Probability Test

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Probability Test - 15

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Weekly Quiz Competition

Question 1
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STATEMENT - 1 : Dependent events are those in which the outcome of one does not affect and is not affected by the other.
STATEMENT - 2 : Dependent events are those in which the outcome of one affects and is affected by the other.

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.
Question 2
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If A and B are independent events,then $$P\left ( \dfrac{B}{A} \right )= $$

A
$$P\left ( A \right )$$

B
$$\dfrac{P\left ( B \right )}{P\left ( A \right )}$$

C
$$P\left ( B \right )$$

D
$$P\left ( A \right ).P\left ( B \right )$$

Solution

If A and B are independent events then: $$P(A\cap B)=P(A)P(B)$$
$$P(\dfrac{B}{A})=\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A)P(B)}{P(A)}=P(B)$$
Question 3
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If $$A$$ and $$B$$ are events such that $$P\left( A|B \right) =P\left( B|A \right) $$, then

A
$$A\subset B$$ but $$A\neq B$$

B
$$A = B$$

C
$$A\cap B=\Phi $$

D
$$P\left( A \right) =P\left( B \right) $$

Solution

It is given that, $$P\left( A|B \right) =P\left( B|A \right) $$
$$\Rightarrow \displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } = \displaystyle\frac { P\left( A\cap B \right) }{ P\left( A \right) } $$
$$\Rightarrow P\left( A \right) =P\left( B \right) $$
Thus, the correct answer is (D).
Question 4
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A coin is tossed three times, where
(i) $$E$$ : head on third toss, $$F$$ : heads on first two tosses
(ii) $$E$$ : at least tow heads, $$F$$ : at most two heads
(iii) $$E$$ : at most two tails, $$F$$ : at least one tail
Determine $$ P\left( E|F \right) $$

A
$$0.42,0.50,0.85$$

B
$$0.50,0.42,0.85$$

C
$$0.85,0.42,0.30$$

D
$$.0.42,0.46,0.47$$

Solution

If a coin is tossed three times, then the sample space $$S$$ is
$$S = \left\{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\} $$
It can be seen that the sample space has $$8$$ elements.
(i) $$E = \left\{HHH, HTH, THH, TTH\right\}$$
$$F = \left\{HHH, HHT\right\}$$
$$\therefore E \cap F = \left\{HHH\right\}$$
$$P\left( F \right) = \displaystyle\frac { 2 }{ 8 } = \displaystyle\frac { 1 }{ 4 } $$ and $$P\left( E \cap F \right) = \displaystyle\frac { 1 }{ 8 } $$
$$P\left( E | F \right) = \displaystyle\frac { P\left( E \cap F \right) }{ P\left( F \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 8 } }{ \displaystyle\frac { 1 }{ 4 } } = \displaystyle\frac { 4 }{ 8 } = \displaystyle\frac { 1 }{ 2 } =0.50$$
(ii) $$E = \left\{HHH, HHT, HTH, THH\right\}$$
$$F = \left\{HHT, HTH, HTT, THH, THT, TTH, TTT\right\}$$
$$\therefore E \cap F = \left\{ HHT,HTH,THH \right\} $$
Clearly, $$P\left( E \cap F \right) = \displaystyle\frac { 3 }{ 8 } $$ and $$P\left( F \right) = \displaystyle\frac { 7 }{ 8 } $$
$$P\left( E | F \right) = \displaystyle\frac { P\left( E \cap F \right) }{ P\left( F \right) } = \displaystyle\frac { \displaystyle\frac { 3 }{ 8 } }{ \displaystyle\frac { 7 }{ 8 } } = \displaystyle\frac { 3 }{ 7 }=0.42 $$
(iii) $$E = \left\{ HHH, HHT, HTT, HTH, THH, THT, TTH \right\} $$
$$F = \left\{ HHT, HTT, HTH, THH, THT, TTH, TTT \right\} $$
$$\therefore E \cap F = \left\{ HHT, HTT, HTH, THH, THT, TTH \right\} $$
$$P\left( F \right) = \displaystyle\frac { 7 }{ 8 } $$ and $$P\left( E \cap F \right) = \displaystyle\frac { 6 }{ 8 } $$
Therefore, $$P\left( E | F \right) = \displaystyle\frac { P\left( E \cap F \right) }{ P\left( F \right) } = \displaystyle\frac { \displaystyle\frac { 6 }{ 8 } }{ \displaystyle\frac { 7 }{ 8 } } = \displaystyle\frac { 6 }{ 7 } =0.85$$
Question 5
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If $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } $$, $$P\left( B \right) =0$$, then $$P\left( A|B \right) $$ is

