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

In the given questions, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. You have to determine relationship between two quantities and choose the appropriate option: Quantity I: A box contains 4 blue pens, and 5 green pens and 6 red pens. Two pens are drawn at random from the box. What is the probability that the ball drawn is either red or green? Quantity II: A box contains 4 blue pens, and 5 green pens and 6 red pens. Two pens are drawn at random from the box. What is the probability that the ball drawn is either blue or green?

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≥ Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or no relation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: P(red or green) = (6C2 + 5C2 + 6C1*5C1)/15C2 = (15 + 10 + 30)/105 = 55/105 = 11/21 ≈ 0.524. Quantity II: P(blue or green) = (4C2 + 5C2 + 4C1*5C1)/15C2 = (6 + 10 + 20)/105 = 36/105 = 12/35 ≈ 0.343. Since 11/21 > 12/35, Quantity I > Quantity II is correct.

Multiple choice

State the correct relationship between Q I, Q II and Q III Q I : A bag contains 50 balls which are pink, violet and red .Find the number of violet balls in the bag, if probability of picking pink ball is 3/5 and that of either a pink or violet ball is 4/5. Q II : A bag contains 40 balls pink, violet and red. The probability of picking violet ball is 3/8. If the first ball was violet and without replacement, probability of picking a pink ball is 4/13.Find the number of red balls. Q III : There are 5 Red balls, 4 Yellow balls and 3 black balls in a bag. Four balls are chosen at random. The probability of their being 2 Red and 2 Yellow ball.

  1. Q I > Q II > Q III

  2. Q I < Q II < Q III

  3. Q I = Q II = Q III

  4. Q III < Q I < Q II

  5. Q I = Q II > Q III
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Q I: P(pink)=3/5 so pink=30. P(pink or violet)=4/5 so pink+violet=40, thus violet=10. Q II: P(violet)=3/8, violet=15. After removing violet: 39 balls left, P(pink)=4/13 so pink=12. Red=40-15-12=13. Q III: Total=12. P(2 red from 5)*P(2 yellow from 4) = C(5,2)*C(4,2)/C(12,4) = 10*6/495=60/495=4/33 ≈ 0.121. 10 < 13 < 0.121 is meaningless - Q III is probability, not count. Comparing as values: 10 < 13, and 4/33 ≈ 0.12. So Q I=10, Q II=13, Q III≈0.12. Order: Q III < Q I < Q II.

Multiple choice

Given below are two quantities named I and II. Based on the given data, you have to determine the relation between them and mark the answer Probability of picking a red ball out of a box is 3/7 , box contains only red, blue and white ball and white balls are 40% more than blue balls. Quantity I: Probability of picking one white balls. Quantity II: when white balls are removed from box then probability of picking one blue ball.

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≥Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or No relation can be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let total balls = T. Red probability = 3/7, so red balls = (3/7)T. Let blue balls = B. White balls are 40% more than blue, so white = 1.4B. Total balls: Red + Blue + White = T, so (3/7)T + B + 1.4B = T. Solving: (3/7)T + 2.4B = T, so 2.4B = (4/7)T, giving B = T/4.2 = (10/42)T = (5/21)T. White = 1.4B = 1.4 × (5/21)T = (7/21)T = (1/3)T. Quantity I: Probability of white = White / Total = (1/3)T / T = 1/3. Quantity II: After removing white balls, probability of blue = Blue / (Red + Blue) = ((5/21)T) / ((3/7)T + (5/21)T) = (5/21)T / (14/21)T = 5/14. Comparing 1/3 vs 5/14: 1/3 ≈ 0.333, 5/14 ≈ 0.357. So Quantity I < Quantity II, answer B is correct.

Multiple choice

The question given below contains two statements giving certain data. You have to decide whether the data given in the statements are sufficient for answering the question? Mark answer- Balls are drawn randomly from a bag and what is the probability of that no ball is yellow. I. There are 5 red, 4 yellow and 2 green balls. II. Two balls are drawn from the bag.

