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1. A dice is thrown. What is the probability that the number shown on the dice is:
(i) An even number.
(ii) An odd number.
(ii) A number divisible by 2.
(iv) A number divisible by 3.
(v) A number less than 4.
(vi) A number less than or equal to 4.
(vii) A number greater than 6.
(viii) A number less than or equal to 6.
  A.  (i): 1/3; (ii): 1/2; (iii): 5/2; (iv): 1/3; (v): 1/2; (vi): 2/3; (vii): 1; (viii): 1
  B.  (i): 1/2; (ii): 1/2; (iii): 1/2; (iv): 1/3; (v): 1/2; (vi): 2/3; (vii): 0; (viii): 1
  C.  (i): 5/2; (ii): 1/2; (iii): 3/2; (iv): 1/3; (v): 1/2; (vi): 2/3; (vii): 1; (viii): 1
  D.  (i): 1/2; (ii): 3/2; (iii): 1/2; (iv): 4/3; (v): 1/2; (vi): 2/3; (vii): 1; (viii): 3
     
   
View Answer

Shortcut:
Probability of an event (E) is denoted by P(E) and is defined as
P(E) =
n(E) / n(S)
=
No. of desired events / total number of events (ie no. of sample space)

In all the above cases, S = {1,2,3,4,5,6}
n(S) = 6

(i) E(an even no.) = {2,4,6}, n(E) = 3
∴ P(E) =
n(E) / n(S)
=
3 / 6
=
1 / 2


(ii) E(an odd no.) = {1,3,5}, n(E) = 3
∴ P(E) =
n(E) / n(S)
=
3 / 6
=
1 / 2


(iii) E(a no. divisible by 2) = {2,4,6}, n(E) = 3
∴ P(E) =
n(E) / n(S)
=
3 / 6
=
1 / 2


(iv) E(a no. divisible by 3) = {3,6}, n(E) = 2
∴ P(E) =
n(E) / n(S)
=
2 / 6
=
1 / 3


(v) E(a no. less than 4) = {1,2,3}, n(E) = 3
∴ P(E) =
n(E) / n(S)
=
3 / 6
=
1 / 2


(vi) E(a no. less than or equal to 4) = {1, 2, 3, 4}, n(E) = 4
∴ P(E) =
n(E) / n(S)
=
4 / 6
=
2 / 3


(vii) E(a no. greater than 6) = {}, i.e., there is no number greater than 6 in the sample space.
∴ P(E) =
n(E) / n(S)
=
0 / 6
= 0
Probability of an impossible event = 0

(viii) E(a no. less than or equal to 6) = {1,2,3,4,5, 6}, n(E) = 6
∴ P(E) =
n(E) / n(S)
=
6 / 6
= 1
Probability of a certain event = 1
Note: 0 ≤ P(E) ≤ 1

2. In a box carrying one dozen of oranges, one third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good? (SBI)
  A.  
54 / 55
  B.  
52 / 57
  C.  
60 / 61
  D.  
53 / 57
     
   
View Answer

(i) n(S) = 12C3
=
12 x 11 x 10 / 3 x 2
= 2 x 11 x 10 = 220
No.of selection of 3 oranges out of the total 12 oranges = 12C3 = 2 x 11 x 10 = 220
No. of selection of 3 bad oranges out of the total 4 bad oranges = 4C3 = 4
∴ n(E) = no. of desired selection of oranges = 220 − 4 = 216
∴ P(E) =
n(E) / n(S)
=
216 / 220
=
54 / 55

3. Out of 15 students studying in a class, 7 are from Maharashtra, 5 are from Karnataka and 3 are form Goa. Four students are to be selected at random. What are the chances that at least one is form Karnataka? (BSRB)
  A.  
13 / 11
  B.  
13 / 15
  C.  
17 / 11
  D.  
11 / 13
     
   
View Answer

Total possible ways of selecting 4 students out of 15 students = n(S) = 15C4
=
15 x 14 x 13 x 12 / 4 x 3 x 2 x 1
= 1365
The no. of ways of selecting 4 studetns in which no student belongs to Karnataka = 10C4
∴ Number of ways of selecting at least one student from Karnataka
= 15C4 − 10C4 = 1155
∴ P(E) =
1155 / 1365
=
77 / 91
=
11 / 13

4. The probability that a teacher will give one surprise test during any class meeting in a week is
1 / 5
. If a student is absent twice, what is the probability that he will miss at least one test. (BSRB)
  A.  
3 / 14
  B.  
4 / 13
  C.  
1 / 15
  D.  
2 / 11
     
   
View Answer

The probability of absenting of the student in the class =
2 / 6
=
1 / 3

∴ the proability of missing his test =
1 / 5
x
1 / 3
=
1 / 15

5. In a throw of a coin, the probability of getting a head is:
  A.  
2 / 7
  B.  
1 / 2
  C.  
2 / 1
  D.  
3 / 2
     
   
View Answer

Here S = {H, T} and E = {H}
∴ P(E) =
n(E) / n(S)
=
1 / 2

6. In a simultaneous throw of two coins, the probability of getting at least one head is:
  A.  
4 / 7
  B.  
4 / 5
  C.  
3 / 4
  D.  
5 / 7
     
   
View Answer

Here S = {HH, HT, TT, TH}
and E = {HH, HT, TH}
∴ P(E) =
n(E) / n(S)
=
3 / 4

7. Three unbiased coins are tossed, what is the probability of getting exactly two heads?
  A.  
5 / 3
  B.  
2 / 7
  C.  
8 / 3
  D.  
3 / 8
     
   
View Answer

Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
and E (Event of getting exactly two heads) = {HHT, HTH, THH}
∴ P(E) =
n(E) / n(S)
=
3 / 8

8. Three unbiased coins aer tossed. What is the probability of getting at most 2 heads?
  A.  
3 / 5
  B.  
8 / 7
  C.  
5 / 9
  D.  
2 / 1
     
   
View Answer

Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
and E (getting 0 or 1 or 2 heads) = {TTT, TTH, THT, HTT, HHT, HTH, THH}
∴ P(E) =
n(E) / n(S)
=
7 / 8

9. A bag contains 6 black balls and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
  A.  
4 / 7
  B.  
7 / 2
  C.  
9 / 5
  D.  
8 / 5
     
   
View Answer

Total no. of balls = 6 + 8 = 14
Probility of drawing 1 ball out of 14 = 14C1
or,
14! / 1!(14 − 1)!
=
14 x 13! / 13!
= 14
No. of white balls = 8
Probility of drawing 1 white ball out of 8 white balls = 8C1
or,
8! / 1!(8 − 1)!
=
8 x 7! / 7!
= 8
∴ P(drawing 1 white ball) =
8 / 14
=
4 / 7

10. A bag contains 8 red and 5 white balls, 2 balls are drawn at random. What is the probability that both are white?
  A.  
2 / 35
  B.  
7 / 31
  C.  
5 / 39
  D.  
8 / 35
     
   
View Answer

n(S) = No. of ways of drawing 2 balls out of 13 = 13C2
=
13 x 12 x 11! / 2!(13 − 2)!

=
13 x 12 x 11! / 2! x 11!

=
13 x 12 / 2
= 78
n(E) = No. of ways of drawing 2 balls out of 5 = 5C2
=
5 x 4 x 3! / 2!(5 − 2)!

=
5 x 4 x 3! / 2! x 3!

=
5 x 4 / 2
= 10
∴ P(E) =
n(E) / n(S)
=
10 / 78
=
5 / 39

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