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 coin of diameter $\dfrac {1}{2}$ units is tossed randomly onto the rectangular cartesian plane, the probability that the coin does not intersect any line whose equation is of the form $x = k, k \epsilon I$ is
  1. $\dfrac {1}{\sqrt 2}$
  2. $1-\dfrac {1}{\sqrt 5}$
  3. $\dfrac {1}{4}$
  4. $\dfrac {1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The coin has diameter 1/2, so its radius is 1/4. For the coin not to intersect any line x = k, its center must be at a distance greater than 1/4 from any integer line. This means the center must lie in the interval (n + 1/4, n + 3/4) for any integer n. The length of this interval is 1/2, and the total distance between integer lines is 1, so the probability is 1/2.

Multiple choice

The sides of a rectangle are chosen at random each less than 10 cm, all such lengths being equally likely. The chance that the diagonal of the rectangle is less than 10 cm is
  1. $\dfrac{1}{10}$
  2. ${1}{20}$
  3. $\displaystyle \dfrac{\pi}{4}$
  4. $\displaystyle \dfrac{\pi}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sides x, y are in [0, 10]. Diagonal d = sqrt(x^2 + y^2) < 10 implies x^2 + y^2 < 100. This is the area of a quarter circle of radius 10 inside a 10x10 square. Probability = (pi * 10^2 / 4) / 10^2 = pi/4.

Multiple choice

Fifty seeds were selected at random from each of $5$ bags $A, B, C, D, E$ of seeds, and were kept under standardised conditions equally favourable to germination. After $20$ days, the number of seeds which had germinated in each collection were counted and recorded as follow : Bag $A$ $B$ $C$ $D$ $E$ Number of seeds germinated $40$ $48$ $42$ $39$ $41$ What is the probability of germination of $(i)$ more than $40$ seeds in a bag? $(ii)$ $49$ seeds in a bag? $(iii)$ more than $35$ seeds in a bag?
  1. $(i)\, 0.690$
    $(ii)\, 0.09$
    $(iii)\, 1$
  2. $(i)\, 0.80$
    $(ii)\, 0.006$
    $(iii)\, 1$
  3. $(i)\, 0.70$
    $(ii)\, 0.001$
    $(iii)\, 1$
  4. $(i)\, 0.60$
    $(ii)\, 0$
    $(iii)\, 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total bags = 5. (i) Bags with > 40 seeds are B(48), C(42), E(41), so 3/5 = 0.6. (ii) No bag has 49 seeds, so 0/5 = 0. (iii) All 5 bags have > 35 seeds, so 5/5 = 1.

Multiple choice

There are $\mathrm{n}$ urns each containing $\mathrm{n}+1$ balls such that the ith urn contains $\mathrm{i}$ white balls and $(\mathrm{n}+1-\mathrm{i})$ red balls. Let $\mathrm{u}{\mathrm{i}}$ be the event of selecting ith urn, $\mathrm{i}=1,2,3\ldots,\ \mathrm{n}$ and $\mathrm{w}$ denotes the event of getting a white ball. $\mathrm{I}\mathrm{f}\mathrm{P}(\mathrm{u}{\mathrm{i}})=\mathrm{c}$, where $\mathrm{c}$ is a constant then $\mathrm{P}(\mathrm{u}_{\mathrm{n}}/\mathrm{w})$ is equal to
  1. $\displaystyle \frac{2}{\mathrm{n}+1}$
  2. $\displaystyle \frac{1}{\mathrm{n}+1}$
  3. $\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Bayes' theorem: P(Un/W) = P(W/Un) * P(Un) / sum(P(W/Ui) * P(Ui)). Since P(Ui) is constant, P(Un/W) = P(W/Un) / sum(P(W/Ui)). P(W/Ui) = i/(n+1). Sum = (1/(n+1)) * sum(i from 1 to n) = (1/(n+1)) * (n(n+1)/2) = n/2. P(W/Un) = n/(n+1). So P(Un/W) = (n/(n+1)) / (n/2) = 2/(n+1).

