Probability Test 2

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Probability Test 2

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are two mutually exclusive and exhaustive events with $$P(B)=3P(A),$$ then what is the value of $$P(\overline{B})$$?

A
$$\dfrac{3}{4}$$

B
$$\dfrac{1}{4}$$

C
$$\dfrac{1}{3}$$

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

Solution

Since P(A) and P(B) are two mutually exclusive and exhaustive events, $$P(A)+P(B)=1$$ and $$P(B)=3P(A)$$
$$\Rightarrow\:P(A)+3P(A)=1$$
$$\Rightarrow\:4P(A)=1$$
$$\Rightarrow\:P(A)=\dfrac{1}{4}$$
$$\Rightarrow\:P(B)=1-\dfrac{1}{4}=\dfrac{3}{4}$$
$$\Rightarrow\:$$ $$P(\overline{B})$$ $$=1-\dfrac{3}{4}=\dfrac{1}{4}$$.
Question 2
1 / -0

Choose the correct alternative for each of the following. If $$ \displaystyle P\left ( E_{1} \right )=\frac{1}{6} $$, $$ \displaystyle P\left ( E_{2} \right )=\frac{1}{3} $$, $$ \displaystyle P\left ( E_{3} \right )=\frac{1}{6} $$, where $$ E_{1} $$, $$ E_{2} $$, $$ E_{3} $$, $$ E_{4} $$ are elementary events of a random experiment, then P($$ E_{4} $$) is equal to

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

B
$$ \displaystyle \frac{2}{3} $$

C
$$ \displaystyle \frac{1}{3} $$

D
None

Solution

Given, probabilities of different elements of the event:
$$ \displaystyle P\left ( E_{1} \right )=\frac{1}{6} $$, $$ \displaystyle P\left ( E_{2} \right )=\frac{1}{3} $$, $$ \displaystyle P\left ( E_{3} \right )=\frac{1}{6} $$
The sum of individual probabilities $$= 1$$
$$P (E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) = 1$$
$$\displaystyle \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + P(E_{4}) = 1$$
$$\displaystyle \frac{2}{3} + P(E_{4}) = 1$$
$$P(E_{4}) = \dfrac{1}{3}$$
Question 3
1 / -0

If $$P(A) = P(B)$$, then the two events $$A$$ and $$B$$ are -

A
Independent

B
Dependent

C
Equally likely

D
Both (A) and (C)

Solution

Events are said to be equally likely if there is no preference for a particular event over the other.

Examples

When a coin is tossed, Head (H) or Tail is equally likely to occur.
If $$P(A)=P(B),$$ then the two events A and B are equally likely
Question 4
1 / -0

If an unbiased coin is tossed once, then the two events of getting a Head and a Tail are -

A
Mutually exclusive

B
Exhaustive

C
Equally likely

D
All of these

Solution

If an unbiased coin is tossed once, then the two events of getting a Head and a Tail are Mutually exclusive,Exhaustive,Equally likely

1.Events are said to be equally likely if there is no preference for a particular event over the other.example:
When a coin is tossed, Head (H) or Tail is equally likely to occur.

2.Two or more than two events are said to be mutually exclusive if the occurrence of one of the events excludes the occurrence of the other
example:When a coin is tossed, we get either Head or Tail. Head and Tail cannot come simultaneously. Hence occurrence of Head and Tail are mutually exclusive events.

3.Exhaustive Event is the total number of all possible outcomes of an experiment.
Examples:When a coin is tossed, we get either Head or Tail. Hence there are 2 exhaustive events.
Question 5
1 / -0

Write the sample space when a coin is tossed.

