Mathematics

Probability Distributions

457 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

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

A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8. from house B, 5 from  house C, 2 from house 0 and rest from house E. A single student is selected at random ,to be the class monitor. The probability that the selected student is not from A, Band C is?

  1. $\displaystyle \frac{4}{23}$
  2. $\displaystyle \frac{6}{23}$
  3. $\displaystyle \frac{8}{23}$
  4. $\displaystyle \frac{17}{23}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Total number of students, n(S) = 23

Number of students in houses A,B and C 

                                = 4+8+5 = 17 

∴ Remaining  students = 23 - 17 = 6 n(E) = 6

So, probability that the selected students is not from A,B and C

$P(E)=\dfrac{6}{23}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

An artillery target may be either at point $I$ with probability $\cfrac{8}{9}$ or at point $II$ with probability $\cfrac{1}{9}$. We have $21$ shells each of which can be fired at point $I$ or $II$. Each shell may hit the target independently of the other shell with probability $\cfrac{1}{2}$. How many shells must be fired at point $I$ to hit the target with maximum probability?

  1. $P(A)$ is maximum where $x=11$.
  2. $P(A)$ is maximum where $x=12$.
  3. $P(A)$ is maximum where $x=14$.
  4. $P(A)$ is maximum where $x=15$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $A$ denote the event that the target is hit when $x$ shells are fired at point $I$.
Let ${ E } _{ 1 }$ and ${ E } _{ 2 }$ denote the events hitting $I$ and $II$, respectively
$\displaystyle \therefore P\left( { E } _{ 1 } \right) =\frac { 8 }{ 9 } ,P\left( { E } _{ 2 } \right) =\frac { 1 }{ 9 } $
Now $\displaystyle P\left( \frac { A }{ { E } _{ 1 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x }$ and $\displaystyle P\left( \frac { A }{ { E } _{ 2 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }$
Hence $\displaystyle P\left( A \right) =\frac { 8 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x } \right] +\frac { 1 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x } \right] $
$\displaystyle \therefore \frac { dP\left( A \right)  }{ dx } =\frac { 8 }{ 9 } \left[ { \left( \frac { 1 }{ 2 }  \right)  }^{ x }\log { 2 }  \right] +\frac { 1 }{ 9 } \left[ -{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }\log { 2 }  \right] $
For maximum probability $\displaystyle \frac { dP\left( A \right)  }{ dx } =0$
$\therefore x=12$   $\left[ \because { 2 }^{ 3-x }={ 2 }^{ x-21 }\Rightarrow 3-x=x-21 \right] $
Since $\displaystyle \frac { { d }^{ 2 }P\left( A \right)  }{ dx^{ 2 } } <0$ for $x=12$
$\therefore P\left( A \right) $ is maximum for $x=12$

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

In an entrance test, there are multiple choice questions. There are four possible options of which one is correct. The probability that a student knows the answer to a question is $90$%. If he gets the correct answer to a question, then the probability that he was guessing is

  1. $\displaystyle \frac { 1 }{ 37 } $
  2. $\displaystyle \frac { 36 }{ 37 } $
  3. $\displaystyle \frac { 1 }{ 4 } $
  4. $\displaystyle \frac { 1 }{ 49 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We define the following events
${ A } _{ 1 }:$ He knows the answer
${ A } _{ 2 }:$ He does not know the answer
$E:$ He gets the correct answer
Thus $\displaystyle P\left( { A } _{ 1 } \right) =\frac { 9 }{ 10 } ,P\left( { A } _{ 2 } \right) =1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } ,P\left( \frac { E }{ { A } _{ 1 } }  \right) =1,P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 4 } $
$\therefore$ required probability $\displaystyle =P\left( \frac { { A } _{ 2 } }{ E }  \right) =\frac { P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \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 =\frac { \dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  }{ \dfrac { 9 }{ 10 } .1+\dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  } =\frac { 1 }{ 36+1 } =\frac { 1 }{ 37 } $

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

$A$ is one of $6$ horses entered for a race, and is to be ridden by one of two jockeys $B$ and $C$. It is $2$ to $1$ that $B$ rides $A$, in which case all the horses are equally likely to win; if $C$ rides $A$, his chance is trebled; what are the odds against his winning?

  1. $13:5$
  2. $13:18$
  3. $18:13$
  4. $5:13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $E _{1}$ be the event that $B$ rides $A$, $E _{2}$, the event that $C$ rides $A$ and $E$ the event that $A$ wins. 
Then according to the question, $\displaystyle P(E _{1})=\dfrac{2}{3}, P(E _{2})=1-\dfrac{2}{3}=\dfrac{1}{3} P(E/E _{1})=\dfrac{1}{6}$ 
(since all the $6$ horses are equally likely to win when $B$ rides $A$)
$P(E/E _{2})=3\times \dfrac{1}{6}=\dfrac{1}{2}$ 
(since $A$'s chance of winning is trebled when $C$ rides $A$) 
$\displaystyle \therefore P(E)=P(E _{1})P(E/E _{1})+P(E _{2})P(E/E _{2})=\dfrac{2}{3}\cdot \dfrac{1}{6}+\dfrac{1}{3}\cdot \dfrac{1}{2}=\dfrac{58}{18}$ 
so that odds against $A$'s win are as $ (18-5):5$, that is $13:5$.
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

An employer sends a letter to his employee but he does not receive the reply (It is certain that employee would have replied if he did receive the letter). It is known that one out of $n$ letters does not reach its destination. Find the probability that employee does not receive the letter.