Multiple choice

A fair die is rolled once. STATEMENT - 1 : The probability of getting a composite number is 1/3 STATEMENT - 2 : There are three possibilities for the obtained number (i) the number is a prime number (ii) the number is a composite number (iii) the number is 1, and hence probability of getting a prime number $=$1/3.

  1. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

  2. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

  3. Statement - 1 is True, Statement - 2 is False

  4. Statement - 1 is False, Statement - 2 is True
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Statement 1 is true: composite numbers on a die are 4 and 6, so probability is 2/6 = 1/3. Statement 2 is false: the probability of getting a prime number (2, 3, 5) is 3/6 = 1/2, not 1/3.

Multiple choice

A bad contain $m$ white and $3$ black balls. Balls are drawn one bye one without replacement till all the black balls are drawn. The probability that this procedure for drawing balls will come to an end at the $rth$ draw is
  1. $\displaystyle \frac { \left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $
  2. $\displaystyle \frac { 3\left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $
  3. $\displaystyle \frac { 2\left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $

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

The procedure ends on the rth draw when the first r - 1 draws contain exactly two black balls and the rth draw is black. Counting the possible positions of the three black balls gives 3(r - 1)(r - 2)/((m + 1)(m + 2)(m + 3)).

Multiple choice

$(A)$A die is thrown once. Find the probability of getting $(i)$ a multiple of $2$ $(ii)$ A number lying between $1$ and $5$ $(iii)$ A child has a die whose six faces show the letters as shown below A B C D E A $(B)$The die is thrown once. What is the probability of getting $(i) A,: (ii) D$?
  1. $(A)\space (i)\space \displaystyle\frac{1}{3}\quad (ii) \space \displaystyle\frac{1}{2}\quad (iii)\space \displaystyle\frac{1}{5} \\ (B)\space (i)\space \displaystyle\frac{1}{2}\quad (ii) \displaystyle\frac{1}{6}$
  2. $(A)\space (i)\space \displaystyle\frac{1}{3}\quad (ii) \space \displaystyle\frac{1}{2}\quad (iii)\space \displaystyle\frac{1}{2} \\ (B)\space (i)\space \displaystyle\frac{1}{2}\quad (ii) \displaystyle\frac{1}{3}$
  3. $(A)\space (i)\space \displaystyle\frac{1}{2}\quad (ii) \space \displaystyle\frac{1}{2}\quad (iii)\space \displaystyle\frac{1}{2} \\ (B)\space (i)\space \displaystyle\frac{1}{3}\quad (ii) \displaystyle\frac{1}{6}$

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

A: (i) Multiples of 2 in {1,2,3,4,5,6} are {2,4,6}, prob = 3/6 = 1/2. (ii) Between 1 and 5 are {2,3,4}, prob = 3/6 = 1/2. (iii) Letters A,B,C,D,E,A. Prob(A) = 2/6 = 1/3. B: (i) Prob(A) = 2/6 = 1/3. (ii) Prob(D) = 1/6.

