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 maths probability - iii theorem of total probability bayes theorem probability and probability distribution

The term law of total probability is sometimes taken to mean the ____

  1. Law of total expectation

  2. Law of alternatives

  3. Law of variance

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

The term law of total probability is sometimes taken to mean the law of alternatives, which is a special case of the law of total probability applying to discrete random variables.

Hence, option B is correct.

Multiple choice maths probability - iii theorem of total probability bayes theorem probability and probability distribution

The events $E _1, E _2, ........$ represents the partition of the sample space $S$, if they are:

  1. pairwise disjoint

  2. exhaustive

  3. have non-zero probabilities

  4. All are correct
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A set of events $E _1 , E _2 ,...$  is said to represent a partition of a sample space $S$, if 


$(a)$  $E _i \cap E _j = \phi, i\neq j; i, j = 1, 2, 3,..., n$  (pairwise disjoint)

$(b)$ $E _i \cup E _2 \cup ... \cup E _n = S$ (exhaustive)

$(c)$ Each $E _i \neq \phi, i.e, P(E _i) > 0$ for all $i = 1, 2, ..., n$ (have non-zero probabilities)

That is the events should be pairwise disjoint, exhaustive and should have non zero probabilities.

Hence, option D is correct.

Multiple choice maths probability - iii theorem of total probability bayes theorem probability and probability distribution

In a construction job, following are some probabilities given:
Probability that there will be strike is $0.65$, probability that the job will be completed on time if there is no strike is $0.80$, probability that the job will be completed on time if there is strike is $0.32$. Determine probability that the construction job will get complete on time.

  1. $0.438$
  2. $0.538$
  3. $0.488$

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

Let $A$ be the event that the construction job is completed on time and $B$ be the event that there is strike.


$\Rightarrow P(B)=0.65$

Hence probability that there will be no strike $=P(B')$
                                                                            $=1-P(B)$
                                                                            $=1-0.65$
                                                                            $=0.35$

$\therefore P(B')=0.35$

By the Law of Total Probability we have $P(A)=P(B) \times P(A|B)+P(B') \times P(A|B')$

Given, $P(construction : job: is: completed: with: no: strike)=P(A|B)=0.80$ 
and $P(construction : job: is: completed: with:  strike)=P(A|B')=0.32$

$\therefore P(A)=0.65 \times 0.80+0.35 \times 0.32=0.488$

Hence the probability that the construction job will get complete on time is $0.488$

Multiple choice maths probability - iii theorem of total probability bayes theorem probability and probability distribution

There are three boxes, each containing a different number of light bulbs. The first box has 10 bulbs, of which four are dead, the second has six bulbs, of which one is dead, and the third box has eight bulbs of which three are dead. What is the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes?

  1. $\dfrac{115}{330}$
  2. $\dfrac{113}{360}$
  3. $\dfrac{113}{330}$

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

Let $𝐴 _1, 𝐴 _2, 𝐴 _3$ denotes the events of selecting bulbs from bags $1,2: and: 3$ respectively. 


Let $𝐵$ denotes the event the bulbs selected are dead.

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

Also $P(B|A _1)=\dfrac{4}{10}, P(B|A _2)=\dfrac{1}{6}, P(B|A _3)=\dfrac{3}{8}$

By law of total probability,

$P(B)=P(A _1)P(B|A _1)+P(A _2)P(B|A _2)+P(A _3)P(B|A _3)$

Substituting the values we get,

$P(B)=\dfrac{1}{3} \times \dfrac{4}{10}+ \dfrac{1}{3} \times \dfrac{1}{6}+ \dfrac{1}{3} \times \dfrac{3}{8}$

$\Rightarrow P(B)=\dfrac{113}{360}$

Thus the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes is $\dfrac{113}{360}$.


Multiple choice business economics and quantitative methods measures of dispersion and skewness shortcut method for calculating mean deviation about mean mean deviation about mean and median range and mean deviation

Probability is expressed as _______.

  1. percentage

  2. ratio

  3. proportion

  4. all (a), (b), (c)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Probability shows the relationship between two variables in the form of ratio, percentage or proportion where there the chances of occurrence of one variable is expressed in terms of other variable. Since the value of one variable belongs to the range of value of another variable, the range o probability varies from 0 to 1.
Multiple choice biology mendel's law of inheritance determination of sex sex determination sex determination in humans

If the first seven children born to a particular pair of parents are all males, what is the probability that the eighth child will also be a male?

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{8}$
  4. $\dfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The sex of an individual depends on the combination of sex chromosomes in its cells, that is, the combination of X and Y chromosomes. An embryo with XX genotype would develop into a female, while an embryo with an XY would develop into a male. The mother has a genotype and would always produce gametes with an X chromosome. The father, however, would produce half the gametes having an X chromosome and half carrying a Y chromosome. The sex of the embryo in humans would hence depend on the chromosome present in the gamete of the father. For every child conceived, there is an equal probability of the child being a male or a female, as the gamete received from the father has a 50% chance of carrying an X chromosome and  50% chance of carrying a Y chromosome.
So, the correct answer is $\dfrac{1}{2}$
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, $P(E)=0.62, P(F)=0.5, P(E\cap F)=0.35$.
$\therefore P(E\cup F)=P(E)+P(F)-P(E\cap F)=0.62+0.5-0.35=.77$