Tag: maths

Questions Related to maths

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

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


Which of the following numbers lie between $1$ and $3$?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\dfrac13$

  2. $\sqrt{2}$

  3. $\sqrt{10}$

  4. $\dfrac{8}3$

Show answer
Correct Option: B,D
Explanation:

Converting all option in decimal format (approximately)

A. $\frac{1}{3}=0.33$
B.$\sqrt {2  } =1.41$
C.$ \sqrt { 10 }=3.16 $
D. $\frac{8}{3}=2.67$
It is clear that option B and D lies between 1 and 3
So correct answer will be option B and D

Between any $2$ real numbers, __________ can always be represented on a number line.

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. an integer

  2. an irrational number

  3. a natural number

  4. a rational number

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Correct Option: B,D
Explanation:

Between any two real number, there is always many rational number.

And between any two rationals there is always an irrational numbers.
Thus, we can represent rationals and irrationals between any two reals.
Hence, option B and D is correct.

Following are the steps to represent $\sqrt5$  on number line.
Arrange them in order.
1) Draw OC on line with $l(OC)=l(OB)$,
2) Draw $AB \perp OA\ and\ l(AB) =1$
3) Take $l(OA)=2$
4) $l(OC)=\sqrt5$, C is required point on real line.

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $1,2,3,4$

  2. $2,4,1,3$

  3. $3,2,4,1$

  4. $3,2,1,4$

Show answer
Correct Option: D
Explanation:

The correct order of representing $\sqrt { 5 } $ on number line is 

Step 1. Take $l(OA) = 2$.
Step 2. Draw $AB$$\perp $$OA\ and\  l(AB)$ $= 1$
Step 3. Draw $OC$ on line with $l(OC) = l(OB)$
Step 4. $l(OC) =$ $\sqrt { 5 } $, $C$ is the required point on real line
Therefore, option(D) is correct.

Which of the following irrational numbers lie between $6$ and $8$?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt{49}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{62}$

Show answer
Correct Option: C,D
Explanation:

$6^{2} = 36$

$7^{2} = 49$
$8^{2} = 64$

$\Rightarrow \sqrt47$ and $\sqrt62$ are only irrational numbers from the stated that lie in between $6$ and $8$.

$\sqrt49 = 7$ which isn't irrational.

The number $\sqrt{10}$ lies between $2$ integers $a$ and $b$ such that $b-a = 1$. Then $b+a = \, ?$

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $4$

  2. $5$

  3. $7$

  4. None of these

Show answer
Correct Option: C
Explanation:

$1^{2} = 1$

$2^{2} = 4$
$3^{2} = 9$
$4^{2} = 16$

$\Rightarrow \sqrt10$ lies between $3$ and $4$

$4 = a , 3 = b$ 

$\therefore a+b = 7$