  1. If statement I alone is sufficient but statement II alone is not sufficient.

  2. If statement II alone is sufficient but statement I alone is not sufficient.

  3. If each statement alone (either I or II) is sufficient.

  4. If statement I and II together are not sufficient.

  5. If both statements I and II together are sufficient, but neither statement alone is sufficient.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Statement I gives the ball composition (5 red, 4 yellow, 2 green = 11 total). Statement I alone is insufficient without knowing how many balls are drawn. Statement II says two balls are drawn but gives no composition information. Together: we can calculate P(no yellow) = (7/11)×(6/10) = 42/110 = 21/55. Both statements needed.

Multiple choice

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is
  1. $\displaystyle \frac{2}{7}$
  2. $\displaystyle \frac{1}{21}$
  3. $\displaystyle \frac{2}{23}$
  4. $\displaystyle \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 9. Ways to pick 3 different colors = (3 red * 4 blue * 2 green) = 24. Total ways to pick 3 balls = 9C3 = (9*8*7)/(3*2*1) = 84. Probability = 24/84 = 2/7.

Multiple choice

A fair six-faced die is rolled $12$ times. The probability that each face turns up twice is equal to
  1. $\dfrac { 12! }{ 6!6!{ 6 }^{ 12 } } $
  2. $\dfrac { { 2 }^{ 12 } }{ { 2 }^{ 6 }{ 6 }^{ 12 } } $
  3. $\dfrac { 12! }{ { 2 }^{ 6 }{ 6 }^{ 12 } } $
  4. $\dfrac { 12! }{ { 6 }^{ 2 }{ 6 }^{ 12 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total number of outcomes is 6^12. The number of ways to arrange 12 items where each of the 6 faces appears twice is given by the multinomial coefficient 12! / (2! 2! 2! 2! 2! 2!) = 12! / (2^6). Thus, the probability is 12! / (2^6 * 6^12).

Multiple choice

Cards are drawn one by one without replacement from a well shuffled pack of $52$ cards. Then, the probability that a face card (jack,queen or king) will appear for the first time on the third turn is equal to
  1. $\cfrac{300}{2197}$
  2. $\cfrac{36}{85}$
  3. $\cfrac{12}{85}$
  4. $\cfrac{4}{51}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total cards = 52, face cards = 12, non-face cards = 40. Probability = (P(non-face on 1st) * P(non-face on 2nd) * P(face on 3rd)) = (40/52) * (39/51) * (12/50) = (10/13) * (13/17) * (6/25) = (10 * 6) / (17 * 25) = 60 / 425 = 12 / 85.

Multiple choice

Three boxes contain respectively $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, from each of the boxes one ball is drawn at random. The probability that $2$ white and $1$ black balls will be drawn is
  1. $\dfrac {13}{32}$
  2. $\dfrac {1}{4}$
  3. $\dfrac {1}{32}$
  4. $\dfrac {3}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The three possible arrangements are WWB, WBW, and BWW. Their probabilities are 9/32, 3/32, and 1/32, respectively, giving a total of 13/32.

Multiple choice

A box $B_{1}$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box $B_{2}$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box $B_{3}$ contains $3$ white balls, $4$ red balls and $5$ black balls. If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$, the probability that all 3 drawn balls are of the same colour is
  1. $\displaystyle \frac{82}{648}$
  2. $\displaystyle \frac{90}{648}$
  3. $\displaystyle \frac{558}{648}$
  4. $\displaystyle \frac{566}{648}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls in B1=6, B2=9, B3=12. Probability of white = (1/6)(2/9)(3/12) = 6/648. Probability of red = (3/6)(3/9)(4/12) = 36/648. Probability of black = (2/6)(4/9)(5/12) = 40/648. Sum = (6+36+40)/648 = 82/648.