A
S = [H, G]

B
S = [H, B]

C
S = [H, T]

D
S = [N.N)

Solution

When a coin is tossed, there are two possible outcomes: a head ($$H$$) or a tail ($$T$$). The sample space of this experiment is $$S = [H, T]$$
Question 6
1 / -0

A pack of playing cards was found to contain only $$51$$ cards. If the first $$13$$ cards which are examined are all red, then the probability thatthe missing card is black, is

A
$$\displaystyle \frac{1}{3}$$

B
$$\displaystyle \frac{2}{3}$$

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

D
$$\displaystyle \frac{^{25} C_{13}}{^{51} C_{13}}$$

Solution

$$Total\quad number\quad of\quad cards=52\\ Number\quad of\quad lost\quad cards=1\\ 13\quad cards\quad are\quad surely\quad red\quad therfore,\quad from\quad the\quad remaining\quad 39\quad cards\quad 26\quad are\quad black\quad and\quad 13\quad are\quad red\\ So\quad probability\quad of\quad lost\quad card\quad being\quad black\quad =\frac { \left( 26\\ 1 \right) }{ \left( 39\\ 1 \right) } =\frac { 26 }{ 39 } =\frac { 2 }{ 3 } $$
Question 7
1 / -0

The probability of a certain event is

A
$$0$$

B
$$1$$

C
greater than $$1$$

D
less than $$0$$

Solution

An event which always happens is called a sure event or a certain event. So the probability of a certain event is $$1$$.
For example, when we throw a die, then the event "getting a number less than $$7$$" is a certain event.
Question 8
1 / -0

There are 50 marbles of 3 colors: blue yellow and black The probability of picking up a blue marble is 3/10 and that of picking up a yellow marble is 1/2 The probability of picking up a black ball is

A
1/5

B
1/10

C
1/4

D
4/5

Solution

P(blue)=$$\displaystyle \frac{3}{10}=\frac{15}{50},$$i.e., there are 15 blue marbles
P(yellow)=$$\displaystyle \frac{1}{2}=\frac{25}{50},$$i.e., there are 25 yellow marbles
$$\displaystyle \therefore $$ Number of black marbles=10
$$\displaystyle \therefore $$ P(black)=$$\displaystyle \frac{10}{50}=\frac{1}{5}$$
Question 9
1 / -0

The probability of an impossible event is

A
$$1$$

B
$$0$$

C
less than $$0$$

D
greater than $$1$$

Solution

An event that has no chance of occurring is called an impossible event.
So, the probability of an impossible event is always zero.
Question 10
1 / -0

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 $$P\left( { E }_{ 1 }|X \right) $$.
$$P\left( { E }_{ 1 }|X \right) =\displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( X|{ E }_{ 2 } \right) } $$
$$=\displaystyle\frac { \displaystyle\frac { 4 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 4 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { 1 }{ 5 } \cdot \displaystyle\frac { 1 }{ 2 } } $$
$$= \displaystyle\frac { \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 4 }{ 5 } }{ \displaystyle\frac { 1 }{ 2 } \left( \displaystyle\frac { 4 }{ 5 } + \displaystyle\frac { 1 }{ 5 } \right) } $$
$$= \displaystyle\frac { \displaystyle\frac { 4 }{ 5 } }{ 1 } $$
$$= \displaystyle\frac { 4 }{ 5 }=0.80 $$

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Re-Attempt Weekly Quiz Competition

Probability Test 2

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Probability Test 2

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SHARING IS CARING

Question 1

$$\dfrac{3}{4}$$

$$\dfrac{1}{4}$$

$$\dfrac{1}{3}$$

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

Solution

Question 2

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

$$ \displaystyle \frac{2}{3} $$

$$ \displaystyle \frac{1}{3} $$

None

Solution

Question 3

Independent

Dependent

Equally likely

Both (A) and (C)

Solution

Question 4

Mutually exclusive

Exhaustive

Equally likely

All of these

Solution

Question 5

S = [H, G]

S = [H, B]

S = [H, T]

S = [N.N)

Solution

Question 6

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{2}{3}$$

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

$$\displaystyle \frac{^{25} C_{13}}{^{51} C_{13}}$$

Solution

Question 7

$$0$$

$$1$$

greater than $$1$$

less than $$0$$

Solution

Question 8

1/5

1/10

1/4

4/5

Solution

Question 9

$$1$$

$$0$$

less than $$0$$

greater than $$1$$

Solution

Question 10

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

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

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

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

Solution

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