Mathematics

Probability Distributions

488 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 computer and ms office mathematical methods for economics economics

A has three children in his family, the probability of all the children being a girl child is _____

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

For each child, the probability of being a girl is 1/2. For three children, the probability of all being girls is (1/2) * (1/2) * (1/2) = 1/8.

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

In an experiment, there are exactly three elementary events. The probability of two of them are $\displaystyle\frac{2}{7}$ and $\displaystyle\frac{1}{7}$. What is the probability of third event?

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

We know that the sum of all the elementary events = 1
$\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7} + P _3 = 1$ 

$\Rightarrow \quad P _3 = 1 - \left(\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7}\right) = \displaystyle\frac{4}{7}$ 

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

Let A be a set of $4$ elements. From the set of all functions from A to A, the probability that it is an into function is?

  1. $\dfrac{3}{32}$
  2. $0$
  3. $\dfrac{29}{32}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let set A be of n element and from A to A then no of onto element is n! and total function are $n^{2}$

$P(E)=\dfrac{n!}{n}=\dfrac{4!}{4\times 4\times 4\times 4}$

$=\dfrac{4\times 3\times 2}{4\times 4\times 4\times }=\dfrac{3}{32}$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

If a coin is tossed twice, then the events 'occurrence of one head',  'occurrence of $2$ heads' and 'occurrence of no head' are -

  1. Independent

  2. Equally likely

  3. Not equally likely

  4. Both (A) and (B)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Possible outcomes when a coin is tossed $=$ H or T
Possible outcomes when a coin is tossed twice$=$ HH,TT,TH,HT
$=4$ outcomes
P(occurrence of $1H$) $=$ TH,HT
$=2/4=1/2$
P(occurrence of $2H$) $=$ HH
$=1/4$
P(occurrence of no H) $=$ TT
$=1/4$
Therefore, the probability of occurrence of no head and occurrence of no head and occurrence of two head is same but probability of occurrence of one Head is $1/2$. Thus the events are:-
Not equally likely.

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

If events A and B are independent and P(A) $=$ 0.15, P(A $\cup $ B) $=$ 0.45, Then P(B) $=$ ..............

  1. $\displaystyle \frac{6}{13}$
  2. $\displaystyle \frac{6}{17}$
  3. $0.315$
  4. $0.352$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation
Events A and B are independent then $P\left( A\bigcap { B }  \right) =0$
$P\left( A\bigcup { B }  \right) =P(A)+P(B)-P(A).P(B)$
Given: $P(A)=0.15$
$P\left( A\bigcup { B }  \right) =0.45$
$\therefore P(B)=P\left( A\bigcup { B }  \right) -P(A)+P\left( A \right) P(B)$
$P(B)=0.45-0.15+1.15P(B)$
$P(B)=0.3+0.15P(B)$
$P(B)(1-0.15)=0.3$
$P(B)(0.85)=0.3$
$P(B)=\cfrac { 0.3 }{ 0.85 } =0.352=\cfrac { 6 }{ 17 } $
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Assume that the birth of a boy or girl to a couple to be equally likely,mutually exclusive exhaustive and independent of the other children in the family for a couple having $6$ children the probability that their 'three oldest are boy'is

  1. $\dfrac{{20}}{{64}}$
  2. $\dfrac{{1}}{{64}}$
  3. $\dfrac{2}{{64}}$
  4. $\dfrac{8}{{64}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that the birth of a boy or girl to a couple to be equally likely, 
$(1)$ mutually exclusive
$(2)$ exhaustive
$(3)$ independent
$\Rightarrow$ One event does not affect the other 
$P(E)=P(B).P(B).P(B), P(B\ or\ G). P(B\ or\ G)$
$P(B)=$ probability of boy is $\dfrac{1}{2}$
$P(G)=$ probability of girl is $\dfrac{1}{2}$
$P(E)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times 2$ either boy or girls
$=\dfrac{2}{64}$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

In a survey conducted among 400 students of X standard in Pune district, 187 offered to join Science faculty after X std. and 125 students offered to join Commerce faculty after X, If a student is selected at random from this group. Find the probability that student prefers Science or Commerce faculty.

  1. $\displaystyle \frac{39}{50}$
  2. $\displaystyle \frac{4}{5}$
  3. $\displaystyle \frac{41}{50}$
  4. $\displaystyle \frac{43}{50}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total number of students $= 400$
No. of science students $= 187$
No. of commerce students $ = 125$
Thus, Probabilty of the student being either science or commerce = $\dfrac{187}{400} + \dfrac{125}{400}$
= $\dfrac{312}{400}$
= $\dfrac{39}{50}$

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

A random variable $X$ has the probability distribution:

$x$ 1 2 3 4 5 6 7 8
$P(X=x)$ 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events $E=\left {x|x \text {is prime}\right }$ and $F=\left {x|x < 4\right }$, the probability $P(E\cup F)$ is

  1. $0.35$
  2. $0.77$
  3. $0.87$
  4. $0.50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using given table, $P(E)=0.62, P(F)=0.5, P(E\cap F)=0.35$.
$\therefore P(E\cup F)=P(E)+P(F)-P(E\cap F)=0.62+0.5-0.35=.77$

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 $\dfrac{1}{4}.$