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41. A and B stand in a ring with 9 other persons. If the arrangement of th 11 persons is at random, then the probability that there are exactly 4 persons between A and B is:
  A.  
1 / 5
  B.  
1 / 5
  C.  
1 / 5
  D.  
1 / 5
     
   
View Answer

Shortcut:
X and Y stand in a ring with 'p' other persons. If the arrangement of all the persons is at random, then the probability that there are exactly 'q' persons between X and Y is given by
2 / p + 1
where q < p.
Here, p = 9, q = 4
Using these values in the shortcut, we get:
P(exactly 4 persons between A and B)
=
2 / 9 + 1
=
2 / 10
=
1 / 5

42. 10 persons are seated at a round table. What is the probability that two particular persons sit together?
  A.  
2 / 9
  B.  
2 / 9
  C.  
2 / 9
  D.  
2 / 9
     
   
View Answer

Shortcut:
If 'p' persons are seated at a round table then the probability that 'q' particular persons sit together is given by
(p − q)!q! / (p − 1)!
.
Here, p = 10, q = 2
Using these values in the shortcut, we get:
P(2 particular persons sit together) =
(10 − 2)!2! / (10 − 1)!

=
8! x 2! / 9!
=
8! x 2! / 9 x 8!
=
2 / 9

43. 12 persons are seated at a round table. Whast is the probability that 4 particular persons sit together?
  A.  
16 / 165
  B.  
4 / 165
  C.  
2 / 165
  D.  
4 / 135
     
   
View Answer

Shortcut:
If 'p' persons are seated at a round table then the probability that 'q' particular persons sit together is given by
(p − q)!q! / (p − 1)!
.
Here, p = 12, q = 4
Using these values in the shortcut, we get:
P(4 particular persons sit together) =
(12 − 4)!4! / (12 − 1)!

=
8! x 4! / 11!

=
8! x 4 x 3 x 2 x 1 / 11 x 10 x 9 x 8!

=
4 / 11 x 5 x 3

=
4 / 165

44. An unbiased coin is tossed 7 times, find the chance that exactly 5 times head will appear.
  A.  
21 / 128
  B.  
21 / 64
  C.  
21 / 32
  D.  
7 / 128
     
   
View Answer

Shortcut:
If an event is repeated, under similar conditions, exactly 'y' times, then the probability that event hpaaens exactly 'z' times is yCz x Pz x Qy − z, provided that
P = probability of happening
Q = Probability of not happening
i.e., P + Q = 1

Here, y = 7, z = 5
Using these values in the shortcut, we get:
P(exactly 5 times head) = 7C5 x
(
1 / 2
)
5 x
(
1 / 2
)
7 − 5
=
7! / 5!(7 − 5)!
x
(
1 / 2
)
5 x
(
1 / 2
)
2
=
7 x 6 x 5! / 5! x 2!
x
1 / 32
x
1 / 4

= 7 x 3 x
1 / 32
x
1 / 4

=
21 / 128

45. An unbiased coin is tossed 6 times, find the chance that exactly 4 times tail will appear.
  A.  
17 / 64
  B.  
15 / 256
  C.  
15 / 128
  D.  
15 / 64
     
   
View Answer

Shortcut:
If an event is repeated, under similar conditions, exactly 'y' times, then the probability that event hpaaens exactly 'z' times is yCz x Pz x Qy − z, provided that
P = probability of happening
Q = Probability of not happening
i.e., P + Q = 1

Here, y = 6, z = 4
Using these values in the shortcut, we get:
P(exactly 4 times tail) = 6C4 x
(
1 / 2
)
4 x
(
1 / 2
)
6 − 4
=
6! / 4!(6 − 4)!
x
(
1 / 2
)
4 x
(
1 / 2
)
2
=
6! / 4! x 2!
x
1 / 16
x
1 / 4

=
6 x 5 x 4! / 4! x 2
x
1 / 64

=
15 / 64

46. A basket contsins 5 white and 9 black balls. There is another basket which contains 7 white and 7 black balls. One ball is to be drawn from either of the two baskets.
(i) What is the probability of drawing a white ball?
(ii) What is the probability of drawing a b lack ball?
  A.  (i): 3/7; (ii): 4/7
  B.  (i): 3/7; (ii): 4/7
  C.  (i): 3/7; (ii): 4/7
  D.  (i): 3/7; (ii): 4/7
     
   
View Answer

Shortcut:
A box contains 'p1' white balls, 'q1' black balls. There is another box which contains 'p2' white balls and 'q2' black balls if one ball is to be drawn from either of the two baskets, then the probability of drawing (i) a white ball is given by
1 / 2
x
(
p1 / p1+q1
+
p2 / p2 + q2
)

