Quantitative Aptitude

Probability

1,860 Questions
All Sample Space Setup 558 Permutation & Combination Counting 288 Multiplication Theorem (AND) 234 Addition Theorem (OR) 151 At Least One Probability 134 Complement Rule (P(Not A)) 130 Drawing Without Replacement 126 Conditional Probability 75 Bayes' Theorem 69 Drawing With Replacement 49 Success and Failure Rates 46

Probability Questions

Multiple choice

A bag contains coins of one rupee, two rupee & five rupees. The total money in the bag is Rs. 120. If the total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5. Find the probability of getting a 5 rupee coin if a coin is randomly picked from the bag?
  1. $12/47$
  2. $21/46$
  3. $12/37$
  4. $21/37$

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

One rupee and two rupee coins total 35 in ratio 2:5, so one rupee = 10 coins and two rupee = 25 coins. Their value = 10 + 50 = Rs 60. Remaining Rs 60 is from five rupee coins, so 12 five rupee coins. Total coins = 10 + 25 + 12 = 47. Probability of 5 rupee coin = 12/47. The key is working backwards from the ratio and total value.

Multiple choice

In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red?

  1. 29/35

  2. 7/15

  3. 23/35

  4. 2/5

  5. 19/35
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total balls = 6 red + 9 green = 15. P(at least one red) = 1 - P(both green). P(both green) = (9/15) × (8/14) = 72/210 = 12/35. Therefore, P(at least one red) = 1 - 12/35 = 23/35. Alternatively, we can calculate directly: cases are (1 red, 1 green) or (2 red). P(1 red, 1 green) = (6/15)×(9/14)×2 = 108/210 = 18/35. P(2 red) = (6/15)×(5/14) = 30/210 = 5/35. Total = 18/35 + 5/35 = 23/35.

Multiple choice

Two boxes contain 8 and 32 balls respectively. Four balls in the first box and eight in the second, are black. If a box is chosen randomly and two balls are drawn at random from it, what is the probability that at least one ball is black if the ball is not replaced?
  1. $\frac{567}{1736}$
  2. $\frac{1067}{1736}$
  3. $\frac{347}{1736}$
  4. $\frac{682}{1736}$

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Box 1: 8 balls (4 black). P(at least 1 black) = 1 - P(no black) = 1 - (4/8 * 3/7) = 1 - 12/56 = 44/56 = 11/14. Box 2: 32 balls (8 black). P(at least 1 black) = 1 - (24/32 * 23/31) = 1 - 552/992 = 440/992 = 55/124. Total P = (1/2)(11/14) + (1/2)(55/124) = 11/28 + 55/248 = (11*8 + 55)/248 = 143/248 = 1067/1736 (converting numerator). Option A gives 567/1736 which is too small. Option C (347/1736) is far too small.

Multiple choice

5 cards are drawn successively from a well-shuffled pack of 52 cards with replacement. Determine the probability that all the five cards should be spades?
  1. $1/1225$
  2. $1/1125$
  3. $1/125$
  4. $1/1025$
  5. $1/1050$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Drawing 5 cards with replacement means each draw is independent. The probability of drawing a spade in one draw is 13/52 = 1/4. For 5 successive draws with replacement, the probability is (1/4)^5 = 1/1024. Therefore, 1/1025 is the closest option. The calculation assumes a standard deck with 13 spades out of 52 cards.

Multiple choice

A box contains 21 balls numbered 1 to 21. A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?
  1. $\(\frac{5}{8}\)$
  2. $\(\frac{5}{7}\)$
  3. $\(\frac{3}{14}\)$
  4. $\(\frac{3}{16}\)$

  5. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From 1 to 21, there are 10 even numbers (2, 4, 6, 8, 10, 12, 14, 16, 18, 20). Probability of first ball being even = 10/21. Without replacement, 9 even balls remain out of 20 total. Probability of second ball being even = 9/20. Combined probability = (10/21) × (9/20) = 90/420 = 3/14.

