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21. In a sinultaneous throw of two dice, what is the probability of getting a total of 10 or 11?.
  A.  
9 / 43
  B.  
7 / 41
  C.  
5 / 36
  D.  
3 / 31
     
   
View Answer

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

22. In a single throw of two dice what is the probability of not getting the same number on both the dice?
  A.  
2 / 9
  B.  
3 / 7
  C.  
5 / 7
  D.  
5 / 6
     
   
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

So, P(not-E) = 1 −
1 / 6
=
5 / 6

23. A card is drawn from a pack of cards. What is the probability that it is:
(i) A card of balck suit?
(ii) A spade card?
(iii) An honous card of red suit?
(iv)An honours card of club?
(v) A card having the number less than 7?
(vi)A card having the number a multiple of 3?
(vii)A king or a queen?
(viii)A digit-card of heart?
(ix)A jack of black suit?
  A.  (i):
1 / 2
; (ii):
1 / 4
; (iii):
2 / 13
; (iv):
1 / 13
; (v):
5 / 13
; (vi)
3 / 13
(vii): P(a king or a queen):
2 / 13
; (viii):
9 / 52
; (ix):
1 / 26
  B.  (i):
1 / 2
; (ii):
1 / 4
; (iii):
2 / 13
; (iv):
1 / 13
; (v):
5 / 13
; (vi)
3 / 13
(vii): P(a king or a queen):
2 / 13
; (viii):
9 / 52
; (ix):
1 / 26
  C.  (i):
1 / 2
; (ii):
1 / 4
; (iii):
2 / 13
; (iv):
1 / 13
; (v):
5 / 13
; (vi)
3 / 13
(vii): P(a king or a queen):
2 / 13
; (viii):
9 / 52
; (ix):
1 / 26
  D.  (i):
1 / 2
; (ii):
1 / 4
; (iii):
2 / 13
; (iv):
1 / 13
; (v):
5 / 13
; (vi)
3 / 13
(vii): P(a king or a queen):
2 / 13
; (viii):
9 / 52
; (ix):
1 / 26
     
   
View Answer

Following chart will be helpful to solve the problems based on cards.
Cart: A pack of cards has a total of 52 cards.

Red Suit (26)
Black Suit (26)
Diamond (13)
Heart (13)
Spade (13)
Club (13)

The numbers in the brackets show the respective number of cards in that category.
Each of Diamond, Heart, Spade and Club contains nine digit-cards 2, 3, 4, 5, 6, 7, 8, 9 and 10 (a total of 9 x 4 = 36 digit- cards) along with four Honour cards Ace, King, Queen and Jack (a total of 4 x 4 = 16 Honour cards)

For all the above cases n(S) = 52C1 = 52
(i)
26 / 52
=
1 / 2


(ii)
13 / 52
=
1 / 4


(iii)
4 x 2 / 52
=
2 / 13


(iv)
4 / 52
=
1 / 13


(v)
5 x 4 / 52
=
5 / 13


(vi)
3 x 4 / 52
=
3 / 13


(vii) P(a king) =
4 / 52
=
1 / 13


P( a Queen) =
4 / 52
=
1 / 13

∴ P(a king or a queen) =
1 / 13
+
1 / 13
=
2 / 13


(viii)
9 / 52


(ix)
2 / 52
=
1 / 26

24. From a pack of 52 cards, 2 cards are drawn at rondom. What is the probability that it has
(i) both the aces?
(ii) exactly one queen?
(iii) no honours card? (iv) no digit-card? (v) One king and one Queen?
  A.  (i):
221 / 20
; (ii):
663 / 8
; (iii):
105 / 221
; (iv):
1 / 221
; (v):
32 / 221
  B.  (i):
1 / 221
; (ii):
32 / 221
; (iii):
105 / 221
; (iv):
20 / 221
; (v):
8 / 663
  C.  (i):
20 / 221
; (ii):
8 / 663
; (iii):
31 / 220
; (iv):
53 / 221
; (v):
8 / 663
  D.  (i):
105 / 221
; (ii):
1 / 221
; (iii):
103 / 221
; (iv):
6 / 661
; (v):
20 / 221
     
   
View Answer

Following chart will be helpful to solve the problems based on cards.

Cart: A pack of cards has a total of 52 cards.

