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# Quant Quiz On Probability Day 25 Bag

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## Quant Quiz On Probability Day 25 Bag

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• There are 100 tickets in a box numbered 1 to 100. 3 tickets are drawn at one by one. Find the probability that the sum of number on the tickets is odd.
A) 2/7
B) 1/2
C) 1/3
D) 2/5
E) 3/7
Option B
Solution:
There will be 4 cases
Case 1: even, even, odd
Prob. = 1/2 × 1/2 × 1/2
Case 2: even, odd, even
Prob. = 1/2 × 1/2 × 1/2
Case 3: odd, even, even
Prob. = 1/2 × 1/2 × 1/2
Case 4: odd, odd, odd
Prob. = 1/2 × 1/2 × 1/2
Add all the cases, required prob. = 1/2
• There are 4 green and 5 red balls in first bag. And 3 green and 5 red balls in second bag. One ball is drawn from each bag. What is the probability that one ball will be green and other red?
A) 85/216
B) 34/75
C) 95/216
D) 35/72
E) 13/36
Option D
Solution:
Case 1:first green, second red
Prob. = 4/9 × 5/8 = 20/72
Case 2:first red, second green
Prob. = 5/9 × 3/8 = 15/72
• A bag contains 2 red, 4 blue, 2 white and 4 black balls. 4 balls are drawn at random, find the probability that at least one ball is black.
A) 85/99
B) 81/93
C) 83/99
D) 82/93
E) 84/99
Option A
Solution:
Prob. (At least 1 black) = 1 – Prob. (None black)
So Prob. (At least 1 black) = 1 – (8C4/12C4) = 1 – 14/99
• Four persons are chosen at random from a group of 3 men, 3 women and 4 children. What is the probability that exactly 2 of them will be men?
A) 1/9
B) 3/10
C) 4/15
D) 1/10
E) 5/12
Option B
Solution:
2 men means other 2 woman and children
So prob. = 3C2 × 7C2 /10C4 = 3/10
• Tickets numbered 1 to 120 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 8/15
B) 5/16
C) 7/15
D) 3/10
E) 13/21
Option C
Solution:
Multiples of 3 up to 120 = 120/3 = 40
Multiples of 5 up to 120 = 120/5 = 24 (take only whole number before the decimal part)
Multiple of 15 (3×5) up to 120 = 120/15 = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = 56/120 = 7/15
• There are 2 people who are going to take part in race. The probability that the first one will win is 2/7 and that of other winning is 3/5. What is the probability that one of them will win?
A) 14/35
B) 21/35
C) 17/35
D) 19/35
E) 16/35
Option D
Solution:
Prob. of 1st winning = 2/7, so not winning = 1 – 2/7 = 5/7
Prob. of 2nd winning = 3/5, so not winning = 1 – 3/5 = 2/5
So required prob. = 2/7 * 2/5 + 3/5 * 5/7 = 19/35
• Two cards are drawn at random from a pack of 52 cards. What is the probability that both the cards drawn are face card (Jack, Queen and King)?
A) 11/221
B) 14/121
C) 18/221
D) 15/121
E) 14/221
Option A
Solution:
There are 52 cards, out of which there are 12 face cards.
So probability of 2 face cards = 12C2/52C2 = 11/221
• A committee of 5 people is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have less number of boys than girls?
A) 7/12
B) 7/15
C) 6/13
D) 5/12
E) 7/13
Option D
Solution:
Case 1: 1 boy and 4 girls
Prob. = 5C1 × 4C4/9C5 = 5/146
Case 2: 2 boys and 3 girls
Prob. = 5C2 × 4C3/9C5 = 40/126
Add the two cases = 45/126 = 5/12
• A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at random. What is the probability that they are not of same color?
A) 5/11
B) 14/33
C) 2/5
D) 6/11
E) 2/3
Option E
Solution:
Three cases
Case 1: one red, 1 blue
Prob = 2C1 × 4C1 / 12C2 = 4/33
Case 2: one red, 1 white
Prob = 2C1 × 6C1 / 12C2 = 2/11
Case 3: one white, 1 blue
Prob = 6C1 × 4C1 / 12C2 = 4/11
• A bag contains 5 blue balls, 4 black balls and 3 red balls. Six balls are drawn at random. What is the probability that there are equal numbers of balls of each color?
A) 11/77
B) 21/77
C) 22/79
D) 13/57
E) 15/77
Option E
Solution:
5C2× 4C2× 3C212C6

Directions (1-3): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is 2/5. There are 10 blue balls.

