# Quantitative Aptitude: Probability Set 1

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 â€“ 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 â€“ 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 (3Ã—8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 â€“ 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 â€“ 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}â€¦â€¦â€¦â€¦â€¦â€¦.{6,6})
For a total of 7 on dices, we have â€“ {1,6}, {6,1}, {2,5}, {5,2}, {3,4}, {4,3} â€“ 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}â€¦â€¦â€¦â€¦â€¦â€¦.{6,6})
For sum of number to be a multiple of 5, we have â€“ {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4}, {5,5} â€“ 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 â€“ 1 = 8 â€“ 1 = 7) â€“ 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) â€“ 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 â€“ 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)

## 52 Thoughts to “Quantitative Aptitude: Probability Set 1”

1. ^^^Jaga^^^....

mam… i think q4 ans is wrong….

2. ^^^Jaga^^^....

q4…………………..4c2+5c2+4c1*5c1/12c2 =6/11

1. Corrected

3. Blak

Q9. there are 7 mutiples 14,28,42,56,70,84,98

4. Ayushi

mam plz can you give the whole data or info related to cards ?plz

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