Tag: probability and probability distribution

Questions Related to probability and probability distribution

The experiment is to randomly select a human and measure his or her length. Identify the type of the sample space.

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

  1. Finite sample space

  2. Continuous sample space

  3. Infinite discrete sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

The experiment is to randomly select a human and measure his or her length. 

Depending of how far reaching our means of selection is it is possible to consider a sample space of about $6.6$ billion humans inhabiting the planet Earth. 
In this case, the height of the selected person becomes a random variable. 
However, it is also possible to consider the sample space consisting of all possible values of height measurements of the world population. 
The tallest man ever measured lived in the United States and had a height of $272$ cm $ (8'11'')$. The height of the shortest person is more difficult to determine. Zero is clearly the low bound, but, for a living adult, it may be safely raised to, say, $40 $ cm. 
This suggests a sample space which is a line segment $[40,272]$ in centimeters. 
While at all times the human population is discrete, we may assume that in some height range near the normal average, all possible heights are realized making a continuous classification. 
We know that, "A continuous sample space is one which takes values in one or more intervals."
Thus the sample space is a continuous sample space.

Given a circle of radius $R$, the experiment is to randomly select a chord in that circle. Identify the type of the sample space.

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

  1. Finite sample space

  2. Continuous sample space

  3. Infinite discrete sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

Given a circle of radius $R$, the experiment is to randomly select a chord in that circle.

There are many ways to accomplish such a selection. 

However the sample space is always the same:

$\{AB: A \:and\: B\: are\: points\: on\: a\: given\: circle\}$.

One natural random variable defined on this space is the length of the chord.

Here the length of the chord takes more values depending on the point we choose in the circle.

We know that, "A continuous sample space is one which takes values in one or more intervals."

Therefore, this is a continuous sample space.

Sample space for experiment in which a dice is rolled is

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

  1. $4$

  2. $8$

  3. $12$

  4. None of these

Show answer
Correct Option: D
Explanation:

A dice is rolled.

Thus $S={1,2,3,4,5,6}$
$\Rightarrow n(S)=6$.
That is, sample space for experiment in which a dice is rolled is $6$.
Since $6$ is not listed in the given options we choose D.

Choosing a birthdate is an example of .........

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

  1. Infinite discrete sample space

  2. Finite sample space

  3. Continuous sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

Let us see the sample space for choosing a birthdate.

A person can be born in any date of a month and a month has maximum of $30$ or $31$ days.
Therefore, $S={1,2,3,4,5,6,7,8,9,10,11,12,13,...,30,31}$ which is a finite set.
Thus choosing a birthdate is an example of finite sample space.

Sample space for experiment in which two coins are tossed is

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

  1. $8$

  2. $4$

  3. $2$

  4. None of these

Show answer
Correct Option: B
Explanation:

Two coins are tossed.

Number of outcomes in sample space when two coins are tossed is given by $2^2=4$.
Hence, sample space for experiment in which two coins are tossed is $4$.

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.

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

  1. $0.438$

  2. $0.538$

  3. $0.488$

  4. None of these

Show answer
Correct Option: C
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 $0.488$

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?

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

  1. $\dfrac{115}{330}$

  2. $\dfrac{113}{360}$

  3. $\dfrac{113}{330}$

  4. None of these

Show answer
Correct Option: B
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 $\dfrac{113}{360}$.


Suppose that two factories supply light bulbs to the market. Factory X's bulbs work for over $5000$ hours in $99\%$ of cases, whereas factory Y's bulbs work for over $5000$ hours in $95\%$ of cases. It is known that factory X supplies $60\%$ of the total bulbs available. What is the chance that a purchased bulb will work for longer than $5000$ hours?

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

  1. $\dfrac{876}{1000}$

  2. $\dfrac{544}{1000}$

  3. $\dfrac{974}{1000}$

  4. None of these

Show answer
Correct Option: C
Explanation:
Let $X$ be the event "comes from factory $X$" and $Y$ be the event "comes fom factory $Y$ " and Let $H$ be the event "works over $5000$ hours." 

Therefore $P(X)=60 \%=0.60 \Rightarrow P(Y)=1-0.60=0.40$

Given that, Factory $X's$ bulbs work for over $5000$ hours in $99 \%$ of cases.

$\therefore P(H|X)=99 \%=0.99$

Also given, factory Y's bulbs work for over $$000$5000$ hours in %$95\%$  of cases.

$\therefore P(H|Y)=95 \%=0.95$

Then by the Law of Total Probability we have

 $P(H) = P (H | X) P(X) + P (H | Y ) P(Y ) $

            $= (.99) (.6) + (.95) (.4)$

            $ = .974$


Thus $P(H)=0.974=\dfrac{974}{1000}$.


Box I contains $2$ white and $3$ red balls and box II contains $4$ white and $5$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

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

  1. $\cfrac{54}{44}$

  2. $\cfrac{54}{14}$

  3. $\cfrac{54}{104}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Probability that the ball drawn is red and from ! =$P(R/A)$

$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$
$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$
$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$
$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$

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 :

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

  1. $\dfrac {1253}{6480}$

  2. $\dfrac {513}{4651}$

  3. $\dfrac {2881}{6480}$

  4. $\dfrac {1320}{4651}$

Show answer
Correct Option: C