Probability Test 6

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Probability Test 6

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Question 1
1 / -0

Probability of getting a prime (or) composite is ________.

A
Mutually exclusive

B
Likely

C
$0$

D
None

Solution

Since if prime occur composite can never occur and vice versa.
So both events occurring of prime number and composite number are mutually exclusive.
Option $A$ is correct.
Question 2
1 / -0

$A, B, C$ are three mutually independent with probabilities $0.3, 0.2$ and $0.4$ respectively.
What is $P(A \cap B\cap C)$ ?

A
$0.400$

B
$0.240$

C
$0.024$

D
$0.500$

Solution

Given: $A, B, C$ are mutually independent event so this means
$P(A\cap B\cap C)=P(A)P(B)P(C)=0.3\times 0.2\times 0.4=0.024$
Question 3
1 / -0

Two events A and B will be independent if

A
$P(A' \cap B') = (1 - P(A)) (1 - P(B))$

B
$P(A) + P(B) = 1$

C
$P(A) = P(B)$

D
$A$ and $B$ are mutually exclusive

Solution

Two events A & B will be independent if
$P\left( A\cap { B } \right) =P\left( A \right) \cap { P\left( B \right) }$
If A & B are independent, then ${ A }^{ ' }\& \ { B }^{ ' }$ are also independent
$\therefore P\left( { A }^{ ' }\cap { { B }^{ ' } } \right) =P\left( { A }^{ ' } \right) P\left( { B }^{ ' } \right)$
$=\left( 1-P(A) \right) \left( 1-P(B) \right)$
Question 4
1 / -0

"The occurrence of one event excludes the occurrence of another event". In a random experiment of probability theory, it is called

A
Complementray event

B
Impossible event

C
Mutually exclusive event

D
Certain event

Solution

"The occurrence of one event excludes the occurrence of the other event."
In a random experiment of probability theory, this means that the occurence of one event does not affect the occurrence of another. Hence, they are called mutually exclusive events.
Therefore, $C$ is the correct option.
Question 5
1 / -0

If three events $A$ , $B$ , $C$ are mutually exclusive, then which one of the following is correct?

A
$P\left( A\cup B\cup C \right) =0$

B
$P\left( A\cup B\cup C \right) =1$

C
$P\left( A\cap B\cap C \right) =0$

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

Solution

Three events $A.B.C$ are mutually exclusive if they are disjoint or cannot be true at the same time.
Thus, the events are mutually exclusive if $A\cap B\cap C=\phi$
Thus, $P(A\cap B\cap C)=0$
Hence, C is correct.
Question 6
1 / -0

Which one of the following is correct?

A
An event having no sample point is called an elementary event

B
An event having one sample point is called an elementary event

C
An event having two sample point is called an elementary event

D
An event having many sample point is called an elementary event

Solution

An elementary event is an event which contains only a single element in the sample space. So, it will have only $1$ sample point.
Hence, option B is true
Question 7
1 / -0

Consider the following statements:
1. If $A$ and $B$ are exhaustive events, then their union is the sample space.
2. If $A$ and $B$ are exhaustive events, then their intersection must be an empty event.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Exhaustive events are those events, whose union covers the whole sample space. The probability of occurring at least one of them is $1$ . So, their intersection may or may not be empty.
Hence, A is correct.
Question 8
1 / -0

Identify and write the like terms in each of the following groups.
(i) $a^2, b^2, -2a^2 , c^2 , 4a$

A
$(a^6,2a^2)$

B
$(a^2,-2a^2)$

C
$(a^3,2a^2)$

D
$(a^2,2a^3)$

Solution

In $a^{2},b^{2},-2a^{2},c^{2},4a.$
$a^{2}$ and $-2a^{2}$ are like terms because $-2a^{2}$ is a factor of $a^{2}$
$B$ is correct.
Question 9
1 / -0

Box I contains $2$ white and $3$ red balls and box II contains $4$ white and $5$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

A
$\cfrac{54}{44}$

B
$\cfrac{54}{14}$

C
$\cfrac{54}{104}$

D
None of these

Solution

Probability that the ball drawn is red and from ! = $P(R/A)$
$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$
$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$
$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$
$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$
Question 10
1 / -0

If $A = \left\{ {1,2,3} \right\},\,B = \left\{ {2,4,5,6} \right\}$ and $S = \left\{ {1,2,3,4,5,6} \right\}$ , then $A$ and $B$ is called as __________

A
Complete mentary event

B
Exhaustive event

C
Mutually exclusive events

D
Not defined

Solution

Since all the elements of A and B and in S, Hence A and B is called Exhaustive event

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Re-Attempt Weekly Quiz Competition

Probability Test 6

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Probability Test 6

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Question 1

Mutually exclusive

Likely

000

None

Solution

Question 2

0.4000.4000.400

0.2400.2400.240

0.0240.0240.024

0.5000.5000.500

Solution

Question 3

P(A′∩B′)=(1−P(A))(1−P(B))P(A' \cap B') = (1 - P(A)) (1 - P(B))P(A′∩B′)=(1−P(A))(1−P(B))

P(A)+P(B)=1P(A) + P(B) = 1P(A)+P(B)=1

P(A)=P(B)P(A) = P(B)P(A)=P(B)

AAA and BBB are mutually exclusive

Solution

Question 4

Complementray event

Impossible event

Mutually exclusive event

Certain event

Solution

Question 5

P(A∪B∪C)=0P\left( A\cup B\cup C \right) =0P(A∪B∪C)=0

P(A∪B∪C)=1P\left( A\cup B\cup C \right) =1P(A∪B∪C)=1

P(A∩B∩C)=0P\left( A\cap B\cap C \right) =0P(A∩B∩C)=0

P(A∩B∩C)=1P\left( A\cap B\cap C \right) =1P(A∩B∩C)=1

Solution

Question 6

An event having no sample point is called an elementary event

An event having one sample point is called an elementary event

An event having two sample point is called an elementary event

An event having many sample point is called an elementary event

Solution

Question 7

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 8

(a6,2a2)(a^6,2a^2)(a6,2a2)

(a2,−2a2)(a^2,-2a^2)(a2,−2a2)

(a3,2a2)(a^3,2a^2)(a3,2a2)

(a2,2a3)(a^2,2a^3)(a2,2a3)

Solution

Question 9

5444\cfrac{54}{44}4454​

5414\cfrac{54}{14}1454​

54104\cfrac{54}{104}10454​

None of these

Solution

Question 10

Complete mentary event

Exhaustive event

Mutually exclusive events

Not defined

Solution

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$0$

$0.400$

$0.240$

$0.024$

$0.500$

$P(A' \cap B') = (1 - P(A)) (1 - P(B))$

$P(A) + P(B) = 1$

$P(A) = P(B)$

$A$ and $B$ are mutually exclusive

$P\left( A\cup B\cup C \right) =0$

$P\left( A\cup B\cup C \right) =1$

$P\left( A\cap B\cap C \right) =0$

$P\left( A\cap B\cap C \right) =1$

$(a^6,2a^2)$

$(a^2,-2a^2)$

$(a^3,2a^2)$

$(a^2,2a^3)$

$\cfrac{54}{44}$

$\cfrac{54}{14}$

$\cfrac{54}{104}$