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

$(A)$ A box contains $3$ blue, $2$ white and $4$ red marbles. If a marble is drawn from the box. What is the probability that it will be $(i)$ White $\quad (ii)$ Blue $\quad (iii)$ Red? $(B)$ A bag contains $5$ red, $8$ green and $7$ white balls. One ball is drawn at random from the bag, find the probability of getting $(i)$ A white ball or a green ball $(ii)$ Neither a green ball nor a red ball.
  1. $(A)\space (i)\quad \displaystyle\frac{1}{9} \quad (ii)\quad \displaystyle\frac{2}{3} \quad (iii)\quad \displaystyle\frac{5}{9} \\ (B)\space (i)\quad \displaystyle\frac{6}{13}\quad (ii)\quad \displaystyle\frac{11}{20}$
  2. $(A)\space (i)\quad \displaystyle\frac{5}{9} \quad (ii)\quad \displaystyle\frac{1}{4} \quad (iii)\quad \displaystyle\frac{2}{11} \\ (B)\space (i)\quad \displaystyle\frac{1}{2}\quad (ii)\quad \displaystyle\frac{5}{19}$
  3. $(A)\space (i)\quad \displaystyle\frac{2}{9} \quad (ii)\quad \displaystyle\frac{1}{3} \quad (iii)\quad \displaystyle\frac{4}{9} \\ (B)\space (i)\quad \displaystyle\frac{3}{4}\quad (ii)\quad \displaystyle\frac{7}{20}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For (A), total marbles = 3+2+4 = 9. P(White) = 2/9, P(Blue) = 3/9 = 1/3, P(Red) = 4/9. For (B), total balls = 5+8+7 = 20. P(White or Green) = (7+8)/20 = 15/20 = 3/4. P(Neither Green nor Red) = P(White) = 7/20.

Multiple choice

A box contains $20$ balls bearing numbers $1, 2, 3, ... , 20$ respectively. A ball is drawn at random from the box what is the probability that the number on the ball is $(i)$ An odd number $(ii)$ Divisible by $2$ or $3$ $(iii)$ A prime number $(iv)$ Not divisible by $10$
  1. $(i)\quad \displaystyle\frac{1}{3}\\ (ii)\quad \displaystyle\frac{12}{19}\\(iii)\quad \displaystyle\frac{2}{5}\\ (iv)\quad \displaystyle\frac{9}{10}$
  2. $(i)\quad \displaystyle\frac{1}{2}\\ (ii)\quad \displaystyle\frac{13}{20}\\(iii)\quad \displaystyle\frac{2}{5}\\ (iv)\quad \displaystyle\frac{9}{10}$
  3. $(i)\quad \displaystyle\frac{1}{3}\\ (ii)\quad \displaystyle\frac{18}{37}\\(iii)\quad \displaystyle\frac{2}{5}\\ (iv)\quad \displaystyle\frac{2}{5}$
  4. $(i)\quad \displaystyle\frac{1}{2}\\ (ii)\quad \displaystyle\frac{15}{29}\\(iii)\quad \displaystyle\frac{2}{5}\\ (iv)\quad \displaystyle\frac{2}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For (i), odd numbers are 10 out of 20, so 1/2. For (ii), multiples of 2 (10) plus multiples of 3 (6) minus multiples of 6 (3) equals 13/20. For (iii), primes are 2, 3, 5, 7, 11, 13, 17, 19 (8 total), so 8/20 = 2/5. For (iv), numbers divisible by 10 are 10 and 20 (2 total), so 18/20 = 9/10.

Multiple choice

The probability of getting a total of $9$ exactly twice in $6$ tosses of a pair of dice is
  1. $_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9} \right) }^{ 2}+_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 3 }$
  2. $_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
  3. $1-_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
  4. ${ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Probability of getting 9 with two dice: (3,6), (4,5), (5,4), (6,3) = 4/36 = 1/9. Probability of not getting 9 = 8/9. Binomial distribution for 6 tosses, 2 successes: 6C2 * (1/9)^2 * (8/9)^4.

Multiple choice

While watching a game of Champions League football in a cafe, you observe someone who is clearly supporting Manchester United in the game. What is the probability that they were actually born within 25 miles of Manchester? Assume that: $\implies $the probability that a randomly selected person in a typical local bar environment is born within $25$ miles of Manchester is $\dfrac{1}{20}$, and; $\implies$the chance that a person born within $25$ miles of Manchester actually supports United is $\dfrac{7}{10}$; $\implies$the probability that a person not born within $25$ miles of Manchester supports United with probability $\dfrac{1}{10}$
  1. $\dfrac{7}{26}$
  2. $\dfrac{8}{26}$
  3. $\dfrac{9}{26}$
  4. $\dfrac{10}{26}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use Bayes' Theorem. Let M be born in Manchester, S be support United. P(M|S) = [P(S|M) * P(M)] / [P(S|M) * P(M) + P(S|not M) * P(not M)]. P(M|S) = [(7/10)(1/20)] / [(7/10)(1/20) + (1/10)*(19/20)] = 7 / (7 + 19) = 7/26.