Multiple choice

$(A)$ From a pack of $52$ playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is $(i)$ a black queen $(ii)$ a red card $(iii)$ a ten $(iv)$ a picture card [jacks, queens and kings are picture cards] $(B)$ All cards of ace, jack and queen are removed from a deck of playing cards. One card is drawn at random from the remaining cards. Find the probability that the card drawn is $(i)$ A face card $(ii)$ Not a face card.
  1. $(A)(i)\quad \displaystyle\frac{1}{22}, \quad (ii)\quad \displaystyle\frac{9}{22}, \quad (iii) \quad \displaystyle\frac{1}{11}, \quad (iv)\quad \displaystyle\frac{3}{22} \\ (B)(i)\quad \displaystyle\frac{1}{10}, \quad (ii)\quad \displaystyle\frac{9}{10}$
  2. $(A)(i)\quad \displaystyle\frac{1}{14}, \quad (ii)\quad \displaystyle\frac{5}{22}, \quad (iii) \quad \displaystyle\frac{1}{11}, \quad (iv)\quad \displaystyle\frac{6}{13} \\ (B)(i)\quad \displaystyle\frac{1}{10}, \quad (ii)\quad \displaystyle\frac{9}{10}$
  3. $(A)(i)\quad \displaystyle\frac{1}{17}, \quad (ii)\quad \displaystyle\frac{9}{22}, \quad (iii) \quad \displaystyle\frac{1}{11}, \quad (iv)\quad \displaystyle\frac{11}{13} \\ (B)(i)\quad \displaystyle\frac{1}{10}, \quad (ii)\quad \displaystyle\frac{9}{10}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

A pair of dice is thrown simultaneously. Find the probability of getting $(i)$ a multiple of $3$ on both dice. $(ii)$ Sum of the numbers on two dice is always less than $7$. $(iii)$ An odd number on the first die and a prime number on the other.
  1. $(i)\quad \displaystyle\frac{1}{5} \\ (ii)\quad \displaystyle\frac{7}{13} \\ (iii)\quad \displaystyle\frac{1}{3}$
  2. $(i)\quad \displaystyle\frac{1}{7} \\ (ii)\quad \displaystyle\frac{11}{18} \\ (iii)\quad \displaystyle\frac{1}{7}$
  3. $(i)\quad \displaystyle\frac{1}{9} \\ (ii)\quad \displaystyle\frac{1}{12} \\ (iii)\quad \displaystyle\frac{1}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

$(A)$ There are $30$ cards numbered from $1$ to $30$. One card is drawn at random. Find the probability that the number of the selected card is not divisible by $3$. $(B)$ A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1,2,3,4,5,6,7,8$ and these are equally likely outcomes. What is the probability that it will point at $(i)$ $8$ $(ii)$ An odd number $(iii)$ A number greater than $2$ $(iv)$ A number less than $9$
  1. $(A)\quad \displaystyle\frac{1}{3}\\(B)\space (i)\quad \displaystyle\frac{1}{8}\\ (ii)\quad \displaystyle\frac{1}{2}\\ (iii)\quad \displaystyle\frac{1}{4} \\ (iv)\quad 1$
  2. $(A)\quad \displaystyle\frac{1}{3}\\(B)\space (i)\quad \displaystyle\frac{10}{19}\\ (ii)\quad \displaystyle\frac{1}{2}\\ (iii)\quad \displaystyle\frac{3}{4} \\ (iv)\quad 1$
  3. $(A)\quad \displaystyle\frac{2}{3}\\(B)\space (i)\quad \displaystyle\frac{11}{16}\\ (ii)\quad \displaystyle\frac{1}{2}\\ (iii)\quad \displaystyle\frac{1}{4} \\ (iv)\quad 1$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

An urn contains 52 balls, numbered from 1 to 52. Let the balls be drawn one at a time and divided among four players $A, B, C, D$. The players $A $ gets the ball drawn on the first, fifth, ninth draws, ...etc., player $B$ gets the balls drawn on the second, sixth, tenth, ....draws and so on.Suppose that the balls numbered 1, 11, 31 and 41 are considered "Lucky". What is the probability that each player will have a 'Lucky' ball?
  1. $\displaystyle \dfrac{4!(13)^{4}}{2.49.50.51.52}$
  2. $\displaystyle \dfrac{2!(13)^{4}}{49.50.51.52}$
  3. $\displaystyle \dfrac{(13)^{4}}{49.50.51.52}$
  4. $\displaystyle \dfrac{4!(13)^{4}}{49.50.51.52}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

There are 52 balls and 4 players. Each player receives 13 balls. The total ways to distribute 52 balls into 4 groups of 13 is 52! / (13!)^4. The number of ways to distribute the 4 lucky balls such that each player gets one is 4! * (ways to distribute the remaining 48 balls). This leads to the probability calculation involving combinations, which simplifies to the provided answer D.