(ii) a black ball is given by
1 / 2
x
(
q1 / p1+q1
+
q2 / p2 + q2
)

Here, p1 = 5, q1 = 9, p2 = 7, q2 = 7
Using these values in the shortcut, we get:
P(white ball) =
1 / 2
x
(
5 / 5 + 9
+
7 / 7 + 7
)

=
1 / 2
x
(
5 / 14
+
7 / 14
)

=
1 / 2
x
5 + 7 / 14

=
1 / 2
x
12 / 14
=
3 / 7

P(black ball) =
1 / 2
x
(
9 / 5 + 9
+
7 / 7 + 7
)

=
1 / 2
x
(
9 / 14
+
7 / 14
)

=
1 / 2
x
9 + 7 / 14

=
1 / 2
x
16 / 14

=
4 / 7

47. A bag contsins 9 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability thqt they are alternatively of different colours?
  A.  
18 / 65
  B.  
9 / 13
  C.  
9 / 65
  D.  
19 / 65
     
   
View Answer

Shortcut:
A box contains 'p' red balls, 'q' white balls. four balls are drawn out one by one and not replaced. Then the probability that they are alternatively of different colours is given by
2p(p − 1)q(q − 1) / (p + q)(p + q − 1)(p + q − 2)(p + q − 3)

Here, p = 9, q = 7
Using these values in the shortcut, we get:
P(alternatively of different colour) =
2 x 9(9 − 1)7(7 − 1) / (9 + 7)(9 + 7 − 1)(9 + 7 − 2)(9 + 7 − 3)

=
2 x 9 x 8 x 7 x 6 / 16 x 15 x 14 x 13
=
9 / 65

Note: Wherever we find the word AND between two events, we use multiplication. Mark that both also means first and second. On the other hand, if the two events are joined with OR, we use addition as in the above question.

48. A bag contsins 5 white and 5 red balls. Two draws of one ball each are made without replacement. What is the probability that one is red and other white?
  A.  
5 / 9
  B.  
15 / 18
  C.  
5 / 18
  D.  
7 / 9
     
   
View Answer

Shortcut:
A box contains 'p' white balls, 'q' red balls. If two draws of one ball each are made without replacement, then the probability that one is red and the other white is given by
2pq / (p + q)(p + q − 1)

Note: The above shortcut may be asked as: 'A bag contains 'p' white and 'q' red balls. If two balls are drawn in succession at random, then the probability that one of them is white and the other red, is given by
2pq / (p + q)(p + q − 1)

Here, p = 5, q = 5
Using these values in the shortcut, we get:
P(one is read and the other white) =
2 x 5 x 5 / (5 + 5)(5 + 5 − 1)

=
2 x 5 x 5 / 10 x 9
=
5 / 9

49. A bag a contains 7 red and 8 black balls. Two draws of three balls each are made, the ball being replaced after the first draw. What is the chance that the balls were red in the first draw and black in the second?
  A.  
8 / 645
  B.  
8 / 845
  C.  
4 / 855
  D.  
24 / 745
     
   
View Answer

Shortcut:
A box contains 'p' red balls, 'q' black balls. If two draws of three balls each are made, the ball being replaced after the first draw, then the chance that the balls were red in the first draw and black in the second draw is given by
pC3 x qC3 /
(
(p + q)C3
)
2

Here, p = 7 , q = 8
Using these values in the shortcut, we get:
7C3 x 8C3 /
(
(7 + 8)C3
)
2

=
35 x 56 / (5 x 7 x 13)2

=
35 x 56 / 5 x 7 x 13 x 5 x 7 x 13

=
8 / 13 x 5 x 13

=
8 / 845

50. A box contains 4 black balls, 6 red balls and 8 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are not of the same colour?
  A.  
114 / 153
  B.  
104 / 155
  C.  
104 / 153
  D.  
52 / 153
     
   
View Answer

Shortcut:
A box contains p black balls, q red balls and r green balls. 'z' balls are drawn from the box at random. The probability that all the balls are of the same colour is given by
pCz + qCz + rCz / (p + q + r)Cz
, where z < p, q, r
Here, p = 4 , q = 6, r = 8, z = 2
Using these values in the shortcut, we get:
4C2 + 6C2 + 8C2 / (4 + 6 + 8)C2

=
6 + 15 + 28 / 18C2

=
49 / 153

P(all the 2 balls are not of the same color) = 1 −
49 / 153

=
153 − 49 / 153
=
104 / 153

Note: The probability that both the balls are not of the same color is given by 1 − P (Probability of same color).

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