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
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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
• Report
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
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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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