Multiple choice

In a series of n independent trials for an event of constant probability p, the most probable number r of successes is given by $\left ( n+1 \right )p-1< r< \left ( n+1 \right )p$. Hence, the most probable number of successes is the integral part of $\left ( n+1 \right )p$. But if $\left ( n+1 \right )p$ is an integer, the chance of r successes is equal to that of $r+1$ successes and both $r,r+1$ are most probable numbers of successes. A bag contains 2 white balls and 1 black ball. A ball is drawn at random and returned to the bag. The experiment is done 10 times. The probability that a white ball is drawn exactly 5 times is
  1. $\dfrac{10!}{5!5!}\left ( \dfrac{2}{3} \right )^5$
  2. $\dfrac{10!}{5!5!}\left ( \dfrac{1}{3} \right )^5$
  3. $\dfrac{10!}{\left(5! \right)^2}\left(\dfrac{2}{9} \right )^5$
  4. $\dfrac{10!}{5!}\left(\dfrac{2}{9} \right )^5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a binomial distribution problem with n=10, p=2/3 (probability of white ball), and q=1/3. The probability of exactly 5 successes is C(10, 5) * (2/3)^5 * (1/3)^5 = (10! / (5!5!)) * (2^5 / 3^10). Option C is (10! / (5!5!)) * (2^5 / 9^5) = (10! / (5!5!)) * (2^5 / 3^10), which matches.

Multiple choice

Let n positive integers taken randomly & multiplied together.Let $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$,$A_{5}$, $A_{6}$, $A_{7}$, be the events define as $A_{1}$ = last digit of the product is not the number 1,3,7 or 9 $A_{2}$ = last digit of the product is among the number 1,3,5,7 or 9, $A_{3}$ = last digit of the product is among the number 1, 3, 7 or 9 $A_{4}$ = last digit of lhe product 'is among the number 2, 4, 6 or 8. $A_{5}$ = last digit of the product is the number 5. $A_{6}$ = last digit of the product is the number 0 $A_{7}$ = last digit of the product is 1,2,3,4,6,7,8 or 9.On the basis of above information answer the following questions. The chance that last digit of the product is $A_{3}$.
  1. $\displaystyle\left (\frac{3}{5}\right )^{n}$
  2. $\displaystyle\left (\frac{2}{5}\right )^{n}$
  3. $\displaystyle\left (\frac{2}{3}\right )^{n}$
  4. $\displaystyle\left (\frac{4}{5}\right )^{n}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

Which of the following has less than 50% chances of occurrence?

  1. Throwing a multiple of 2 in a 6 face dice.

  2. Throwing a multiple of 5 in case of a 6 face dice.

  3. Getting both heads while throwing two coins.

  4. Getting a prime no. or an even no, in throwing a 6 face dice.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A: Multiples of 2 are 2, 4, 6 (3/6 = 50%). B: Multiples of 5 is 5 (1/6 = 16.67%). C: Getting HH is 1/4 (25%). D: Prime (2, 3, 5) or Even (2, 4, 6) are {2, 3, 4, 5, 6} (5/6 = 83.33%). Option B is the only one clearly less than 50%.

Multiple choice

Harshad and Amit have thrown two dices. If Harshad has thrown a sum 10 on the two dices, what is Amit`s chance of throwing a sum higher than the sum thrown by the Harshadr?

  1. 1/12

  2. 1/10

  3. ΒΌ

  4. 1/16

  5. 1/8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Harshad's sum is 10. Possible pairs for 10 are (4,6), (5,5), (6,4) - 3 outcomes. Total outcomes for two dice = 36. Amit needs a sum > 10 (i.e., 11 or 12). Sum 11: (5,6), (6,5) - 2 outcomes. Sum 12: (6,6) - 1 outcome. Total favorable = 3. Probability = 3/36 = 1/12.

Multiple choice

A bag contains 6 yellow balls and 5 red balls. A ball is selected and a coin is tossed. What is the probability of selecting a blue ball and getting heads?

  1. 1

  2. -1

  3. 0

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The question asks for the probability of selecting a blue ball. Since the bag only contains yellow and red balls, the number of blue balls is 0, making the probability of selecting one 0. Regardless of the coin toss result, the joint probability is 0.