A
$$0$$

B
$$\displaystyle\frac{1}{2}$$

C
Not defined

D
$$1$$

Solution

It is given that, $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } $$ and $$P\left( B \right) =0$$
$$P\left( A|B \right) = \displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } = \displaystyle\frac { P\left( A\cap B \right) }{ 0 } $$
Therefore, $$P\left( A|B \right) $$ is not defined.
Thus, the correct answer is (C).
Question 6
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Probability that $$A$$ speaks truth is $$\displaystyle\frac { 4 }{ 5 } $$. A coin is tossed. A reports that a head appears. The probability that actually there was head is

A
$$\displaystyle\frac { 4 }{ 5 } $$

B
$$\displaystyle\frac { 1 }{ 2 } $$

C
$$\displaystyle\frac { 1 }{ 5 } $$

D
$$\displaystyle\frac { 2 }{ 5 } $$

Solution

Let $${ E }_{ 1 }$$ and $${ E }_{ 2 }$$ be the events such that
$${ E }_{ 1 } : A$$ speaks truth
$${ E }_{ 2 } : A$$ speaks false
Let $$X$$ be the event that a head appears.
$$P\left( { E }_{ 1 } \right) =\displaystyle\frac { 4 }{ 5 } $$
$$\therefore P\left( { E }_{ 2 } \right) =1-P\left( { E }_{ 1 } \right) =1-\displaystyle\frac { 4 }{ 5 } =\displaystyle\frac { 1 }{ 5 } $$
If a coin is tossed, then it may result in either head $$\left( H \right) $$ or tail $$\left( T \right) $$.
The probability of getting a head is $$\displaystyle\frac { 1 }{ 2 } $$ whether $$A$$ speaks truth or not.
$$\therefore P\left( X|{ E }_{ 1 } \right) =P\left( X|{ E }_{ 2 } \right) =\displaystyle\frac { 1 }{ 2 } $$
The probability that there is actually a head is given by $$P\left( { E }_{ 1 }|X \right) $$.
$$P\left( { E }_{ 1 }|X \right) =\displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( X|{ E }_{ 2 } \right) } $$
$$=\displaystyle\frac { \displaystyle\frac { 4 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 4 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { 1 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } } $$
$$= \displaystyle\frac { \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 4 }{ 5 } }{ \displaystyle\frac { 1 }{ 2 } \left( \displaystyle\frac { 4 }{ 5 } + \displaystyle\frac { 1 }{ 5 } \right) } $$
$$= \displaystyle\frac { \displaystyle\frac { 4 }{ 5 } }{ 1 } $$
$$= \displaystyle\frac { 4 }{ 5 }=0.80 $$
Question 7
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Two events $$A$$ and $$B$$ will be independent, if

A
$$A$$ and $$B$$ are mutually exclusive

B
$$P\left( A\prime B\prime \right) =\left[ 1-P\left( A \right) \right] \left[ 1-P\left( B \right) \right] $$

C
$$P\left( A \right) =P\left( B \right) $$

D
$$P\left( A \right) +P\left( B \right) =1$$

Solution

Two events $$A$$ and $$B$$ are said to be independent, if $$P\left( AB \right) =P\left( A \right) \times P\left( B \right) $$
Consider the result given in alternative $$B$$.
$$P\left( A\prime B\prime \right) =\left[ 1-P\left( A \right) \right] \left[ 1-P\left( B \right) \right] $$
$$\Rightarrow P\left( A\prime \cap B\prime \right) =1-P\left( A \right) -P\left( B \right) +P\left( A \right) \cdot P\left( B \right) $$
$$\Rightarrow 1-P\left( A\cup B \right) =1-P\left( A \right) -P\left( B \right) +P\left( A \right) \cdot P\left( B \right) $$
$$\Rightarrow P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A \right) \cdot P\left( B \right) $$
$$\Rightarrow P\left( A \right) +P\left( B \right) -P\left( AB \right) =P\left( A \right) +P\left( B \right) -P\left( A \right) \cdot P\left( B \right) $$
$$\Rightarrow P\left( AB \right) =P\left( A \right) \cdot P\left( B \right) $$
This implies that $$A$$ and $$B$$ are independent, if $$P\left( A\prime B\prime \right) =\left[ 1-P\left( A \right) \right] \left[ 1-P\left( B \right) \right] $$
Question 8
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If $$P\left( A \right) = \displaystyle\frac { 6 }{ 11 }, P\left( B \right) = \displaystyle\frac { 5 }{ 11 }$$ and $$ P\left( A \cup B \right) = \displaystyle\frac { 7 }{ 11 }$$, find
(i) $$ P\left( A \cap B \right)$$
(ii) $$ P\left( A | B \right)$$
(iii) $$ P\left( B | A \right) $$