Multiple choice

A card is drawn from a pack of 52 cards. The card is drawn at random; find the probability that it is neither club nor queen?
  1. $\(\frac{4}{13}\)$
  2. $\(\frac{5}{13}\)$
  3. $\(\frac{7}{13}\)$
  4. $\(\frac{9}{13}\)$

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Neither club nor queen = Total - (clubs or queens). Clubs: 13. Queens: 4. But queen of clubs counted twice, so clubs or queens = 13+4-1 = 16. Cards that are neither club nor queen = 52-16 = 36. Probability = 36/52 = 9/13. The correct answer is D.

Multiple choice

If two dice are rolled simultaneously find the probability of getting their sum a multiple of 3 or 4. यदि दो पांसो को एक साथ फेका जाता है तो उनका योग 3 या 4 से विभाजित होने की संभावना ज्ञात कीजिये ?

  1. 5/9

  2. 4/9

  3. 5/18

  4. 7/36

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When two dice are rolled, there are 36 total outcomes. Sums that are multiples of 3: 3 (2 outcomes), 6 (5 outcomes), 9 (4 outcomes), 12 (1 outcome). Sums that are multiples of 4: 4 (3 outcomes), 8 (5 outcomes), 12 (1 outcome). Using inclusion-exclusion: (2+3+5+5+4+1) - 1 (12 counted twice) = 19 outcomes. Probability = 19/36 = 5/9.

Multiple choice

Six coins are tossed simultaneously, find the probability of getting 4 tails or 3 heads. छह सिक्कों को एक साथ उछाला जाता है, 4 पूंछ या 3 शीर्ष प्राप्त होने की संभावना ज्ञात कीजिये।

  1. 1/8

  2. 1/4

  3. 17/64

  4. 35/64

  5. 1/2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes when tossing 6 coins = 2^6 = 64. For 4 tails: C(6,4) = 15 outcomes. For 3 heads: C(6,3) = 20 outcomes. Since these are mutually exclusive events (4 tails means exactly 2 heads, not 3 heads), total favorable outcomes = 15 + 20 = 35. Probability = 35/64.

Multiple choice

A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random, what is the probability that at least one is Black? एक टोकरी में 6 सफ़ेद 4 काली 2 पिंक और 3 हरी गेंद हैं। यदि चार गेंदों को यादृच्छिक रूप से चुना जाता है, तो क्या संभावना है कि कम से कम एक काली हो?

  1. 69/91

  2. 56/91

  3. 23/91

  4. 17/91

  5. 19/91
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 6+4+2+3 = 15. Ways to choose 4 from 15 = 15C4 = 1365. Ways with no black balls = choose from 11 non-black = 11C4 = 330. Probability of at least one black = 1 - (330/1365) = 1035/1365 = 69/91. This uses complementary probability.

Multiple choice

Read the following information carefully and answer the questions given below. निम्नलिखित जानकारियों का ध्यानपूर्वक अध्ययनकर नीचे पूछे गये प्रश्नों के उत्तर दीजिये। A bag contains 9 red, 7 white and 4 black balls. एक बैग में 9 लाल, 7 सफेद और 4 काली गेंदें है। If 2 balls are drawn randomly then what is the probability that both are of same colour? यदि 2 गेंदे बेतरतीब ढंग से बाहर निकाली जाती हैं तो दोंनों के समान रंग के होने की प्रायिकता क्या होगी?

  1. 31/95

  2. 63/190

  3. 123/380

  4. 7/19

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total balls = 9 + 7 + 4 = 20. Ways to draw 2 balls = 20C2 = 190. Ways both are same color: 2 red from 9 = 9C2 = 36, 2 white from 7 = 7C2 = 21, 2 black from 4 = 4C2 = 6. Total favorable = 36 + 21 + 6 = 63. Probability = 63/190.

Multiple choice

A bag contains 8 black and 7 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? एक बैग में 8 काली और 7 सफेद गेंदें हैं। एक गेंद यादृच्छिक निकाली गई है। क्या संभावना है कि निकाली गई गेंद सफेद है?

