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 business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

The chance of throwing a total of $3$ or $5$ or $11$ with two dice is:

  1. $\dfrac{5}{36}$
  2. $\dfrac{1}{9}$
  3. $\dfrac{2}{9}$
  4. $\dfrac{19}{36}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The number of possible combinations with two dice is $6 \times 6=36$.

The number of ways of getting $3$ as sum is $2$.
The number of ways of getting $5$ as sum is $4$.
The number of ways of getting $11$ as sum is $2$.
The number of ways of getting $3$ or $5$ or $11$ as sum is $8$.
The probability is $\cfrac{8}{36}=\cfrac{2}{9}$.
Therefore option $C$ is correct.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Two dice each numbered from $1$ to $6$ are thrown together. Let $A$ and $B$ be two events given by
$A:$ even number on the first die
$B:$  number on the second die is greater than $4$

What is $P(A\cup B)$ equal to?

  1. $1/2$
  2. $1/4$
  3. $2/3$
  4. $1/6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given:

Two dice are thrown, hence the total number of all possible ways, n(S) = $6\times 6=36$

A:  even number on the first die
B: the number on the second die is greater than 4

To find:
$P(A\cup B) $
favourable ways of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Hence n(A)=18
$\therefore, P(A)=\dfrac {n(A)}{n(S)}=\dfrac {18}{36}=\dfrac 12$
favourable ways of event B = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}
Hence n(B) = 12
$\therefore, P(B)=\dfrac {n(B)}{n(S)}=\dfrac {12}{36}=\dfrac 13$
Hence, $P(\cup B)=P(A)+P(B)-P(A)(P(B)=\dfrac 12+\dfrac 13-\dfrac 12\times \dfrac 13=\dfrac {3+2-1}6=\dfrac 46=\dfrac 23$

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Find the probability of getting a total of $7$ or $11$ when a pair of dice is tossed.

  1. $\dfrac{1}{9}$
  2. $\dfrac{7}{9}$
  3. $\dfrac{2}{9}$
  4. $\dfrac{5}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Total number of all possible outcomes, $n(S)=6\times 6=36$
Let A: be the event that the sum is $7$.
Hence the favourable outcomes are ${(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$, i.e., $n(A)=6$
Hence the probability of getting a total of $7$. $P(E)=\dfrac {n(A)}{n(S)}=\dfrac {6}{36}$
Similarly, let B: be the event that the sum is $11$
Hence the favourable outcomes are {(5, 6), (6, 5)}, i.e., $n(E)=2$
Hence the probability of getting a total of 11 $P(B)=\dfrac {n(B)}{n(S)}=\dfrac {2}{36}$
Now, the probability of getting a total of $7$ or $11$ is
$P(A)+P(B)\\\implies =\dfrac 6{36}+\dfrac 2{36}=\dfrac 8{36}=\dfrac 29$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either $3$ or $13$?

  1. $\dfrac{2}{5}$
  2. $\dfrac{1}{9}$
  3. $\dfrac{11}{27}$
  4. $\dfrac{9}{27}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: first $30$ natural numbers

To find: the probability of getting a multiple of either $3$ or $13$
According to the question, 
$n(S)=30$
Multiple of 3 in first 30 natural numbers are ${3, 6, 9, 12, 15, 18, 21, 24, 27, 30} = 10$
Multiple of $13$ in first $30$ natural numbers are ${13, 26}$
Hence the probability of getting a multiple of either $3$ or $13$ = probability of getting a multiple of 3 or probability of getting a multiple of 13
$\implies \dfrac {10}{30}+\dfrac 2{30}=\dfrac {12}{30}=\dfrac 25$
is the required probability.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A, B and C in order toss a coin. the first one to throw a head wins. If A starts to toss, then 

  1. $P(A)=\dfrac{4}{7}$
  2. $P(B)=\dfrac{2}{7}$
  3. $P(C)=\dfrac{1}{7}$
  4. $P(C)=\dfrac{2}{7}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
P(A)=H+TTTH+TTTTTTH+.....
        =$1/2+{1 (1/2 })^{ 4 }+{ (1/2 })^{ 7 }+....$
        $=\frac { 1/2 }{ 1-1/8 } =4/7$
P(B)=TH+TTTTH+TTTTTTTH+.....
        =${ (1/2 })^{ 2 }+{ (1/2 })^{ 5 }+....$
        $=\frac { 1/4 }{ 1-1/8 } =2/7$
P(C)=TTH+TTTTTH+TTTTTTTTH+.....
        =${ (1/2 })^{ 3 }+{ (1/2 })^{ 6 }+....$
        $=\frac { 1/8 }{ 1-1/8 } =1/7$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A bag contain $5$ balls of unknown colors. A ball is drawn at random from it and is found to be white. The probability that bag contains only white ball is

