  A) 1 1   P ( B  A)  P ( A) 2 8

1 1  A  P( A  B) Now , P      B P ( B ) 8 P ( B ) 4  

1 2 A and B are independent events

6 1  k (15)(16)(31) 1240

15

(a)

 PB   1 n

n

15

(b) P ( E ) 

n

n 0

15



 E   n 

 P(B )P  B

 n2  n

( P ( A  B )  P ( A) P ( B ))



1

n 0



15(16)   2 

  1240   15   (1240)(15) 

s c i t a        m e h t a a m  r M   a E e h   -JE S v . i IIT ct .L.K je Er   b O

 P( B) 

2

14400 24  (1240)(15) 31

P A B  P A  P B  P A B

3 1  3  1  7        4 2  4  2  8

 A P A B 3 P    P A  B P ( B ) 4  

B 1 P    P B  2  A

27. P ( A) 

1 1 1 ; P (b )  , P (c )  3 2 4

(a) P( A  B)  P( B  C )  P(C  A)  3P( A  B  C )

 1  1   1   1   1  1   1   1  1                3       3  2   2   4   4  3   3   2  4 

 E P  Bs  P   B  Bs  (c) P  s   15 E  E    P  Bn  P    Bn  n 0 25  5    25  31 1240  15    24 (24)(1240)(3) 31 

29. For mutually exclusive and exhaustive events , P ( A  B )  0 , P ( A  B )  P( A)  P( B)  1



     

(d) P  A  B  C   P( A  B)  P( B  C ) 



E  E  P ( B3 ) P  1   P ( B4 ) P  1   B4   B3 

2  1  3 1 4  1  1    (1)          10  2  10  3  10  4   10 

P (C  A)  2 P ( A  B  C ) 

Mathematics for IIT-JEE (Hints/Solutions)



E  E  30. (a) P ( E1 )  P( B1 ) P  1   P ( B2 ) P  1    B1   B2 

 2  1  3  3  1        3  2  4  4  1  1  1  1 (c) P ( A  B  C )        3  2  4  24



 P A  B  0 or P A  B  0

1  . 4

(b) P ( A  B  C )  1  P A P B P C

5 576

4 2  10 5

[48]

E  P ( B3 ) P  2   B3   B3  (b) P    4 E   E2  P( Bi ) P  2   Bi  i 1

Now





3 1 1   10  3    10 1 2 1 3 1 4 1 3 (0)          10 10  2  10  2  10  4  10

B  1  P 3    E2  3

1 p p

2

 1  p2  p 1  0

5 1  p 1 2

 Least value of p 

   2sin    10  Statuments (1) and (2) are true and the reasoning is appropriate. o

 E  P  B2  E3  (c) P  3   P  B2   B2 

(d) P  E3   0  0  P  E3  

1 5

n

1 3 n 33. 1  C0      0.90 4 4

2 (0)  10 0 2 10

n

3     0.10 4

s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

3 1 4 1     10  3  10  4 

n



4    10  n  log 4 /310 3

2log 4  12  log 4  144  log 4 10    3

31. (0)4 k  2   (0) (1)4 k  2   (1) (2)4 k  2   (4) (3)4 k  2   (9) (4)4 k  2   (6) (5)

4k  2

  (5)



4 2  10 5 Statement (1) is false and statement (2) is true.



P AB  A  P A B 32. P     B P B P B  



 

Let P ( A  B )  p  A  p   ( p)  B 2

1 3



P ( A  B )  1

( Statement (2) is true)

 2 1  P ( A  B ) lies in  ,  15 3  Statement (1) and (2) are true and the reasoning is also appropriate.

P( E ) 

1 p

2 15

Now , P ( A  B ) 

(9)4 k  2   (1)



4 1   P( A  B) 5 3

 P( A  B ) 

(8)4 k  2   (4)

 

3

34. P ( A  B )  P( A)  P ( B )  P ( A  B )

(7) 4 k  2   (9)



  3

Statements (1) and (2) are true and the reasoning is appropriate.

(6)4 k  2   (6)



5 1 2

  p 

2

 

p B

Mathematics for IIT-JEE (Hints/Solutions)

35. For ellipse , 2a = 2(radius of cricle)



 16  a  r  b  r 2 1    25 



ar ;b

3r 5

 3r   r 2   (r )    5  1 3  2 Now , P ( E )  2 5 5  (r )

Statement (1) and (2) are true but the reasoning is not appropriate.

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1. P ( H ) 

There are 10 equiprobable events E1 to E10.

2 1 and P(T )  3 3

If n  2 then the last two throws must be heads and the previous all throws must be heads and tail or tail and head alternatively. 

In fact P ( Ei ) 

( HT )(TH )( HT )( HT )....(TH )( HH )

Probability of HT or TH 

4 9

4 4  9  4 5 1 9

2 number of throws    3

1  3  4 C1  6  3C1  2 C1  2 C1  2 C1  2 C1  7   6 10  C3  25

E  that one of each color was picked  P  1   A 4

2

 A P    P ( E1 ) E   1  P ( A)

C1 

1 10

6

C3 P ( A)



1 14

6. The bag contains equal number of white and green

4 5 Required probability  9  4 9 5

balls in events E1 and E6

3. For An , the game ends with either two heads on two tails

E1 E2 E3 E4 E5 E6 E7 E8 E9 E10

i

i

i 1

Probability that the bag contained 4 white balls , given

If n  2 , probability that experiment ends in minimum

4.