  1. 8/15

  2. 8/7

  3. 7/15

  4. 1/7

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total balls = 8 black + 7 white = 15 balls. Number of white balls = 7. Probability of drawing a white ball = Number of favorable outcomes / Total outcomes = 7/15. Option C is correct.

Multiple choice

A box contains 4 red, 5 black and 6 green balls. 3 balls are drawn at random. What is the probability that all the balls are of same colour? एक बॉक्स में 4 लाल, 5 काली और 6 हरी गेंदें होती हैं। 3 गेंदों को यादृच्छिक रूप से निकाला गया है। क्या संभावना है कि सभी गेंदें एक ही रंग की हैं?

  1. 34/455

  2. 17/455

  3. 9/455

  4. 8/455

  5. None of these इनमे से कोई नही|
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 15. Drawing 3 same-colored balls means either all 3 red (from 4), all 3 black (from 5), or all 3 green (from 6). Total favorable outcomes = C(4,3) + C(5,3) + C(6,3) = 4 + 10 + 20 = 34. Total possible outcomes = C(15,3) = 455. Probability = 34/455. Option B (17/455) would only account for one color category.

Multiple choice

A bag contains 7 red, 5 green and 6 blue balls. Three balls are drawn one by one at random without replacement. Find the number of ways that the first ball is red, second ball is green and third ball is blue? एक बैग में 7 लाल, 5 हरे और 6 नीले रंग की गेंदें होती हैं। तीन गेंदों को बिना प्रतिस्थापन के एक-एक करके एक-एक करके खींचा जाता है। पहली गेंद लाल है, दूसरी गेंद हरे रंग की है और तीसरी गेंद नीली है?

  1. 31/512

  2. 35/816

  3. 33/713

  4. 34/812

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

P(R first, G second, B third) = (7/18) × (5/17) × (6/16) = 35/816. The question asks for 'number of ways' but gives fractions - it should ask for probability. If it meant actual count, the answer would be 7×5×6 = 210, not matching any option. Assuming probability, option B is correct.

Multiple choice

A bag contains X red balls , 2 white and 3 green balls. 2 balls are picked up from the bag at random, one after the othe, without replacement, the probability of being both red is 4/7. What is the value of X? एक बैग में x लाल गेंद, 2 सफेद गेंद और 3 हरी गेंद हैं। 2 गेंदों को बैग से यादृच्छिक रूप से, एक के बाद एक, प्रतिस्थापन के बिना, निकाला जाता है दोनों लाल गेंद होने की प्रायिकता 4/7 है। X का मान क्या है?

  1. 10

  2. 15

  3. 13

  4. 20

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Total balls = X + 2 + 3 = X + 5. Probability of both red (without replacement) = (X/(X+5)) × ((X-1)/(X+4)) = X(X-1)/((X+5)(X+4)) = 4/7. Cross-multiplying: 7X(X-1) = 4(X+5)(X+4). This gives 7X² - 7X = 4X² + 36X + 80, so 3X² - 43X - 80 = 0. Using quadratic formula: X = (43 ± √(1849 + 960))/6 = (43 ± √2809)/6 = (43 ± 53)/6. X = 96/6 = 16 (valid) or X = -10/6 (invalid). Since 16 is not in the options, 'None of these' is correct.

Multiple choice

A dice is thrown twice what is the probability of getting a sum more than 9 from the both throws? एक पासे को दो बार फेंका जाता है। दोनों बार में आने वाले अंको का योग 9 से अधिक होने की क्या सम्भावना होगी?

  1. 5/36

  2. 1/9

  3. 1/6

  4. 1/3

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When throwing two dice, there are 36 total outcomes. Sums greater than 9 are: 10, 11, 12. Count: (4,6), (5,5), (6,4) for 10; (5,6), (6,5) for 11; (6,6) for 12. That's 6 favorable outcomes. Probability = 6/36 = 1/6. Option C is correct.