Red Suit (26)
Black Suit (26)
Diamond (13)
Heart (13)
Spade (13)
Club (13)

The numbers in the brackets show the respective number of cards in that category.

Each of Diamond, Heart, Spade and Club contains nine digit-cards 2, 3, 4, 5, 6, 7, 8, 9 and 10 (a total of 9 x 4 = 36 digit- cards) along with four Honour cards Ace, King, Queen and Jack (a total of 4 x 4 = 16 Honour cards)

For all the above cases n(S) = 52C2 =
52x51 / 2
= 26 x 51

(i) Total number of aces = 4
∴ n(E) = 4C2 =
4 x 3 / 2
= 6
∴ P(E) =
6 / 26x51
=
1 / 221


(ii) Total number of Queens = 4
Selection of 1 Queen card out of 4 can be done in 4C1 = 4 ways.
He can select the remaining 1 card from the remaining (52 − 4 = 48) cards, Now, cards in 48C1 = 48 ways
∴ n(E) = 4 x 48
∴ P(E) =
4 x 48 / 26 x 51
=
32 / 221


(iii) Total no. of honous card = 16
To have no honours card, he has to select two cards out of the remaining 52 − 16 = 36 cards which he can do in 36C2 =
36 x 35 / 2
= 18 x 24 ways
∴ P(E) =
18 x 35 / 26x51
=
105 / 221


(iv) P(E) =
16C2 / 26 x 51
=
8 x 15 / 26 x 51
=
20 / 221


(v) n(E) = 4C1 x 4C1 = 4 x 4 = 16
∴ P(E) =
16 / 26 x 51
=
8 / 663

25. From a pack of 52 cards, 3 cards are drawn. What is the probability that it has
(i) all three aces?
(ii) no queen?
(iii) one ace, one king and one queen?
(iv) one ace and two jacks?
(v) two digit-cards and one honours card of black suit?
  A.  (i):
252 / 1105
; (ii):
4324 / 5525
; (iii)
6 / 5525
; (iv)
16 / 5525
; (v):
2 / 5525
  B.  (i):
16 / 5525
; (ii):
252 / 525
; (iii)
5 / 5525
; (iv)
1 / 5525
; (v):
6 / 5525
  C.  (i):
6 / 5525
; (ii):
252 / 1105
; (iii)
1 / 5525
; (iv)
16 / 5525
; (v):
4324 / 5524
  D.  (i):
1 / 5525
; (ii):
4324 / 5525
; (iii)
16 / 5525
; (iv)
6 / 5525
; (v):
252 / 1105
     
   
View Answer

Following chart will be helpful to solve the problems based on cards.

Cart: A pack of cards has a total of 52 cards.

Red Suit (26)
Black Suit (26)
Diamond (13)
Heart (13)
Spade (13)
Club (13)


For all the above cases n(S) = 52C3 =
52 x 51 x 50 / 3 x 2
= 26 x 17 x 50

(i) Total number of aces = 4
∴ n(E) = 4C3 =
4 x 3 / 3
= 4
∴ P(E) =
4 / 26 x 17 x 50
=
1 / 5525


(ii) n(E) = 48C3 = 8 x 47 x 46
∴ P(E) =
8 x 47 x 46 / 26 x 17 x 50
=
4324 / 5525


(iii) n(E) = 4C1 x 4C1 x 4C1 = 4 x 4 x 4
∴ P(E) =
4 x 4 x 4 / 26 x 17 x 50
=
16 / 5525


(iv) n(E) = 4C1 x 4C2 = 4 x 6
∴ P(E) =
4 x 6 / 26 x 17 x 50
=
6 / 5525


(v) n(E) = 36C2 x 8C1 = 18 x 35 x 8
∴ P(E) =
18 x 35 x 8 / 26 x 17 x 50
=
252 / 1105

26. A card is drawn from a pack of 52 cards. A card is drawn at random. What is the probability that it is neither a heart nor a king?
  A.  
9 / 13
  B.  
4 / 13
  C.  
1 / 26
  D.  
2 / 13
     
   
View Answer

Following chart will be helpful to solve the problems based on cards.

Cart: A pack of cards has a total of 52 cards.

Red Suit (26)
Black Suit (26)
Diamond (13)
Heart (13)
Spade (13)
Club (13)


There are 13 hearts and 3 more kings.