1. What is the probability of choosing one blue ball?
A) 2/7
B) 1/4
C) 1/3
D) 2/5
E) 3/7
Option C
Solution:
Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3
2. What is the total number of balls in the urn?
A) 45
B) 34
C) 40
D) 30
E) 42
Option D
Solution:
Probability of choosing one blue ball is 1/3
And total blue balls are 10. So with 10/30 we get probability as 1/3
So total balls must be 30
3. If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 2 or 3?
A) 3/4
B) 4/5
C) 1/4
D) 1/3
E) 2/3
Option E
Solution:
There are 30 balls in the urn.
Multiples of 2 up to 30 = 30/2 = 15
Multiples of 3 up to 30 = 30/3 = 10 (take only whole number before the decimal part)
Multiples of 6 (2×3) up to 30 = 30/6 = 5
So total such numbers are = 15 + 10 – 5 = 20
So required probability = 20/30 = 2/3
4. There are 2 brothers A and B. Probability that A will pass in exam is 3/5 and that B will pass in exam is 5/8. What will be the probability that only one will pass in the exam?
A) 12/43
B) 19/40
C) 14/33
D) 21/40
E) 9/20
Option B
Solution:
Only one will pass means the other will fail
Probability that A will pass in exam is 3/5. So Probability that A will fail in exam is 1 – 3/5 = 2/5
Probability that B will pass in exam is 5/8. So Probability that B will fail in exam is 1 – 5/8 = 3/8
So required probability = P(A will pass)*P(B will fail) + P(B will pass)*P(A will fail)
So probability that only one will pass in the exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40
5. If three dices are thrown simultaneously, what is the probability of having a same number on all dices?
A) 1/36
B) 5/36
C) 23/216
D) 1/108
E) 17/216
Option A
Solution:
Total events will be 6*6*6 = 216
Favorable events for having same number is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4}, {5,5,5}, {6,6,6} – so 6 events
Probability of same number on all dices is 6/216 = 1/36
6. There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?
A) 52/125
B) 53/150
C) 17/50
D) 37/150
E) 32/75
Option E
Solution:
Multiples of 3 up to 150 = 150/3 = 50
Multiples of 7 up to 150 = 150/7 = 21 (take only whole number before the decimal part)
Multiples of 21 (3×7) up to 150 = 150/21 = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = 64/150 = 32/75
7. There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a ticket which has a prime number on it?
A) 3/55
B) 5/58
C) 8/21
D) 16/55
E) 4/13
Option D
Solution:
Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 41, 47, 53.
So probability = 16/55
8. A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue?
A) 61/110
B) 59/108
C) 45/134
D) 53/108
E) 57/110
Option D
Solution:
Case 1: Ball from first bag is white, from another is blue
So probability = 4/9 * 7/12 = 28/108
Case 1: Ball from first bag is blue, from another is white
So probability = 5/9 * 5/12 = 25/108
So required probability = 28/108 + 25/108 = 53/108
9. The odds against an event are 2 : 3 and the odds in favor of another independent event are 3 : 4. Find the probability that at least one of the two events will occur.
A) 11/35
B) 27/35
C) 13/35
D) 22/35
E) 18/35
Option B
Solution:
Let 2 events A and B
Odds against A are 2 : 3
So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5
Odds in favor of B are 3 : 4
So probability of occurrence of B = 3/(3+4) = 3/7. And non-occurrence of B = 4/7
Probability that at least one occurs
Case 1: A occurs and B does not occur
So probability = 3/5 * 4/7 = 12/35
Case 2: B occurs and A does not occur
So probability = 3/7 * 2/5 = 6/35
Case 3: Both A and B occur
So probability = 3/5 * 3/7 = 9/35
So probability that at least 1 will occur = 12/35 + 6/35 + 9/35 = 27/35
10. The odds against an event are 1 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.
A) 17/28
B) 5/14
C) 11/25
D) 9/14
E) 19/28
Option A
Solution:
Let 2 events A and B
Odds against A are 1 : 3
So probability of occurrence of A = 3/(1+3) = 3/4. And non-occurrence of A = 1/4
Odds in favor of B are 2 : 5
So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7
Case 1: A occurs and B does not occur
So probability = 3/4 * 5/7 = 15/28
Case 2: B occurs and A does not occur
So probability = 2/7 * 1/4 = 2/28
So probability that one will occur = 15/28 + 2/28 = 17/28
1. From a pack of 52 cards, 1 card is chosen at random. What is the probability of the card being diamond or queen?
A) 2/7
B) 6/15
C) 4/13
D) 1/8
E) 17/52
Option C
Solution:

In 52 cards, there are 13 diamond cards and 4 queens.
1 card is chosen at random
For 1 diamond card, probability = 13/52
For 1 queen, probability = 4/52
For cards which are both diamond and queen, probability = 1/52
So required probability = 13/52 + 4/52 – 1/52 = 16/52 = 4/13
2. From a pack of 52 cards, 1 card is drawn at random. What is the probability of the card being red or ace?
A) 5/18
B) 7/13
C) 15/26
D) 9/13
E) 17/26
Option B
Solution:

