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11. A bag contains 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black?.
  A.  
4 / 7
  B.  
1 / 6
  C.  
5 / 9
  D.  
3 / 8
     
   
View Answer

Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is black.
Then, n(S) = Number of ways of drawing 3 balls out of 9 = 9C3 =
9 x 8 x 7 / 3 x 2 x 1
= 84
n(E) = No. of ways of drawing 2 balls out of 5 and 1 ball out of 4 = 5C2 + 4C1
=
5 x 4 / 2
+
4 x 3! / 3!
= 10 + 4 = 14
∴ P(E) =
n(E) / n(S)
=
14 / 84
=
1 / 6

12. An urn contains 9 red, 7 white and 4 black balls. A ball is drawn at random. What is the probability that the ball drawn is not red?
  A.  
11 / 20
  B.  
13 / 23
  C.  
17 / 11
  D.  
21 / 11
     
   
View Answer

n(S) = Number of ways of drawing 1 red balls out of 20 = 20C1 =
20 x 19! / 1!(20 − 1)!
=
20 x 19! / 19!
= 20
n(E) = No. of ways of drawing 1 red ball out of 9 red ball = 9C1 =
9! / 1!(9 − 1)!
=
9 x 8! / 8!
= 9
∴ P(E) =
n(E) / n(S)
=
9 / 20

∴ P(no red ball) = 1 −
9 / 20
=
11 / 20

13. Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a nukmber which is a multiple of 3?
  A.  
13 / 4
  B.  
11 / 3
  C.  
4 / 9
  D.  
3 / 10
     
   
View Answer

S = {1, 2, 3,......20}
n(S) = 20
E = {3, 6, 9, 12, 15, 18}
n(E) = 6
∴ P(E) =
n(E) / n(S)
=
6 / 20
=
3 / 10

14. Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a nukmber which is a multiple of 3 or 7?
  A.  
2 / 9
  B.  
7 / 3
  C.  
2 / 5
  D.  
5 / 7
     
   
View Answer

S = {1, 2, 3, ......., 20}
n(S) = 20
E = {3, 6, 9, 12, 15, 18, 7, 14}
n(E) = 8
∴ P(E) =
n(E) / n(S)
=
8 / 20
=
2 / 5

15. (i) What is the chance that a leap year selected randomly will have 53 Sundays?
(ii) What is the chance, if the year selected is a not a leap year?
  A.  (i):
2 / 7
; (ii):
3 / 5
  B.  (i):
3 / 5
; (ii):
6 / 1
  C.  (i):
5 / 2
; (ii):
7 / 2
  D.  (i):
8 / 9
; (ii):
4 / 3
     
   
View Answer

(i) A leap year has 366 days so it has 52 complete weeks and 2 moe days.
The two days can be {Sunday & Monday, Monday & Tuesday, Tuesday & Wednesday, Wednesday & Thursday, Thursday & Friday, Friday & Saturday, Saturday & Sunday}, i.e., n(S) = 7
Out of these 7 cases, cases favourable for more Sundays are {Sunday & Monday, Saturday & Sunday}, i.e., n{E} = 2
∴ P(E) =
n(E) / n(S)
=
2 / 7


(ii) When the year is not a leap year, it has 52 complete weeks and 1 more day
and that can be {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} i.e., n(S) = 7
Out of these 7 cases favourable for one more Sunday is {Sunday}, n(E) = 1
∴ P(E) =
n(E) / n(S)
=
1 / 7

16. What is the probability that a leap year selected randomly will have 53 Mondays?
  A.  
3 / 8
  B.  
5 / 2
  C.  
2 / 7
  D.  
5 / 7
     
   
View Answer

(i) A leap year has 366 days so it has 52 complete weeks and 2 more days. The two days can be {Sunday & Monday, Monday & Tuesday, Tuesday & Wednesday, Wednesday & Thursday, Thursday & Friday, Friday & Saturday, Saturday & Sunday}, i.e., n(S) = 7
Out of these 7 cases, cases favourable for more Mondays are {Sunday & Monday, Monday & Tuesday}, i.e., n{E} = 2
∴ P(E) =
n(E) / n(S)
=
2 / 7

