Mathematics

Probability

303 Questions

Probability measures the likelihood of an event occurring, such as rolling a specific number on a die or drawing a colored ball. Questions cover simple events, mutually exclusive outcomes, and dice or coin combinations. This topic is consistently asked in mathematics and reasoning sections of competitive exams.

dice probabilitycoin toss eventsdrawing balls probabilitiesplaying card problemsmutually exclusive events

Probability Questions

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

A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occasions. The chance that the die chosen was a rigged die, is

  1. $\displaystyle \frac {216}{217}$
  2. $\displaystyle \frac {215}{219}$
  3. $\displaystyle \frac {216}{219}$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $A$ is the event which selects the rigged one and $B$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $P(A/E)=\dfrac{1}{4}(1)^3=\dfrac{1}{4}$ ($\dfrac{1}{4}$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $P(B/E)=\dfrac{3}{4}(\dfrac{1}{6})^3=\dfrac{1}{1152}$ (since probability of getting 5 in fair dice case=$\dfrac{1}{5}$)
By Baye's theorem,Probability of selecting the rigged case among both=$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$

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

Box $I$ contains $5$ red and $4$ blue balls, while box $II$ contains $4$ red and $2$ blue balls. A fair die is thrown. If it turns up a multiple of $3$, a ball is drawn from the box $I$ else a ball is drawn from box $II$. Find the probability of the event ball drawn is from the box $I$ if it is blue.

  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{15}{19}$
  3. $\displaystyle \frac{4}{19}$
  4. $\displaystyle \frac{10}{27}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Box $I$ contains $:5$ red $+{4}$ blue
Box $II$ contains $:4$ red $+{2}$ blue
A ball is taken from Box $I$ if a multiple of $3$ comes up i.e, $3$ and $6.$

Ball is taken from Box $II$ when $1,2,4$ and $5$ turns up.

$\Rightarrow$ Event of picking up from Box $I=P(A _1)=\dfrac{2}{6}=\dfrac{1}{3}.$

$\Rightarrow$ Event of picking up from Box $II=P(A _2)=\dfrac{4}{6}=\dfrac{2}{3}.$

$\Rightarrow R=$ event of drawing a blue ball

$=P(A _1)P(\dfrac R{A _1})+P(A _2)P(\dfrac{R}A _2)$

$=\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{3}\times\dfrac{2}{6}$

$=\dfrac{4}{27}+\dfrac{4}{18}$

$=\dfrac{10}{27}.$
Hence, the answer is $\dfrac{10}{27}.$

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

There are three different Urns, Urn-I, Urn-II and Urn-III containing 1 Blue, 2 Green, 2 Blue, 1 Green, 3 Blue, 3 Green balls respectively. If two Urns are randomly selected and a ball is drawn from each Urn and if the drawn balls are of different colours then the probability that chosen Urn was Urn-I and Urn-II is

  1. $\dfrac {1}{7}$
  2. $\dfrac {5}{13}$
  3. $\dfrac {5}{14}$
  4. $\dfrac {5}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Required probability$\displaystyle =\dfrac {\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {1}{3}+\dfrac {2}{3}.\dfrac {2}{3}\right )}{\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {2}{3}.\dfrac {2}{3}\right )+\dfrac {1}{3}\left (\dfrac {2}{3}.\dfrac {3}{6}+\dfrac {1}{3}.\dfrac {3}{6}\right )+\dfrac {1}{3}\left (\dfrac {3}{6}.\dfrac {2}{3}+\dfrac {3}{6}.\dfrac {1}{3}\right )}$

$\displaystyle =\dfrac {\dfrac {5}{9}}{\dfrac {5}{9}+\dfrac {9}{18}+\dfrac {9}{18}}\\ =\dfrac {5}{14}$

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

A man is know to speak the truth $3$ out if $4$ times. He throws a die and reports that it is a six. The probability that it is actually a six is:

  1. $\dfrac{3}{8}$
  2. $\dfrac{1}{5}$
  3. $\dfrac{3}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let E be the event that the man reports that six occurs in the throwing of the die and let $S _1$ be the event that six occurs and $S _2$ be the event that six does not occur.
$P(S _1)=\dfrac 16, P(S _2)= 1-\dfrac 16=\dfrac 56$
$P(E/S _1)$=probability that the man reports that six occurs when 6 has actually occurred on the die.
$P(E/S _1)$=probability that the man speaks the truth=$\dfrac 34$
$P(E/S _2)$=probability that the man reports that six occurs when 6 has not actually occurred on the die.
$P(E/S _2)$=probability that the man does not speak the truth
$\implies = 1−\dfrac 34=\dfrac 14$
Hence by Baye's theorem, we get,
$P(S _1/E)$=Probability that the report of the man that six has occurred is actually a six.
$P(S _1/E)=\dfrac {P(S _1).P(E/S _1)}{P(S _1)P(E/S _1)+P(S _2).P(E/S _2)}\\\implies = \dfrac {\dfrac 16\times \dfrac 34}{\dfrac 16\times \dfrac 34+\dfrac 56\times \dfrac 14}=\dfrac 18\times \dfrac {24}{8}=\dfrac {3}{8}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

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

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

Let $E _1(0\leq i\leq 2)$ denote the event that urn contains $i$ white and $(2-i)$ black balls.
Let $A$ denote the event that a white ball is drawn from the urn.
We have $P(E _i)=1/3$ for $i=0, 1, 2$. and $P(A|E _1)=1/3, P(A|E _2)=2/3, P(A|E _3)=1$.
By the total probability rule,
$P(A)=P(E _1)P(A|E _1)+P(E _2)P(A|E _2)+P(E _3)P(A|E _3)$
$\displaystyle =\frac {1}{3}\left [\frac {1}{3}+\frac {2}{3}+1\right ]=\frac {2}{3}$

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

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is

  1. $\dfrac38$
  2. $\dfrac15$
  3. $\dfrac34$

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

Let $E$ be the event that the man reports that six occurs whle throwing the die and let $S$ be the event that six occurs. Then 
$P(S)=$ Probability that six occurs $ \displaystyle =\frac { 1 }{ 6 }   $
$P\left( { S }^{ 1 } \right) =$ probability that six does not occur $ \displaystyle =1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } $
$ \displaystyle P\left( \frac { F }{ S }  \right) = $ probability that the man reports that six occurs when six has actually occurred 
$=$ probability that the man reports the truth $ \displaystyle =\frac { 3 }{ 4 }  $ 
$ \displaystyle P\left( \frac { E }{ { S }^{ 1 } }  \right) =$ probability that the man report that six occur when six has not actually occurred. 
$=$ probability that the man does not speak the truth 
$ \displaystyle 1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } . $ 
By Bayes' theorem 
$ \displaystyle P\left( \frac { S }{ E }  \right) =$ probability that the man 
reports that six occurs when six has actually occured 
$ \displaystyle =\frac { P\left( S \right) P\left( \frac { F }{ S }  \right)  }{ P\left( S \right) \times P\left( \frac { F }{ S }  \right) +P\left( { S }^{ 1 } \right) \times P\left( \frac { E }{ { S }^{ 1 } }  \right)  }$ 
$ \displaystyle =\frac { \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 }  }{ \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } \times \dfrac { 1 }{ 4 }  } =\frac { 1 }{ 8 } \times \frac { 24 }{ 8 } =\frac { 3 }{ 8 }  $

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

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is $10$. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains $ 1$ white and $9$ black balls is

  1. $\dfrac {14}{55}$
  2. $\dfrac {12}{55}$
  3. $\dfrac {2}{11}$
  4. $\dfrac {8}{55}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $E _i$ denote the event that the bag contains $i$ black and ($10-i$) white balls $(i=0, 1, 2, ...., 10)$. Let $A$ denote the event that the three balls drawn at random from the bag are black. We have
$P(E _i)=\dfrac {1}{11} (i=0, 1, 2, ...., 10)$
$P(A|E _i)=0$ for $i=0, 1, 2$
and $P(A|E _i)=\dfrac {^iC _3}{^{10}C _3}$ for $i\geq 3$
Now, by the total probability rule
$\displaystyle P(A)=\sum _{i=0}^{10}P(E _i)P(A|E _i)$
$=\frac {1}{11}\times \frac {1}{^{10}C _3}[^3C _3+^4C _4+....+^{10}C _3]$
But $^3C _3+^4C _3+^5C _3+....+^{10}C _3$
$=^4C _4+^4C _3+^5C _3+...+^{10}C _3$
$=^5C _4+^5C _3+^6C _3+....+^{10}C _3$
$=^6C _4+^6C _3+....+^{10}C _3=....=^{11}C _4$
Thus, $P(A)=\dfrac {^{11}C _4}{11\times ^{10}C _3}=\dfrac {1}{4}$
By the Bayes' rule
$P(E _9|A)=\dfrac {P(E _9)P(A|E _9)}{P(A)}=\dfrac {\dfrac {1}{11}\dfrac {(^9C _3)}{^{10}C _3}}{\dfrac {1}{4}}=\dfrac {14}{55}$.