  1. $\dfrac{3}{5}$
  2. $\dfrac{1}{5}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $Q$ be the event that the drawn balls are white

$A$ be the event that the bag contains $1$ white ball
$B$ be the event that the bag contains $2$ white balls
$C$ be the event that the bag contains $3$ white balls
$D$ be the event that the bag contains $4$ white balls
$E$ be the event that the bag contains $5$ white balls

$P(A)=1/5,\ P(B)=1/5,\ P(C)=1/5,\ P(D)=1/5,\ P(E)=1/5$
$P(Q/A)=\dfrac {^1C _1}{^5C _1}=\dfrac {1}{5}$
$P(Q/B)=\dfrac {^2C _1}{^5C _1}=\dfrac {2}{5}$
$P(Q/C)=\dfrac {^3C _1}{^5C _1}=\dfrac {3}{5}$
$P(Q/D)=\dfrac {^4C _1}{^5C _1}=\dfrac {4}{5}$
$P(Q/E)=\dfrac {^5C _1}{^5C _1}=1$

using Baye's thorem,

$P(A/Q)=\dfrac {P(E)\ P(Q/E)}{P(A)\ P(Q/A)+P(B)\ P(Q/B)+P(C)\ P(Q/C)+P(D)\ P(Q/D)+P(E)\ P(Q/E)}$

$=\dfrac {1/5. 1}{1/5. 1/5+1/5.2/5+1/5.3/5 +1/5.4/5+1/5.1.}$

$=\dfrac {1/5}{1/5 (1/5)+2/5+3/5+4/5+1}$

$=\dfrac {1}{\dfrac {1+2+3+4+5}{5}}=\dfrac {5}{1+2+3+4+5}=\dfrac {5}{15}=\dfrac {1}{3}$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:

Sum  Frequency
2 14
3 30
4 43
5 55
6 72
7 75
8 70
9 53
10 46
11 28
12 15

If the dice are thrown once more, what is the probability of getting a sum between 8 and 12?

  1. $0.154$
  2. $0.20$
  3. $0.254$
  4. $0.30$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If we rolled dice around 500 times then it can be considered as its expected outcome.
Hence,
Probability of obtaining between 8 and 12 as sum = 53 + 46 +28/500= 0.254

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Box I contains $2$ white and $3$ red balls and box II contains $4$ white and $5$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

  1. $\cfrac{54}{44}$
  2. $\cfrac{54}{14}$
  3. $\cfrac{54}{104}$

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

Probability that the ball drawn is red and from ! =$P(R/A)$

$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$
$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$
$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$
$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A bag contains 12 balls out of which x are white.If one ball is drawn at random, what is the probability it will be a white ball?

  1. $ \displaystyle \frac{x}{2}$
  2. $\displaystyle \frac{x}{12}$
  3. $ \displaystyle \frac{x}{10}$
  4. $ \displaystyle \frac{12}{x}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total number of balls = 12
Number of white balls = x
P (white ball) = $= \displaystyle \frac{x}{12}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bag is found to be red. Find the probability that it was drawn from bag $B$.

  1. $\dfrac{3}{8}$
  2. $\dfrac{25}{52}$
  3. $\dfrac{1}{8}$
  4. $\dfrac{3}{14}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $X$ be the probability of choosing bag $A$,$Y$ be the probability of choosing bag $B$


Let $E$ be the probability of ball drawn is red


Then $P\left( X \right) = \dfrac{1}{2}$

$P\left( Y \right) = \dfrac{1}{2}$

$P\left( {E/X} \right) = \dfrac{3}{5}$

$P\left( {E/Y} \right) = \dfrac{5}{9}$

Apply the Bayes theorem:

$P\left( {Y/E} \right) = \dfrac{{P\left( Y \right) \times P\left( {E/Y} \right)}}{{P\left( X \right) \times P\left( {E/X} \right) + P\left( Y \right) \times P\left( {E/Y} \right)}}$

                 $ = \dfrac{{\dfrac{1}{2} \times \dfrac{5}{9}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{5}{9}}}$

                 $ = \dfrac{{50}}{{104}}$

                 $ = \dfrac{{25}}{{52}}$

Hence, the probability that the red ball is drawn from bag $B$ is  $ = \dfrac{{25}}{{52}}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

An urn contains $10$ balls coloured either black or red When selecting two balls from the urn at random, the probability that a ball of each color is selected is $8/15$. Assuming that the urn contains more black balls then red balls, the probability that at least one black ball is selected, when selecting two balls, is

  1. $\dfrac {18}{45}$
  2. $\dfrac {30}{45}$
  3. $\dfrac {39}{45}$
  4. $\dfrac {41}{45}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Number of black balls $=x$

Number of red balls $=y$

$i)=x+y=10$ (Total)

$ii)\ P$ (selecting exactly $1$ black & one red)

$=^xC _1\times ^yC _1 /^{10}C _2=8/15$

by equation : $x^2-10x+24=0\ ;\ x=4$ or $6$

since $ x > y$ it is $6$

$iii)\ P$ (selecting at least one black)

$=^xC _1\times ^yC _1 +^xC _2 /^{10}C _2$

Above equation reduced to $\Rightarrow \ 2xy+x(x-1)/90$

putting $x=6$, results is $39/45$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than $7$ if $4$ appear on the top face of the first dice is...