A
$$0.58,0.58,0.83$$

B
$$0.42,0.67,0.57$$

C
$$0.36,0.80,0.66$$

D
$$0.80,0.98,0.87$$

Solution

It is given that $$P\left( A \right) = \displaystyle\frac { 6 }{ 11 } , P\left( B \right) = \displaystyle\frac { 5 }{ 11 }$$ and $$ P\left( A \cup B \right) = \displaystyle\frac { 7 }{ 11 }$$
(i) $$ P\left( A \cup B \right) = \displaystyle\frac { 7 }{ 11 }$$
$$ \therefore P\left( A \right) + P\left( B \right) - P\left( A \cap B \right) = \displaystyle\frac { 7 }{ 11 }$$
$$ \Rightarrow \displaystyle\frac { 6 }{ 11 } + \displaystyle\frac { 5 }{ 11 } - P\left( A \cap B \right) = \displaystyle\frac { 7 }{ 11 }$$
$$ \Rightarrow P\left( A \cap B \right) = \displaystyle\frac { 11 }{ 11 } - \frac { 7 }{ 11 } = \displaystyle\frac { 4 }{ 11 }=0.36$$
(ii) It is known that, $$ P\left( A | B \right) = \displaystyle\frac { P\left( A \cap B \right) }{ P\left( B \right) }$$
$$ \Rightarrow P\left( A | B \right) = \displaystyle\frac { \displaystyle\frac { 4 }{ 11 } }{ \displaystyle\frac { 5 }{ 11 } } = \displaystyle\frac { 4 }{ 5 }=0.80$$
(iii) It is known that, $$ P\left( B | A \right) = \displaystyle\frac { P\left( A \cap B \right) }{ P\left( A \right) }$$
$$ \Rightarrow P\left( B | A \right) = \displaystyle\frac { \displaystyle\frac { 4 }{ 11 } }{ \displaystyle\frac { 6 }{ 11 } } = \displaystyle\frac { 4 }{ 6 } = \displaystyle\frac { 2 }{ 3 } =0.66$$
Question 9
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10% of tools produced by a certain manufacturing process turn out to be defective. Assuming binomial distribution, the probability of 2 defective in simple of 10 tools chosen at random, is

A
0.368

B
0.194

C
0.271

D
None of these

Solution

Let the event of a tool being defective be $$X$$.
$$\therefore$$ Probability of $$X$$ occurring = $$p=\dfrac{1}{10}$$
$$\therefore q=1-p=\dfrac{9}{10}$$
Now, as the process follows the binomial distribution, the probability of 2 tools being defective is
$$^{10}C_2p^2q^{10-2}$$

$$={}^{10}C_2p^2q^8$$

$$=\dfrac{10!}{2!8!}\left(\dfrac{1}{10}\right)^2 \left(\dfrac{9}{10}\right)^8$$

$$=0.194$$
This is the required solution.
Question 10
1 / -0

Difference between sample space and subset of sample space is considered as

A
numerical complementary events.

B
equal compulsory events.

C
complementary events.

D
compulsory events.

Solution

The set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S.
Any subset E of the sample space S
Difference between sample space and subset of sample space is considered as complementary events.

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Re-Attempt Weekly Quiz Competition

Probability Test - 15

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Probability Test - 15

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Question 1

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is False

Statement - 1 is False, Statement - 2 is True

Solution

Question 2

$$P\left ( A \right )$$

$$\dfrac{P\left ( B \right )}{P\left ( A \right )}$$

$$P\left ( B \right )$$

$$P\left ( A \right ).P\left ( B \right )$$

Solution

Question 3

$$A\subset B$$ but $$A\neq B$$

$$A = B$$

$$A\cap B=\Phi $$

$$P\left( A \right) =P\left( B \right) $$

Solution

Question 4

$$0.42,0.50,0.85$$

$$0.50,0.42,0.85$$

$$0.85,0.42,0.30$$

$$.0.42,0.46,0.47$$

Solution

Question 5

$$0$$

$$\displaystyle\frac{1}{2}$$

Not defined

$$1$$

Solution

Question 6

$$\displaystyle\frac { 4 }{ 5 } $$

$$\displaystyle\frac { 1 }{ 2 } $$

$$\displaystyle\frac { 1 }{ 5 } $$

$$\displaystyle\frac { 2 }{ 5 } $$

Solution

Question 7

$$A$$ and $$B$$ are mutually exclusive

$$P\left( A\prime B\prime \right) =\left[ 1-P\left( A \right) \right] \left[ 1-P\left( B \right) \right] $$

$$P\left( A \right) =P\left( B \right) $$

$$P\left( A \right) +P\left( B \right) =1$$

Solution

Question 8

$$0.58,0.58,0.83$$

$$0.42,0.67,0.57$$

$$0.36,0.80,0.66$$

$$0.80,0.98,0.87$$

Solution

Question 9

0.368

0.194

0.271

None of these

Solution

Question 10

numerical complementary events.

equal compulsory events.

complementary events.

compulsory events.

Solution

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