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

A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband`s selection is $\dfrac {1}{7}$ and that of wife's selection is $\dfrac {1}{5}$. What is the probability that only one of them will be selected?

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

There would be three cases.

1. Husband is selected while wife doesn't get selected.
2. Wife is selected while husband doesn't get selected.
3. Both of them get selected.

Probability of husband not to get selected $=\dfrac{6}{7}$
Probability of wife not to get selected $=\dfrac{4}{5}$
Probability only one to get selected $=\dfrac{6}{7} \times \dfrac{1}{5} +\dfrac{1}{7} \times \dfrac{4}{5} $
$=\dfrac{10}{35} =\dfrac{2}{7}$

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

A number $x$ is selected from first $100$ natural numbers. Find the probability that $x$ satisfies the condition $x+ \dfrac{100}{x} >50$

  1. $\dfrac{55}{100}$
  2. $\dfrac{45}{100}$
  3. $1$
  4. $\dfrac{50}{100}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x+\dfrac {100}x>50\\implies x^2 +100>50x\\implies x^2-50x+100+(25)^2-(25)^2>0\\implies x^2+(25)^2-50x-525>0\\implies (x-25)^2>525\\implies x-25>\pm\sqrt{525}\\implies x>25+22.91, x<25-22.91\\implies x>47.91, x<2.09\\implies x\ge 48, x\le 2$

As we have to select from 1 to 100 and x is greater than or equal to 48. Hence we have to select from 49 to 100 which is 53 numbers and favorable cases are 1, 2.
Total favorable case = 53+2=55.
Total number of cases = 100
Hence the required probability, $=\dfrac {55}{100}$

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

$A$ speaks the truth in $60\%$ cases and $B$ in $70\%$ cases. The probability that they will say the same thing while describing a single event is:

  1. $0.56$
  2. $0.54$
  3. $0.38$
  4. $0.94$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Both will speak the same thing means both can say truth or both can say false

The probability of truth for $A$ is $0.6$ and for $B$ is $0.7$
The probability of false for $A$ is $0.4$ and for $B$ is $0.3$
The probability that both say same thing is $0.6 \times 0.7 + 0.4 \times 0.3 = 0.42+0.12=0.54$
Therefore option $B$ is correct

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

If A and B are mutually exclusive events such that $P(A)=\frac{3}{5}$ and $ P(B)=\frac{1}{5}$, then find $P(A \cup B)$. 

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

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

$A\quad and\quad B\quad are\quad mutually\quad exclusive\quad events\quad than\quad P(A\cap B)=0\ P(A)=\frac { 3 }{ 5 } \quad and\quad P(B)=\frac { 1 }{ 5 } \ P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \qquad \qquad =\frac { 3 }{ 5 } +\frac { 1 }{ 5 } -0=\frac { 4 }{ 5 } $

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

A is interviewed for $3$ posts. There are $3$ candidates for post $1,4$ for second post and $2$ for post No. three. The probability of A's being selected for at least one post is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$    $P(B)=\dfrac{1}{4}$     $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{4} \right)-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{2} \right)-\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2} \right)+\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4}\times\dfrac{1}{2}\right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{8+6+12-2-4-3+1}{24}$
                                 $=\dfrac{3}{4}.$
Hence, the answer is $\dfrac{3}{4}.$

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

A is interviewed for $3$ posts. There are $3$ candidates for post $1, 4$ for second post and $2$ for post No. three. The probability of A's being selected for none of the posts is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{4}$
  4. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(A\cap C)-P(B\cap C)+P(A\cap B \cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$   $P(B)$  $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac {1  }{  3} \times\dfrac{1}{4}\right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{2}  \right) -\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2}  \right)=\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4} \dfrac{1}{2} \right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{3}{4}$
$\Rightarrow P$ ( not getting selected for any of the post ) $=1-\dfrac{3}{4}$
                                                                                $=\dfrac{1}{4}$
Hence, the answer is $\dfrac{1}{4}.$
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

n men and n women are seated at round table in random order. The probability that they can be divided into n non-interrecting pairs so that each pair consists of a man and a women is

  1. 1/2n
  2. $2(2^n-1)/^{2n}C _n$
  3. $2n/^{2n}C _n$
  4. $1/(^nC _n)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The women can be seated in $^{2n}C _n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left {(M, W), (M, W), ...(M, W)\right }$ of A(B) and $\left {(W, M), (W, M), ..., (W, M)\right }$ of B(A).
Therefore, $P(A\cup B)=P(A)+P(B)-P(A\cup B)$
$\displaystyle =\frac {2^n+2^n-2}{^{2n}C _n}=\frac {2^{n+1}-2}{^{2n}C _n}$.