Multiple choice

Let $a_1, a_2, a_3$ are three consecutive terms of an increasing $A.P.$, where $a_1$ and $a_2$ are prime numbers such that their sum is minimum possible odd prime number. Urn-1: Contains $a_1$ red and $a_3$ green balls, Urn-2 : Contains $a_2$ red and $a_2$ green balls, Urn -3 : Contains $a_3$ red and $a_1$ green balls. P(i) represents the probability of choosing $i^{th}$ urn & $P(R)$ represents probability of choosing red ball & similarly $P(G)$ represents the probability of choosing green ball. On the basis of above information answer the following: If $P(i) \propto i^2$ and one ball is drawn from one of these urns then -
  1. $P(G) = \dfrac{3}{7}$
  2. $P(R) = \dfrac{6}{7}$
  3. $P(R) > P(G)$
  4. $P(R) = P(G)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

a1, a2, a3 are consecutive primes with minimum sum as odd prime. Primes are 2, 3, 5. a1=2, a2=3, a3=5. Urn probabilities are proportional to i^2 (1:4:9). Total weight = 14. P(R) = (1/14 * 2/7) + (4/14 * 3/6) + (9/14 * 5/7) = 2/98 + 2/14 + 45/98 = (2+14+45)/98 = 61/98. P(G) = 1 - 61/98 = 37/98. P(R) > P(G).

Multiple choice

The chance of an event happening is the square of the chance, of a second event but the odds against the first are the cubes of the odds against the first are the cubes of the odds against the second. Find the chance of each.
  1. probability of first event $\displaystyle =\dfrac{1}{9}$ ,probability of second event $\displaystyle =\dfrac{1}{3}$
  2. probability of first event $\displaystyle =\dfrac{1}{3}$, probability of second event $\displaystyle =\dfrac{1}{9}$
  3. probability of first event $\displaystyle =\dfrac{1}{4}$, probability of second event $\displaystyle =\dfrac{1}{2}$
  4. probability of first event $\displaystyle =\dfrac{1}{2}$, probability of second event $\displaystyle =\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the second probability be q, so the first probability is q^2. Using odds against, (1 - q^2)/q^2 = ((1 - q)/q)^3, which gives q = 1/3. Hence the first probability is 1/9 and the second is 1/3.

Multiple choice

Write the probability distribution when three coins are tossed.
  1. $X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$
  2. $X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 7 }{ 8 } \end{matrix}$
  3. $X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 7 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$
  4. $X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For 3 coins, the outcomes are 0 heads (1), 1 head (3), 2 heads (3), 3 heads (1) out of 8 total. The probability distribution is 1/8, 3/8, 3/8, 1/8.

Multiple choice

A bag contains $6$ red balls, $8$ white balls, $5$ green balls and $2$ black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is $(i)$ White $(ii)$ Red or black $(iii)$ Not green $(iv)$ Neither white nor black
  1. $(i)\quad \displaystyle\frac{1}{11}\ \ (ii)\quad \displaystyle\frac{7}{22}\ \ (iii)\quad \displaystyle\frac{13}{22}\ \ (iv)\quad \displaystyle\frac{1}{4}$
  2. $(i)\quad \displaystyle\frac{4}{11}\ \ (ii)\quad \displaystyle\frac{9}{22}\ \ (iii)\quad \displaystyle\frac{17}{22}\ \ (iv)\quad \displaystyle\frac{1}{2}$
  3. $(i)\quad \displaystyle\frac{5}{11}\ \ (ii)\quad \displaystyle\frac{13}{22}\ \ (iii)\quad \displaystyle\frac{19}{22}\ \ (iv)\quad \displaystyle\frac{1}{7}$

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