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11.
A box contains 4 red balls, 5 green balls and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?
  • A.
    		 2  
    5  
  • C.
    		 1  
    5  
  • B.
    		 3  
    5  
  • D.
    		 7  
    15  
  • Answer & Explanation
  • Report
Answer : [B]
Explanation :
 
Total number of balls = (4 + 5 + 6) = 15
 
  n(S) = 15    
Let E1 = event of drawing a red ball
and E2 = event of drawing a green ball
Then, E1 ∩ E2 = 0
  P(E1 or E2) = P(E1) + P(E2) =  
4
15
+
5
15
  =
9
15
=
3
5
   
 
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12.
What is the probability of getting atleast on head from 3 tosses of a coin?
  • A.
    		 11
    8
  • C.
    		 5
    8
  • B.
    		 7
    8
  • D.
    		 3
    8
  • Answer & Explanation
  • Report
Answer : [B]
Explanation :
When 3 coins are tossed the sample space is given by 2 x 2 x 2 = 8
The favourable case is getting atleast one head (i.e) 1,2 or 3 heads (or) tail should not occur in all the 3 coins simultaneously.
 
Favourable number of cases = 7
 
Required probability =
7
8
   
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13.
Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
  • A.
    		 1  
    4  
  • C.
    		 1  
    3  
  • B.
    		 1  
    2  
  • D.
    		 1  
    8  
  • Answer & Explanation
  • Report
Answer : [B]
Explanation :
 
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = Event of getting at least two heads = {THH, HTH, HHT, HHH}
 
  P(E) =
n(E)
n(E)
=
4
8
=
1
2
   
 
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14.
 In a box there are 6 blue, 6 red, 6 white and 6 black balls. Four balls are picked up randomly. What is the probability that all the four balls may not be blue?
  • A.
    		 10625
    10626
  • C.
    		 203
    204
  • B.
    		 3735
    3542
  • D.
    		 3537
    3542
  • Answer & Explanation
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Answer : [D]
Explanation :
Total no. of balls = 24
Total no. of cases of selecting 4 balls = 24C4
Total no. of ways of selecting all blue balls = 6C4
 
Required probability =
24C4 - 6C4
24C4
=
10626 - 15
10626
     
  =
10611
10626
=
3537
3542
 
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15.
Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are kings, is:
  • A.
    		 7  
    13  
  • C.
    		 63  
    221  
  • B.
    		 3  
    26  
  • D.
    		 55  
    221  
  • Answer & Explanation
  • Report
Answer : [D]
Explanation :
 
Clearly, n(S) = 52C2 =
(52 x 51)
2
 = 1326    
Let E1 = event of getting both red cards,
  E2 = event of getting both kings  
Then, E1 ∩ E2 = event of getting 2 kings of red cards.
 
  n(E1) = 26C2 =
(26 x 25)
(2 x 1)
 = 325; n(E2) = 4C2 =
(4 x 3)
(2 x 1)
 = 6    
n(E1 ∩ E2) = 2C2 = 1
 
  P(E1) =
n(E1)
n(S)
 =
325
1326
; P(E2) =
n(E2)
n(S)
 =
6
1326
; P(E1 ∩ E2) =
1
1326
 
 
  P(both red or both kings) = P(E1 ∪ E2)
  = P(E1) + P(E2) = P(E1 ∩ E2)  
  =  
325
1326
+
6
1326
-
1
1326
     
  =
330
1326
   
  =
55
221
   
 
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