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

Let $E$ be the event that employee received the letter and $A$ that employer received the reply, then
$\displaystyle P\left ( E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$

$\displaystyle P\left ( A/E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( A/\bar{E} \right )= 0$
Now $\displaystyle P\left ( A \right )= P\left ( E\cap A \right )+P\left ( \bar{E}\cap A \right )$
$\displaystyle = P\left ( E \right ).P\left ( A/E \right )+P\left ( \bar{E} \right ).P\left ( A/\bar{E} \right )$
$\displaystyle = \left ( \frac{n-1}{n} \right )\left ( \frac{n-1}{n} \right )+\frac{1}{n}.0$
$\displaystyle P\left ( A \right )= \left ( \frac{n-1}{n} \right )^{2}$
$\displaystyle
P\left ( \bar{A} \right )= 1-\left ( \frac{n-1}{n} \right )^{2}=
\frac{n^{2}-n^{2}-1+2n}{n^{2}}= \frac{2n-1}{n^{2}}$
Now the required probability
$\displaystyle P\left ( E/\bar{A} \right )= \frac{P\left ( E\cap \bar{A} \right  )}{P\left ( \bar{A} \right )}= \frac{P\left ( E \right )-P\left ( E\cap A

\right )}{P\left ( \bar{A} \right )}$

$\displaystyle = \frac{P\left ( E \right )-P\left ( E \right ).P\left ( A/E \right )}{P\left ( \bar{A} \right )}$
Putting the values, we get
$\displaystyle = \dfrac{\dfrac{n-1}{n}-\dfrac{n-1}{n}.\dfrac{n-1}{n}}{\dfrac{2n-1}{n^{2}}}$
$\displaystyle \therefore P\left ( E/\bar{A} \right )= \frac{n-1}{2n-1}.$

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

There are two groups of subjects one of which consists of 5 science subjects and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects. An unbaised die is cast. If number 3 or number 5 turns up, a subject is selected at random from the first group, other wise the subject is selected at random from the second group. Find the probability that an engineering subject is selected ultimately.

  1. $\displaystyle \frac{13}{24}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{11}{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let  $\displaystyle E _{1}$ be the event that a subject is selected from first group.
$\displaystyle E _{2}$ the event that a subject is selected from the second group.
$E$ be the event that an engineering subject is selected.
Now the probability that die shows $3$ or $5$  is
$\displaystyle P(E _1)=\frac{2}{6}=\frac{1}{3}$

$\displaystyle P\left ( E _{2} \right )=\frac{1}{3}=\frac{2}{3}.$
Now probability of choosing an engineering subject from first group is 
$\displaystyle P\left ( E|E _{1} \right )=$  $\displaystyle \frac{^{3}C _{1}}{^{8}C _{1}}=\frac{3}{8}$
Similarly, $\displaystyle P\left( E|E _{2} \right )=\frac{^{5}C _{1}}{^{8}C _{1}}=\frac{5}{8}$ 
Hence $\displaystyle P\left ( E \right )=P\left ( E _{1} \right )P\left ( E|E _{1} \right )+P\left ( E _{2}\right )P\left ( E|E _{2} \right )$
$\displaystyle =\frac{1}{3}.\frac{3}{8}+\frac{2}{3}.\frac{5}{8}$

$=\dfrac{13}{24}$ 

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

A signal which can be green or red with probability $\displaystyle \frac{4}{5}$ and $\displaystyle \frac{1}{5}$, respectively, is received at station A and then transmitted to station B. The probability of each station receiving the signal correctly is $\displaystyle \frac{3}{4}$. If the signal received at station B is green, then the probability that the original signal was green is

  1. $\displaystyle \frac{3}{5}$
  2. $\displaystyle \frac{6}{7}$
  3. $\displaystyle \frac{20}{23}$
  4. $\displaystyle \frac{9}{20}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
 Event $G$ = original signal is green
$E _1=A$ receives the signal correct
$E _2=B$ receives the signal correct
E = signal received by B is green
$P(\text{signal received by B is green}) = P(GE _1E _2)+ P(G\cap {E _1}\cap {E _2})+ P(\cap GE _1\cap{E _2})+ P(\cap G\cap {E _1}E _2)$
$P(E)=\dfrac {46}{5\times 16}$
$ P(G/E)=\dfrac {\dfrac {40}5\times 16}{\dfrac {46}5\times16}=\dfrac {20}{23}.$
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If the letters of the word  $"ATTEMPT"$  are written down at random. The probability that all the  $T's$  come together is

  1. $1/21$
  2. $6/7$
  3. $1/7$
  4. $1/42$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of ways of arranging the words keeping all $T's$ together is  $5!$

Number of ways of arranging the words  is  $\dfrac{7!}{3!}$
Probability that all the $T's$  together is $\dfrac{5! }{\dfrac{7!}{3!}}=\dfrac{1}{7}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability of a certain event is 

  1. $0$
  2. $1$
  3. greater than $1$
  4. less than $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
An event which always happens is called a sure event or a certain event. So the probability of a certain event is $1$. 
For example, when we throw a die, then the event "getting a number less than $7$" is a certain event.
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If P(E) = 0 then E is a/an

  1. sure event

  2. impossible event

  3. equally likely event

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

$P(E)=\frac{number  of   outcomes  favorable}{Total   numbers  of  possible  outcomes}$

If P(E)=0 then the event is called impossible event.
For example -
When a dice is thrown the possible outcomes are 1,2,3,4,5 and 6.
then  the probability is to getting the number 7  in a single throw of a dice is 0 then this is called impossible event.
$P(E)=\frac{0}{6}=0$    

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability of an event which is sure to occur at every performance of an experiment is called a ___________.

  1. simple event

  2. compound event

  3. complementary event

  4. certain event
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The probability of an event which is sure to occur at every performance of an experiment is called a certain event.
Example: Head or Tail is a certain event connected with tossing a coin.