  1. $\dfrac {1}{3}$
  2. $\dfrac {1}{2}$
  3. $\dfrac {1}{12}$
  4. $\dfrac {1}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l}\left. \begin{array}{l}{\rm{combinations}} = \left( {4,1} \right)\left( {4,2} \right)\\left( {4,3} \right)\left( {4,4} \right)\end{array} \right} \le 7\\left. {\left( {4,5} \right)\left( {4,6} \right)} \right} > 7\{\rm{P}}\left( { > 7} \right) = \frac{1}{3}\end{array}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A bag contains $6$ red, $4$ white and $8$ blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

  1. $\dfrac {2}{17}$
  2. $\dfrac {3}{17}$
  3. $\dfrac {5}{17}$
  4. $\dfrac {4}{17}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$E\rightarrow$ Event of getting one is red, one is white and one $6$ red $+4$ white $+8$ blue balls $=18$ balls

Total outcomes $=(\ ^{18}C _{3})$

$\underbrace { \bigodot  } _{ R } \underbrace { \bigodot  } _{ W } \underbrace { \bigodot  } _{ B } \rightarrow$ no. of fobourable element

$=\ ^{6}C _{1}\times \ ^{4}C _{1}\times \ ^{8}C _{1}$

$=6\times 4\times 8$

$\therefore P(E)=\dfrac{6\times 4\times 8}{\ ^{18}C _{3}}=\dfrac{6\times 4\times 8\times 3\times 2\times 1}{18\times 17\times 16}$

$=\dfrac{4}{17}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

There are two balls in an urn whose colors are not known ( ball can be either white or black). A white ball is put into the urn. A ball is then drawn from the urn. The probability that it is white is 

  1. $\displaystyle \frac { 1 }{ 4 } $
  2. $\displaystyle \frac { 1 }{ 3 } $
  3. $\displaystyle \frac { 2 }{ 3 } $
  4. $\displaystyle \frac { 1 }{ 6 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\displaystyle { E } _{ i }\left( 0\le i\le 2 \right) $ denotes the event that urn contains $i$ white and $2-i$ black balls.

Let $A$ denotes the event that a white ball is drawn from the urn.
We have $\displaystyle P\left( { E } _{ i } \right) =\frac { 1 }{ 3 } $ for $i=0,1,2$ and $\displaystyle P\left( \frac { A }{ { E } _{ i } }  \right) =\frac { 1 }{ 3 } ,P\left( \frac { A }{ { E } _{ 2 } }  \right) =\frac { 2 }{ 3 } ,P\left( \frac { A }{ { E } _{ 3 } }  \right) =1$
By the total probability rule,
$\displaystyle P\left( A \right) =P\left( { E } _{ 1 } \right) P\left( \frac { A }{ { E } _{ 1 } }  \right) +P\left( { E } _{ 2 } \right) P\left( \frac { A }{ { E } _{ 2 } }  \right) +P\left( { E } _{ 3 } \right) P\left( \frac { A }{ { E } _{ 3 } }  \right) $
$\displaystyle =\frac { 1 }{ 3 } \left[ \frac { 1 }{ 3 } +\frac { 2 }{ 3 } +1 \right] =\frac { 2 }{ 3 } $

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

There are $3$ coins in a box. One is a two-headed coin; another is a fair coin; and third is biased coin that comes up heads $75\%$ of time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that its was the two-headed coin ?

  1. $\dfrac{2}{9}$
  2. $\dfrac{1}{3}$
  3. $\dfrac{4}{9}$
  4. $\dfrac{5}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Formula using Baye's theorem,

$P(E _1|A)=\dfrac{P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)+P(E _3)P(A|E _3)}$

Let E1: event that the coin is 2 headed, 

E2 be the event that its biased with heads 75% of the time and 

E3 be the fair coin.

All these three events are mutually exclusive and exchaustive, and are equally likely.

$\therefore P(E _1)=P(E _2)=P(E _3)=\dfrac{1}{3}$

P( coin shows head given that its 2 headed coin)$ =P(E|E _1)=1$

P(coin shows head given that its 75% biased for heads) $=P(E|E _2)=\dfrac{3}{4}$

P(coin shows head given that its a fair coin) $=P(E|E3)=\dfrac{1}{2}$

$\therefore P(E _1|E)=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{3}{4} +\dfrac{1}{3}\cdot \dfrac{1}{2}}$

$P(E _1|E)=\dfrac{4}{9}$