∴ P(heart or a king) =
13 + 3 / 52
=
4 / 13

∴ P(neither a heart nor a king) = 1 −
4 / 13
=
9 / 13


27. A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a spade or a king?
  A.  
7 / 13
  B.  
5 / 13
  C.  
4 / 13
  D.  
9 / 13
     
   
View Answer

Following chart will be helpful to solve the problems based on cards.

Cart: A pack of cards has a total of 52 cards.

Red Suit (26)
Black Suit (26)
Diamond (13)
Heart (13)
Spade (13)
Club (13)


Let E and F be the event of getting a spade and that of getting a king respectively.
Then E ∩ F is the event of getting a king of spade
∴ n(E) = 13
n(F) = 4
and n(E∩F) = 1
So, P(E) =
13 / 52
=
1 / 4

∴ P(F) =
4 / 52
=
1 / 13

and P(E∩F) =
1 / 52

∴ P(a spade or a king) = P(E ∪ F) = P(E) + P(F) − P(E∩F)
=
1 / 4
+
1 / 13
−
1 / 52
=
4 / 13

28. A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours?
  A.  
2 / 11
  B.  
3 / 11
  C.  
5 / 11
  D.  
6 / 11
     
   
View Answer

Shortcut:
If a bag contains 'p' red, 'q' yellow and 'r' green balls, 3 balls are drawn randomly, then the probability of the balls drawn contain balls of different colour is
6pqr / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or, P(E) =
pC1 x qC1 x rC1 / (p + q + r)C3

Here, p = 3, q = 5, r = 4
Using these values in the shortcut, we get:
P(3 balls of different color) =
6 x 3 x 5 x 4 / (3 + 5 + 4)(3 + 5 + 4 − 1)(3 + 5 + 4 − 2)

=
6 x 3 x 5 x 4 / 12 x 11 x 10
=
3 / 11

29. A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that balls drawn contain exactly two green balls?
  A.  
16 / 55
  B.  
13 / 55
  C.  
23 / 55
  D.  
12 / 55
     
   
View Answer

Shortcut:
If a bag contains 'p' red, 'q' yellow and 'r' green balls, 3 balls are drawn randomly, then the probability of the balls drawn contain (i) exactly 2 green balls is given by
3r(r − 1)(p + q) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(p + q)C1 x rC2 / (p + q + r)C3


(ii) exactly 2 yellow balls is given by:
3q(q − 1)(p + r) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(p + r)C1 x qC2 / (p + q + r)C3

(iii) exactly 2 red balls is given by:
3p(p − 1)(q + r) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(q + r)C1 x pC2 / (p + q + r)C3

Here, p = 3, q = 5, r = 4
Using these values in the shortcut (i), we get:
P(3 balls having 2 exactly green balls) =
3 x 4(4 − 1)(3 + 5) / (3 + 5 + 4)(3 + 5 + 4 − 1)(3 + 5 + 4 − 2)

=
3 x 4 x 3 x 8 / 12 x 11 x 10
=
3 x 4 / 11 x 5
=
12 / 55

30. A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain no yellow ball?
  A.  
3 / 44
  B.  
9 / 44
  C.  
7 / 44
  D.  
5 / 44
     
   
View Answer

Shortcut:
If a bag contains 'p' red, 'q' yellow and 'r' green balls, 3 balls are drawn randomly, then the probability of the balls drawn contain
(i) no yellow ball is given by
(p + r)(p + r − 1)(p + r − 2) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(p + r)C3 / (p + q + r)C3


(ii) no red ball is given by
(q + r)(q + r − 1)(q + r − 2) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(q + r)C3 / (p + q + r)C3


(iii) no green ball is given by
(p + q)(p + q − 1)(p + q − 2) / (p + q + r)(p + q + r − 1)(p + q + r − 2)

or,
(p + q)C3 / (p + q + r)C3


Here, p = 3, q = 5, r = 4
Using these values in the shortcut (i), we get:
P(3 balls having no yellow ball)
=
(3 + 4)(3 + 4 − 1)(3 + 4 − 2) / (3 + 5 + 4)(3 + 5 + 4 − 1)(3 + 5 + 4 − 2)

=
7 x 6 x 5 / 12 x 11 x 10
=
7 / 2 x 11 x 2

=
7 / 44

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