 A

 P  E   P(E )

s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

4 (from solution of part (1)) 5



1 for i  1 to 10 10

10

Then P ( A) 



2. Probability that experment ends with a head



1 10

5. Let A be the event of picking up three balls one of each color.

2 2  4  4    2  P ( E )  1      ....       9  9    3 





Probability of E1 

 2 C1  2 C1  2 C1  1  A P     P ( E6 )  6 E6  C3  E6     10 P    7 P( A)  A 25 E  E  Required probability is P  1   P  6   A  A

 E  P  1  An  White

Green

Red

1 1 1 1 2 2 2 3 3 4

1 2 3 4 1 2 3 1 2 1

4 3 2 1 3 2 1 2 1 1

Mathematics for IIT-JEE (Hints/Solutions)



1 1 3   14 7 14

7. X  3

 5   5   1  25 P( E )          6   6   6  216 8. X  3 2

3

5 1 5 1 P ( E )            .....  6 6 6  6

[50]

3

2 2   5   1    5   5       1        .....    6   6    6   6  

2



 5   1  1  25        6   6  1  5 / 6  36

 3

4

5

6

5 1 5 1 9. x  3 ; P  E1             ....  6 6 6  6 5 1  5 1 x  6 ; P  E2             ....  6 6  6  6

2

2

1 (6)

3



12 (6)

3



13 216

432 p 2 13

13. Let break points B1 and B2 lies at a distance of x and y meters respectively from live initial position 0



( x  y)2  (3)2  R 2

 ( x  y)2  9  25

5

5 1 P ( E1 )  6   6  Required probability   P ( E2 )  5 3  1      6 6 

2

1 1 4 1  4 1  4  P( E )                   6 6 6 6  6  6  6

 | x y|  4

(where 0  y , y  10 )

s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

25 36

10. Total number of ways of which papers of 4 students , can be checked by seven teachers  74

Now , choosing two teachers out of 7 is 7 C2  21 The number of ways in which 4 papers can be checked by exactly two teachers  24  2  14  Favourable ways  (21)(14)

 Required probability  

49 P  49 

(21)(14) 7

4



from graph , p 

6 49

6 =6 49



3

1

2

  3!    2  10 C2      270  2!   

3

 C  C1   C1  C1  C1  27 p1   26    6  C3   C3   200



Case (ii) John pick 1 black and 2 white balls  3 C  3C2 p2   16  C3  Now , p  p1  p2  

25 25 64 p  8 2 2 100

14. Total number of ways of dialing the last three digits such that exactly 2 of them are same

11. Case (i) John pick 2 black and 1 white ball 3

100  (36) 64  100 100

  2 C1  1C1  3C1  27      6 C3    200

p

1 270

 (1080) p  4

15.

27 100

100 p 9 3

12. Sum of 6 in three throws  (2 , 2 , 2) , (3 , 0 , 3) , (0 , 3 , 3) , (3 , 3 , 0) Total number of triangles equals to 8 C3  56 Total number of equilateral triangles whose one vertex

Mathematics for IIT-JEE (Hints/Solutions)

[51]

is A equals to 3. (AFH , AFC and AHC ) . Now total number of equilateral triangles are equal 8 3 8 3 Total number of isosceles triangle in plane ABCD is 4.  Total number of isosceles triangles (not equilateral) are equal to 4  6  24  Now total number of scalene triangles

to

 8C3  24  8  24 Now , all proper isosceles triangle (not equilateral) are right angled triangles. Because scalene triangle is formed by one edge , one force diagonal and one cube diagonal , so all scalene triangle are right angled triangles.  Total right angled triangle  24  24  48

5

3 P (occurrence of 3 , 4 , 5 on all the dice)    6 Now , required probability 5

5

5

5

 4  3  3  2          6 6 6  6 5

5

5

2  1  1     2     3    2 3 5

5

2 1 1         3  3  2

4

16. (a) m = 3 Smallest number appearing on dice is 3 and it may appear on one or more dice

s c i t a m e h t a a m r M a E e h JE iv .S T K t . II c L . je Er b O

P (m  3)  (Probability of occurrence of 3 , 4 , 5 , 6 on all the five dice) – (Probability of occurrence of 4 , 5 , 6 on all the dice) 5

5

4 3 Required probability       6 6



5

5

2 1      3 2 (b) n  4 5

5

5

5

4 3 2 1 P(n  4)             6 6 3 2 (c) 2  m  4

P ( E )  P (m  2)  P (m  3)  P (m  4) 5

5

5

5

5

5

 5  4   2   1   3  2              6 6 3 2 6 6 5

5

5

5

5

5 2 2 1 1 1                  6 3 3 2  2 3 5

 5 1       6   3

5

5

(d) m  2 and n  5

4 P (occurrence of 2, 3, 4, 5 on all the dice)    6 2 P (occurrence of 3 , 4 on all the dice)    6

5

5

5

3 P (occurrence of 2 , 3 , 4 on all the dice)    6

Mathematics for IIT-JEE (Hints/Solutions)

[52]