In 52 cards, there are 26 red cards and 4 ace and there 2 such cards which are both red and ace.
1 card is chosen at random
For 1 red card, probability = 26/52
For 1 ace, probability = 4/52
For cards which are both red and ace, probability = 2/52
So required probability = 26/52 + 4/52 – 2/52 = 28/52 = 7/13
3. There are 250 tickets in an urn numbered 1 to 250. One ticket is chosen at random. What is the probability of it being a number containing a multiple of 3 or 8?
A) 52/125
B) 53/250
C) 67/125
D) 101/250
E) 13/25
Option A
Solution:

Multiples of 3 up to 250 = 250/3 = 83 (take only whole number before the decimal part)
Multiples of 8 up to 250 = 250/3 = 31
Multiples of 24 (3×8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = 104/250 = 52/125
4. There are 4 white balls, 5 blue balls and 3 green balls in a box. 2 balls are chosen at random. What is the probability of both balls being non-blue?
A) 23/66
B) 5/18
C) 8/21
D) 7/22
E) 1/3
Option D
Solution:

Both balls being non-blue means both balls are either white or green
There are total 12 balls (4+3+5)
and total 7 white + green balls.
So required probability = 7C2/12C2 = [(7*6/2*1) / (12*11/2*1)] = 21/66 = 7/22
5. There are 4 white balls, 3 blue balls and 5 green balls in a box. 2 balls are chosen at random. What is the probability that first ball is green and second ball is white or green in color?
A) 1/3
B) 5/18
C) 1/2
D) 4/21
E) 11/18
Option B
Solution:

There are total 4+3+5 = 12 balls
Probability of first ball being green is = 5/12
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or green is 8/12 = 2/3
So required probability = 5/12 * 2/3 = 5/18
6. 2 dices are thrown. What is the probability that there is a total of 7 on the dices?
A) 1/3
B) 2/7
C) 1/6
D) 5/36
E) 7/36
Option C
Solution:

There are 36 total events which can happen ({1,1), {1,2}……………….{6,6})
For a total of 7 on dices, we have – {1,6}, {6,1}, {2,5}, {5,2}, {3,4}, {4,3} – so 6 choices
So required probability = 6/36 = 1/6
7. 2 dices are thrown. What is the probability that sum of numbers on the two dices is a multiple of 5?
A) 5/6
B) 5/36
C) 1/9
D) 1/6
E) 7/36
Option E
Solution:

There are 36 total events which can happen ({1,1), {1,2}……………….{6,6})
For sum of number to be a multiple of 5, we have – {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4}, {5,5} – so 7 choices
So required probability = 7/36
8. There are 25 tickets in a box numbered 1 to 25. 2 tickets are drawn at random. What is the probability of the first ticket being a multiple of 5 and second ticket being a multiple of 3.
A) 5/11
B) 1/4
C) 2/11
D) 1/8
E) 3/14
Option D
Solution:

There are 5 tickets which contain a multiple of 5
So probability of 1st ticket containing multiple of 5 = 5/25 = 1/5
Now:
Case 1: If the ticket chosen contained 15
If there was a 15 on first draw, then there are 7 tickets in box which contain a multiple of 3 out of 24 tickets. (25/3 – 1 = 8 – 1 = 7) – because 15 is already out from the box
So probability = 7/24 (24 tickets remaining after 1st draw)
Case 2: If the ticket chosen contained other than 15 (5 or 10 or 20 or 25)
If 15 was not there on first draw, then there are 8 tickets in box which contain a multiple of 3 out of 24 tickets. (25/3 = 8) – because 15 is already out from the box
So probability = 8/24 (24 tickets remaining after 1st draw)
Add the cases for probability of multiple of 3 on second ticket, so prob. = 7/24 + 8/24 = 15/24 (added the cases because we want one of these cases to happen and not both)
So required probability = 1/5 * 15/24 = 1/8 (multiplied the cases because we want both to happen)
9. What is the probability of selecting a two digit number at random such that it is a multiple of 2 but not a multiple of 14?
A) 17/60
B) 11/27
C) 13/30
D) 31/60
E) 17/30
Option C
Solution:

There are 90 two digit numbers (10-99)
Multiple of 2 = 90/2 = 45
Multiple of 14 = 90/14 = 6
Since all multiples of 14 are also multiple of 2, so favorable events = 45 – 6 = 39
So required probability = 39/90 = 13/30
10. There are 2 urns. 1st urn contains 6 white and 6 blue balls. 2nd urn contains 5 white and 7 black balls. One ball is taken at random from first urn and put to second urn without noticing its color. Now a ball is chosen at random from 2nd urn. What is the probability of the second ball being a white colored ball?
A) 11/13
B) 6/13
C) 5/13
D) 5/12
E) 11/12
Option A
Solution:

Case 1: first was a white ball
Now it is put in second urn, so total white balls in second urn = 5+1 = 6, and total balls in second urn = 12+1 = 13
So probability of white ball from second urn = 6/13
Case 2: first was a blue ball
Now it is put in second urn, so total white balls in second urn remain 5, and total balls in second urn = 12+1 = 13
So probability of white ball from second urn = 5/13
So required probability = 6/13 + 5/13 = 11/13 (added the cases because we want one of these cases to happen and not both)