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

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

  1. Infinite discrete sample space

  2. Finite sample space

  3. Continuous sample space

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

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

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?

  1. $\dfrac{876}{1000}$
  2. $\dfrac{544}{1000}$
  3. $\dfrac{974}{1000}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $P(H)=0.974=\dfrac{974}{1000}$.


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

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

  1. $\dfrac13$
  2. $\sqrt{2}$
  3. $\sqrt{10}$
  4. $\dfrac{8}3$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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

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

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

  1. an integer

  2. an irrational number

  3. a natural number

  4. a rational number
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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.

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

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.

  1. $1,2,3,4$
  2. $2,4,1,3$
  3. $3,2,4,1$
  4. $3,2,1,4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $\sqrt{49}$
  2. $\sqrt{19}$
  3. $\sqrt{47}$
  4. $\sqrt{62}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
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.