Probability Test 14

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Probability Test 14

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Weekly Quiz Competition

Question 1
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The term law of total probability is sometimes taken to mean the ____

A
Law of total expectation

B
Law of alternatives

C
Law of variance

D
None of these

Solution

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.
Question 2
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The experiment is to randomly select a human and measure his or her length. Identify the type of the sample space.

A
Finite sample space

B
Continuous sample space

C
Infinite discrete sample space

D
None of these

Solution

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.

Question 3

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The experiment is to repeatedly toss a coin until first tail shows up. Identify the type of the sample space.

Finite sample space

Continuous sample space

Infinite discrete sample space

None of these

Solution

The experiment is to repeatedly toss a coin until first tail shows up.

A coin is tossed. The possible outcomes are $$head-H, tail-T$$.

Look into the following table:

$$1^{st}$$ toss	$$2^{nd}$$ toss	$$3^{rd}$$ toss	$$4^{th}$$ toss	$$5^{th}$$ toss
$$T$$	$$-$$	$$-$$	$$-$$	$$-$$
$$H$$	$$T$$	$$-$$	$$-$$	$$-$$
$$H$$	$$H$$	$$T$$	$$-$$	$$-$$
$$H$$	$$H$$	$$H$$	$$T$$	$$-$$
$$H$$	$$H$$	$$H$$	$$H$$	$$T$$

If we get $$T$$ at first toss, then our experiment ends.

Otherwise second toss. If we get $$T$$, out experiment ends. If not the process continues till we end up in tail.

Hence the possible outcomes are sequences of $$H$$ that, if finite, end with a single $$T$$, and an infinite sequence of $$H$$.

Therefore, $$ S=\{T,HT,HHT,HHHT,HHHHT,..., \{HHHH....\}\}$$

Thus we get a infinite but countable(depends on the number of toss) sample space.

As we shall see elsewhere, this is a remarkable space that contains a not impossible event whose probability is $$0$$.

We know that, "a discrete sample space is one with a finite or countably infinite number of possible values."

Hence, it is a infinite discrete sample space.

Question 4
1 / -0

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

A
Finite sample space

B
Continuous sample space

C
Infinite discrete sample space

D
None of these

Solution

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.
Question 5
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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.

A
$$0.438$$

B
$$0.538$$

C
$$0.488$$

D
None of these

Solution

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 $$0.488$$
Question 6
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If $$P(A) + P(B) = 1$$; then which of the following option explains the event $$A$$ and $$B$$ correctly?

A
Event $$A$$ and $$B$$ are mutually exclusive, exhaustive and complementary events

B
Event $$A$$ and $$B$$ are mutually exclusive and exhaustive events

C
Event $$A$$ and $$B$$ are mutually exclusive and complementary events

D
Event $$A$$ and $$B$$ are exhaustive and complementary events

Solution

Mutually exclusive events are those events which cannot happen at the same time.
Mutually exhaustive events are those events in which atleast one of the events should occur.
Atleast one of $$A,B$$ should happen , both $$P(A),P(B)$$ cannot be zero at same time because given that $$P(A)+P(B)=1$$
Complementary events means those two events are the only possible events.
Here only $$A$$ and $$B$$ are possible events because $$P(A)+P(B)=1$$
Therefore Event $$A$$ and $$B$$ are mutually exclusive , exhaustive and complementary events.
So the correct option is $$A$$.
Question 7
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The events $$E_1, E_2, ........$$ represents the partition of the sample space $$S$$, if they are:

A
pairwise disjoint

B
exhaustive

C
have non-zero probabilities

D
All are correct

Solution

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.
Question 8
1 / -0

Probability of sure event is

A
$$1$$

B
$$0$$

C
$$\dfrac {1}{2}$$

D
$$2$$

Solution

Probability of sure event is always $$1$$
Question 9
1 / -0

Which one can represent a probability of an event

A
$$\dfrac {7}{4}$$

B
$$-1$$

C
$$-\dfrac {2}{3}$$

D
$$\dfrac {2}{3}$$

Solution

$$P < 1$$

Optoion (D) is correct $$2/3$$
Question 10
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Let $$u_1$$ and $$u_2$$ be two urns such that $$u_1$$ contains 3 white, 2 red balls and $$u_2$$ contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn $$u_1$$ and put into $$u_2$$. However, if tail appears, then 2 balls are drawn at random from $$u_1$$ and put into $$u_2$$. Now, 1 ball is drawn at random from $$u_2$$. Then, probability of the drawn ball from $$u_2$$ being white is