17. What is the probability that an ordinary year has 53 Sundays?
  A.  
7 / 3
  B.  
9 / 2
  C.  
5 / 2
  D.  
1 / 7
     
   
View Answer

A ordinary year has 365 days so it has 52 complete weeks and 1 more day.
So, the probability that this day is Sunday is
∴ P(E) =
n(E) / n(S)
=
1 / 7

18. When two dice are thrown, what is the probability that
(i) Sum of numbers appeared is 6 and 7?
(ii) Sum of numbers appeared <= 8?
(iii) Sum of numbers is an odd number?
(iv) Sum of numbers is a multiple of 3?
(v) Number shown are equal?
(vi) The difference of the numbers is 2?
(vii) Sum of the numbers is at least 5.
  A.  (i):
1 / 3
; (ii):
13 / 18
; (iii):
1 / 2
; (iv):
5 / 4
; (v):
7 / 2
; (vi):
9 / 2
; (vii):
15 / 2
  B.  (i):
1 / 6
; (ii):
13 / 18
; (iii):
1 / 2
; (iv):
1 / 3
; (v):
1 / 6
; (vi):
2 / 9
; (vii):
5 / 18
  C.  (i):
18 / 5
; (ii):
18 / 7
; (iii):
6 / 1
; (iv):
6 / 7
; (v):
8 / 3
; (vi):
19 / 5
; (vii):
53 / 9
  D.  (i):
18 / 13
; (ii):
9 / 2
; (iii):
18 / 5
; (iv):
1 / 6
; (v):
13 / 2
; (vi):
9 / 4
; (vii):
1 / 6
     
   
View Answer

i-(a) For 6, P(E) =
n(E) / n(S)
=
5 / 36

i-(b) For 7, P(E) =
n(E) / n(S)
=
6 / 36
=
1 / 6


(ii) Desired sums of the number are 2, 3, 4, 5, 6, 7 and 8;
n(S) = 1 + 2 + 3 + 4 + 5 + 6 + 5 = 26
∴
n(E) / n(S)
=
26 / 36
=
13 / 18


(iii) Desired sums of he number are 3, 5, 7, 9 and 11;
n(S) = 2 + 4 + 6 + 4 + 2 = 18
∴
n(E) / n(S)
=
18 / 36
=
1 / 2


(iv) Desired sums of he numbers are 3, 6, 9 and 12;
n(S) = 2 + 5 + 4 + 1 = 12
∴
n(E) / n(S)
=
12 / 36
=
1 / 3


(v) Events are {(1,1), (2, 2), (3, 3), (4, 4),(5, 5), (6, 6);
n(S) = 6
∴
n(E) / n(S)
=
6 / 36
=
1 / 6


(vi) Events are {(3,1), (4, 2), (5, 3), (6, 4), (4, 6), (3, 5), (2,4), (1,3)};
n(S) = 8
∴
n(E) / n(S)
=
8 / 36
=
2 / 9


(vii) Events are either 2 or 3 or 4 or 5;
n(E) = 1 + 2 + 3 + 4 = 10
n(S) = 36
∴
n(E) / n(S)
=
10 / 36
=
5 / 18

19. In a throw of a dice, the probability of getting a prime number is:
  A.  
1 / 3
  B.  
3 / 2
  C.  
2 / 5
  D.  
7 / 2
     
   
View Answer

Here S = {1, 2, 3, 4, 5, 6} and E = {2,3,5};
n(S) = 6
∴ P(E) =
n(E) / n(S)
=
3 / 6
=
1 / 2

20. In a simultaneous throw of two dice, what is the probability of getting a doublet?
  A.  
3 / 7
  B.  
7 / 4
  C.  
1 / 6
  D.  
8 / 3
     
   
View Answer

In a simultaneous throw of 2 dice n(S) = 6 x 6 = 36
Let (E) = {(1,1), (2, 2), (3, 3), (4, 4),(5, 5), (6, 6);
n(E) = 6
∴ P(E) =
n(E) / n(S)
=
6 / 36
=
1 / 6

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