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

A is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. The probability of A's being selected for at least one post is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{4}$    $P(B)=\dfrac{1}{3}$     $P(C)=\dfrac{1}{5}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$

                                 $=\dfrac{15+20+12-5-3-4+1}{60}$
                                 $=\dfrac{3}{5}.$
Hence, the answer is $\dfrac{3}{5}.$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

$A$ is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. the probability of $A$'s being selected for none of the posts is:

  1. $\dfrac{3}{4}$
  2. $\dfrac{2}{5}$
  3. $\dfrac{1}{4}$
  4. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$
                                 $=\dfrac{3}{5}.$
$\Rightarrow P$ ( not getting selected for none of the post ) $=1-\dfrac{3}{5}$
                                                                                   $=\dfrac{2}{5}$
Hence, the answer is $\dfrac{2}{5}.$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

An arrangement is selected at random from all possible arrangements of five digits written from the digits $0,1,2,3,\cdots 9$ with repetition. The probability that the randomly selected arrangement will have largest number $'8'$ given that the smallest number is $'4'$ is :

  1. $\dfrac {1253}{6480}$
  2. $\dfrac {513}{4651}$
  3. $\dfrac {2881}{6480}$
  4. $\dfrac {1320}{4651}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a conditional probability problem involving arrangements of digits. The constraints on the smallest and largest digits restrict the sample space significantly.

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

In a class 5% of boys and 10% of girls have an I.Q of more than 150.In this class 60% of students are boys. If a student is selected at random and found to have an I.Q. of more than 150. Find the probability that the student is a boy.

  1. $\dfrac{3}{7}$
  2. $\dfrac{23}{7}$
  3. $\dfrac{3}{5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let us consider the problem
$E _1$ : Event that boys are selected
$E _2$ : Event that girls are selected
$A$ : event that have IQ $150$
Implies that,
\begin{array}{l} P\left( { A/{ E_{ 1 } } } \right) =\dfrac { 5 }{ { 100 } }  \\ P\left( { A/{ E_{ 2 } } } \right) =\dfrac { { 10 } }{ { 100 } }  \\ P\left( { { E_{ 1 } } } \right) =\dfrac { { 60 } }{ { 100 } } ,P\left( { { E_{ 2 } } } \right) =\dfrac { { 40 } }{ { 100 } }  \\ P\left( { A|{ E_{ 1 } } } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right)  } }{ { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { A|{ E_{ 2 } } } \right)  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } }  } }{ { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } +\dfrac { { 40 } }{ { 100 } } \times \dfrac { { 10 } }{ { 100 } }  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 60\times 5 } }{ { 60\times 5+40\times 10 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 300 } }{ { 300+400 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { 3 }{ 7 }  \end{array}

Hence, the probability is $\dfrac {3}{7}$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Suppose that of all used cars of a particular year 30% have bad brakes. You are considering buying a used car of that year. You take the car to a mechanic to have the brakes checked. The chance that the mechanic will give you the wrong report is 20%. Assuming that the car you take to the mechanic is selected at random from the population of cars of that year. The chance that the car's brakes are good, given that the mechanic says its brakes are good, is

  1. $\displaystyle \frac{28}{130}$
  2. $\displaystyle \frac{29}{31}$
  3. $\displaystyle \frac{37}{62}$
  4. $\displaystyle \frac{29}{62}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $30$% of the cars of bad brakes
$P(E _1)=70$%$=\dfrac7{10}$        $P(E _2)=30$%$=\dfrac3{10}$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =0.2\times0.2=0.04$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 2 } }  \right) =0.8\times0.8=0.64$
$\therefore P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =\dfrac { \dfrac { 7 }{ 10 } \times \dfrac { 2 }{ 10 } \times \dfrac { 2 }{ 10 }  }{ \dfrac { 7\times 4 }{ 1000 } +\dfrac { 3 }{ 10 } +\dfrac { 8 }{ 10 } +\dfrac { 8 }{ 10 }  } =\dfrac { 28 }{ 102+28 } =\dfrac { 28 }{ 130 } $
Hence, the answer is $\dfrac { 28 }{ 130}.$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A & B are sharp shooters whose probabilities of hitting a target are $\displaystyle \frac{9}{10}$ & $\displaystyle \frac{14}{15}$ respectively. If it is knownthat exactly one of them has hit the target, then the probability that it was hit by A is equal to

  1. $\displaystyle \frac{24}{55}$
  2. $\displaystyle \frac{27}{55}$
  3. $\displaystyle \frac{9}{23}$
  4. $\displaystyle \frac{10}{23}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E _1$ : only A hits the target
$E _2$ : only B hits the target
$E$ : exactly one hits the target.
$\therefore \displaystyle P(E _1 / E) = \frac{P(E _1). P (E / E _1)}{P (E _1). P (E/ E _1) + P (E _2). P (E/ E _2)}$
$=

\displaystyle \frac{\displaystyle \frac{9}{10} \times \frac{1}{15}}{

\displaystyle \frac{9}{10} \times \frac{1}{15} + \frac{14}{15} \times

\frac{1}{10}}\ = \dfrac{9}{23}$