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

A box contain $N$ coins, $m$ of which are fair are rest and biased. The probability of getting a head when a fair coin is tossed is $1/2$, while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

  1. $\dfrac {9m}{8N+m}$
  2. $\dfrac {m}{8N+m}$
  3. $\dfrac {N}{8n+m}$
  4. $\dfrac {1}{m}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $E _1, E _2$ and $A$ denote the following events:
$E _1$: coin selected is fair
$E _2$: coin selected is biased
$A: $ the first toss results in a head and the second toss results in a tail.
$\displaystyle P(E _1)=\frac {m}{N}, P(E _2)=\frac {N-m}{N}$,
$\displaystyle P(A|E _1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E _2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$.
By Bayes' rule
$\displaystyle P(E _1|A)=\frac {P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)}=\frac {9m}{8N+m}$.

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

A bag contains $(2n+1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\displaystyle \frac{31}{42}$, then $n$ is equal to 

  1. $10$
  2. $11$
  3. $12$
  4. $13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let ${ A } _{ 1 }$ denote the event that a coin having heads on both sides is chosen, and ${ A } _{ 2 }$ denote the vent that a fiar coin is chosen.
Let $E$ denote the vent that head occurs, Then
$\displaystyle P\left( { A } _{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A } _{ 2 } \right) =\frac { n+1 }{ 2n+1 } $
Probability of occurrence of event $E$, if unfair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 1 } }  \right) =1$
Probability of occurrence of event $E$, if fair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 2 } $
$\because P\left( E \right) =P\left( { A } _{ 1 }\cap E \right) +P\left( { A } _{ 2 }\cap E \right) $
$\displaystyle \therefore P\left( E \right) =P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right) $
$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right)  } \ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$

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

The contents of urn I and II are as follows:
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. Find the probability the chosen urn was I.

  1. $\displaystyle \frac{125}{287}$
  2. $\displaystyle \frac{64}{127}$
  3. $\displaystyle \frac{25}{287}$
  4. $\displaystyle \frac{79}{192}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Bayes' theorem, we compare the likelihood of the observed sequence of balls (1 white, 3 black) occurring in Urn I versus Urn II. The calculation involves binomial probabilities for each urn.

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

One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. What is the probability that both the balls will be of the same colour?

  1. $207/625$
  2. $191/625$
  3. $23/625$
  4. $227/625$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Bag $I=$ $3$ White $+$ ${7}$ Red $+$ $15$ Black

Bag $II=$ $10$ White $+$ ${6}$ Red $+$ $9$ Black

Each beg contains total of $25$ balls.

There are three cases for selection of a particular ball :

$1.$ White ball from Bag $I$ and Bag $II=\dfrac{3}{25}\times\dfrac{10}{25}$

$2.$ Red ball from Bag $I$ and Bag $II=\dfrac{7}{25}\times\dfrac{6}{25}$

$3.$ Black ball from Bag $I$ and Bag $II=\dfrac{15}{25}\times\dfrac{9}{25}$

$\therefore$ Total probability $=\dfrac{30}{625}+\dfrac{42}{625}+\dfrac{135}{25}=\dfrac{207}{625}.$

Hence, the answer is $\dfrac{207}{625}.$
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

  1. $0$
  2. $\displaystyle\frac { 1 }{ 3 } $
  3. $\displaystyle\frac { 1 }{ 12 } $
  4. $\displaystyle\frac { 1 }{ 36 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two dice are rolled, the number of outcomes is $36$.
The only even prime number is $2$.
Let $E$ be the event of getting an even prime number on each die.
$\therefore E=\left{ \left( 2,2 \right)  \right} $
$\Rightarrow P\left( E \right) =\displaystyle\frac { 1 }{ 36 } $
Therefore, the correct answer is (D).