A
$$\dfrac{13}{30}$$

B
$$\dfrac{23}{30}$$

C
$$\dfrac{19}{30}$$

D
$$\dfrac{11}{30}$$

Solution

$${\textbf{Step - 1: Finding probabilities of different possible events}}$$

$${\text{Case - 1: Head appears on the coin, }}$$

$${\text{P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = Probability of drawing white ball from urn 2, when head appears on the coin}}$$
  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{(when white is tranferred) }}$$
  $${\text{ + }}\left( {\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{1}}{{\text{C}}_{\text{1}}}}}{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when red is transferred)}}$$

  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = }}\left( {\dfrac{{\text{3}}}{{\text{5}}}{\text{ }} \times {\text{ 1 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}\left( {\dfrac{{\text{2}}}{{\text{5}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = }}\dfrac{{\text{3}}}{{{\text{10}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{10}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{5}}}$$

  $${\text{Case - 2: Tail appears on the coin,}}$$

  $${\text{P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = Probability of drawing white ball from urn 2, when tail appears on the coin}}$$

  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{2}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{2}}}}} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 2 white are transferred) }}$$
     $${\text{+ }}\left( {\dfrac{{^{\text{2}}{{\text{C}}_{\text{2}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{1}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{1}}}}} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 2 red is transferred)}}$$

     $${\text{ + }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}{\text{ }} \times {{\text{ }}^{\text{2}}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 1 red and 1 white is transferred)}}$$

$$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\left( {\dfrac{{{\text{3 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ 1 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}\left( {{\text{ }}\dfrac{{{\text{1 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{3}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$$
$$+\left( {\dfrac{{{\text{3 }} \times {\text{ 2 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$$

$$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\dfrac{{\text{3}}}{{{\text{20}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{60}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{5}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{30}}}}$$

$${\textbf{Step - 2: Finding total probability}}$$

  $${\text{P = P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ + P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right)$$

  $$ \Rightarrow {\text{ P = }}\dfrac{{\text{2}}}{{\text{5}}}{\text{ + }}\dfrac{{{\text{11}}}}{{{\text{30}}}}{\text{ = }}\dfrac{{{\text{23}}}}{{{\text{30}}}}$$

$$\mathbf{{\text{Hence, the probability that a white ball is drawn from Urn 2 is }}\dfrac{{{\text{23}}}}{{{\text{30}}}}}$$

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Re-Attempt Weekly Quiz Competition

Probability Test 14

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Probability Test 14

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SHARING IS CARING

Question 1

Law of total expectation

Law of alternatives

Law of variance

None of these

Solution

Question 2

Finite sample space

Continuous sample space

Infinite discrete sample space

None of these

Solution

Question 3

Finite sample space

Continuous sample space

Infinite discrete sample space

None of these

Solution

Question 4

Finite sample space

Continuous sample space

Infinite discrete sample space

None of these

Solution

Question 5

$$0.438$$

$$0.538$$

$$0.488$$

None of these

Solution

Question 6

Event $$A$$ and $$B$$ are mutually exclusive, exhaustive and complementary events

Event $$A$$ and $$B$$ are mutually exclusive and exhaustive events

Event $$A$$ and $$B$$ are mutually exclusive and complementary events

Event $$A$$ and $$B$$ are exhaustive and complementary events

Solution

Question 7

pairwise disjoint

exhaustive

have non-zero probabilities

All are correct

Solution

Question 8

$$1$$

$$0$$

$$\dfrac {1}{2}$$

$$2$$

Solution

Question 9

$$\dfrac {7}{4}$$

$$-1$$

$$-\dfrac {2}{3}$$

$$\dfrac {2}{3}$$

Solution

Question 10

$$\dfrac{13}{30}$$

$$\dfrac{23}{30}$$

$$\dfrac{19}{30}$$

$$\dfrac{11}{30}$$

Solution

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