aptitude : : probability
 
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   Exercise 2
   Exercise 1
1.
A bag contains 5 black and 7 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that
(i) both the balls drawn were black;
(ii) both were white;
(iii) the first ball was white and the second black;
(iv) the first ball was black and the second white?
 
A.
20/144, 49/144, 35/144, 35/144

B.

25/144, 45/144, 35/144, 35/144
C.
25/144, 49/144, 15/144, 35/144
D.
25/144, 49/144, 35/144, 35/144
   
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Answer: D
Explanation :
The events are independent and capable of simultaneous occurrence.
The rule of multiplication would be applied. The probability that
(i) both the balls were black = (5/12) x (5/12) = (25/144)
(ii)    both the balls were white = (7/12) x (7/12) = (49/144)
(iii)   the first was white and second black = (7/12) x (5/12) = (35/144)
(iv)    the first was black and second white = (5/12) x (7/12) = (35/144) 






 
2.
A bag contains 5 red and 8 black balls. Two draws of three bal!s 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.
140/20449

B.

140/20445
C.
140/20549
D.
120/20449
   
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Answer: A
Explanation :
Required probability = (5c3/13c3) x (8c3/13c3) = 140/20449






 
3.
Two dice are thrown together. What is the ,·rob. c::iny tktt tiw sum of the numbers on the two faces is divisible by 4 or 6 ?
 
A.
9/16

B.

7/18
C.
6/17
D.
3/4
   
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Answer: B
Explanation :
n(S)    =6x6=36
E   = {(1,3)(1,5) (2,2) (2,4) (2,6) (3, I) (3,3) (3,5) (4,2) (4,4) (5, I) (5,3) (6,2) (6,6)}
n(E) = 14
p(E) = 94/36 = 7/18






 
4.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that  none of the balls drawn is blue?
 
A.
10/21

B.

11/21
C.
2/7
D.
5/7
   
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Answer: A
Explanation :
Total balls  = 2 + 3 + 2 = 7
n(S) = 7c2 = (7x6)/(2x1) = 21
n (E) = number of ways of drawing 2 balls out of (2 + 3) balls
=5c2 = (5x4)/(2x1) = 10
p(E) = n(E)/n(S) = 10/21






 
5.
A boy contains 6 black and 8 white balls one ball is drawn at random. What is the probability that the ball drawn is white?
 
A.
3/4

B.

4/7
C.
1/8
D.
3/7
   
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Answer: B
Explanation :
Total balls= 6 + 8 = 14, White balls= 8 
p(drawing white ball) = 8/14 = 4/7






 
6.
The letters of the word 'ARTICLE' are arranged in different ways randomly. What is the chance that the vowels occupy the even places?
 
A.
1/35

B.

2/5
C.
3/4
D.
4/3
   
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Answer: A
Explanation :
The 7 different letters ofthe word ARTICLE can be arranged in 7!
ways, i.e., n(S) = 7!
n(E)=  3 P3  x  4 P4   =3!x4!=6x24
p(E) = (6x24)/7! = 1/35






 
7.
A card is drawn from a pack of cards. What is the probability that it is
(i) a card of black suit?
(ii) a spade card?
(iii) an honours card of red suit?
(iv) an honours card of club?
(v) a card having the number less than 7?
 
A.
1/2,1/4,2/13,1/13,5/13

B.

1,1/4,2/13,1/13,5/13
C.
1/2,1,2/13,1/13,5/13
D.
1/2,1/4,2/11,1/13,5/13
   
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Answer: A
Explanation :
For all the above cases 52c1 = 52
(i) 26/52 = 1/2
(ii)13/52 = 1/4
(iii)(4x2)/52 = 2/13
(iv) 4/52 = 1/13
(v) (5x4)/52 = 5/13






 
8.
When two dice are thrown, what is the probability that
(i) sum of numbers appeared is 6 and 7?
(ii) sum of numbers appeared    8?
(iii) sum of numbers is an odd no.?
(iv) sum of numbers is a multiple of 3?
(v) numbers shown are equal?
(vi) the difference of the numbers is 2?
 
A.
1/3,13/18,1/2,1/3,1/6,2/9

B.

1/6,12/18,1/2,1/3,1/6,2/9
C.
1/6,13/18,1/4,1/3,1/6,2/9
D.
1/6,13/18,1/2,1/3,1/6,2/9
   
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Answer: D
Explanation :
(i)  For 6, required probability = n(E)/n(S) = 5/36
     For 7, required probability = n(E)/n(S) = 6/36 = 1/6
(ii) Desired sums of the numbers are 2, 3, 4, 5, 6, 7 and 8;
     n(S) = 1+2 + 3 + 4 + 5 + 6 + 5 = 26
     required probability = 26/36 = 13/18
(iii)Desired sums ofthe numbers are 3, 5, 7, 9 and 11;
     n(S) = 2 + 4 + 6 + 4 + 2 = 18
     required probability = 18/36 = 1/2
(iv)Desired sums of the numbers are 3, 6, 9 and 12;
    n(S) = 2 + 5 + 4 + 1= 12
    required probability = 12/36 = 1/3
(v)Events= {I, I}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}; n(S) = 6
   p(E) = 6/36 = 1/6
(vi)Events=  {3, 1}, {4, 2}, {5, 3}, {6, 4}, {4, 6}, {3, 5}, {2, 4}, {1,3}or, n(S) =8
    p(E) = 8/36 = 2/9






 
9.
(i)What is the chance that a leap year selected randomly will have 53 Sundays?
(ii)What is the chance, if the year selected is not a leap year?
 
A.
1/7,2/7

B.

2/7,1/7
C.
1,2
D.
2,1
   
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Answer: B
Explanation :
(i)A leap year has 366 days so it has 52 complete weeks and 2 more days. The two days can be {Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday}, i.e. n(E) = 7.
Out of these 7 cases, cases favorable for more Sundays are
{Sunday and Monday, Saturday and Sunday}, i.e., n(E) = 2
p(E) = 2/7
(ii)When the year is not a leap year, it has 52 complete weeks and
1 more day that can be {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}, n(S) = 7
Out of these 7 cases, cases favorable for one more Sunday is
{Sunday}, n(E) = 1
. :. P(E) = 1/7






 
10.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings
 
A.
1/5

B.

25/57
C.
35/256
D.
1/221
   
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Answer: D
Explanation :
n (S) =52, C2 = (52x51)/2 = 1326
E =event of getting 2 kings out of 4
n(E) = 4c2  = (4x3)/(2x1) = 6
p(E) = n(E)/n(S) = 6/1326 = 1/221






 

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