Probability Test

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Probability Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

$$\sim (p\rightarrow \sim q)$$ is equivalent to

A
$$\sim p^{\wedge} \sim q$$

B
$$\sim p^{\wedge} q$$

C
$$p^\wedge q$$

D
$$None\ of\ these$$

Solution

In if-then form, p q means that "If you do not do your homework, then you will flunk", where p (which is equivalent to ~~p ) is "You do not do your homework". By definition, p q is false if, and only if, its hypothesis, p, is true and its conclusion, q, is false.
Question 2
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
Question 3
1 / -0

Consider the following relations
(1)$$A-B=A-\left( A\cap B \right) $$
(2)$$A=A-\left( A\cap B \right) \cup \left( A-B \right) $$
(3)$$A-\left( A\cup C \right) =\left( A-B \right) \cup \left( A-C \right) $$
Which of these is correct

A
1 and 3

B
2 only

C
2 and 3

D
1 and 2

Solution

$$\begin{array}{l} 1)\, \, \, \, A-B=A-\left( { A\cap B } \right) \\ 2)\, \, \, A=A-\left( { A\cap B } \right) \cup \left( { A-B } \right) \\ 3)\, \, \, A-\left( { A\cup C } \right) =\left( { A-B } \right) \cup \left( { A-C } \right) \\ Here,\, \, \left( 1 \right) \, and\, \left( 3 \right) \, are\, true \\ Hence,\, \, option\, \, A\, \, \, is\, the\, correct\, answer. \end{array}$$
Question 4
1 / -0

Result of each experiment is called

A
Sample space

B
Random test

C
Sample point

D
Ordered pair

Solution

C is correct
Question 5
1 / -0

If the probabilities of the events $$A\cap B, A, B$$ and $$A\cup B$$ are in A.P. with second term of A.P. is equal to common difference then A and B are

A
independent

B
equally likely

C
mutually exclusive

D
can't be determined

Solution

$$P(A\cap B), P(A), P(B), P(A\cup B)$$ are in A.P.
So $$D=P(A)-P(A\cap B)=P(B)-P(A)=P(A\cup B)-P(B)$$
and second term of A.P. is equal to common difference: $$P(A)=D$$
$$P(A)=P(A)-P(A\cap B)$$ $$\Rightarrow$$ $$P(A\cap B)=0$$
We also have to find the value of P(B)
$$P(A)=P(B)-P(A)$$ $$\Rightarrow$$ $$P(B)=2P(A)$$
Hence A and B are Mutually Exclusive
Question 6
1 / -0

The probability that at least one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then $$P\left ( \bar{A} \right )+ P\left ( \bar{B} \right )=$$

A
$$0.4$$

B
$$0.8$$

C
$$1.2$$

D
$$1.6$$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=1-P\left ( \bar{A} \right )+ 1 - P\left ( \bar{B} \right )-P(A\cap B)$$
$$0.6=2-P\left ( \bar{A} \right )- P\left ( \bar{B} \right )-0.2$$ Given: $$\left [ {P(A\cup B)=0.6} ; {P(A\cap B)=0.2} \right ] $$
$$P\left ( \bar{A} \right )+ P\left ( \bar{B} \right )=2-0.6-0.2=1.2$$
Question 7
1 / -0

If A, B are two independent events, $$P\left ( A \right )=\dfrac{3}{4}$$ and $$P\left ( B \right )=\dfrac{5}{8}$$ , then $$P\left ( A\cup B \right )=$$

A
$$\dfrac{3}{32}$$

B
$$\dfrac{29}{32}$$

C
$$\dfrac{15}{32}$$

D
$$\dfrac{5}{32}$$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
A, B are two independent events so $$P(A\cap B)=P(A)P(B)$$
$$P(A\cup B)=\dfrac{3}{4}+\dfrac{5}{8}-({\dfrac{3}{4}}\times{\dfrac{5}{8}})$$

$$P(A\cup B)=\dfrac{29}{32}$$
Question 8
1 / -0

If A and B are any two events in a sample space S then $$P\left ( A\cup B \right )$$ is

A
$$\geq P\left ( A \right )+P\left ( B \right )$$

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

C
$$\leq P\left ( A \right )+P\left ( B \right )$$

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

Solution

From De-morgan's laws,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq P(A) + P(B)$$
so, we conclude that $$P(A \cup B) \leq P(A) + P(B)$$
Question 9
1 / -0

If A & B are two events then $$P\left \{ \left ( A\cap \bar{B} \right )\cup\left ( \bar{A}\cap B \right ) \right \}= $$

A
$$P\left ( A\cup B \right )-P\left ( A\cap B \right )$$

B
$$P\left ( A\cup B \right )+P\left ( A\cap B \right )$$

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

D
$$P\left ( A \right )+P\left ( B \right )+P\left ( A\cap B \right )$$

Solution

$$P\left ( A\cup B \right )-P\left ( A\cap B \right )$$
Question 10
1 / -0

If A, B, C are any three events and P(S) denotes the probability of S happening, then $$P\left \{ A\cap \left ( B\cup C \right ) \right \}= $$

A
$$P\left ( A \right )+P\left ( B \right )+P\left ( C \right )-P\left ( A\cap B \right )-P\left ( A\cap C \right )$$

B
$$P\left ( A \right )+P\left ( B \right )+P\left ( C \right )-P\left ( B \right ).P\left ( C \right )$$

C
$$P\left ( A\cap B \right )+P\left ( A\cap C \right )-P\left ( A\cap B\cap C \right )$$

D
$$P\left ( B\cap C \right )+P\left ( A\cap B \right )-P\left ( A\cap B\cap C \right )$$

Solution

We have formula, given by,

$$P\left \{ A\cap (B\cup C) \right \}$$

$$=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)$$

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Probability Test - 21

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Probability Test - 21

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SHARING IS CARING

Question 1

$$\sim p^{\wedge} \sim q$$

$$\sim p^{\wedge} q$$

$$p^\wedge q$$

$$None\ of\ these$$

Solution

Question 2

Complete mentary event

Exhaustive event

Mutually exclusive events

Not defined

Solution

Question 3

1 and 3

2 only

2 and 3

1 and 2

Solution

Question 4

Sample space

Random test

Sample point

Ordered pair

Solution

Question 5

independent

equally likely

mutually exclusive

can't be determined

Solution

Question 6

$$0.4$$

$$0.8$$

$$1.2$$

$$1.6$$

Solution

Question 7

$$\dfrac{3}{32}$$

$$\dfrac{29}{32}$$

$$\dfrac{15}{32}$$

$$\dfrac{5}{32}$$

Solution

Question 8

$$\geq P\left ( A \right )+P\left ( B \right )$$

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

$$\leq P\left ( A \right )+P\left ( B \right )$$

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

Solution

Question 9

$$P\left ( A\cup B \right )-P\left ( A\cap B \right )$$

$$P\left ( A\cup B \right )+P\left ( A\cap B \right )$$

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

$$P\left ( A \right )+P\left ( B \right )+P\left ( A\cap B \right )$$

Solution

Question 10

$$P\left ( A \right )+P\left ( B \right )+P\left ( C \right )-P\left ( A\cap B \right )-P\left ( A\cap C \right )$$

$$P\left ( A \right )+P\left ( B \right )+P\left ( C \right )-P\left ( B \right ).P\left ( C \right )$$

$$P\left ( A\cap B \right )+P\left ( A\cap C \right )-P\left ( A\cap B\cap C \right )$$

$$P\left ( B\cap C \right )+P\left ( A\cap B \right )-P\left ( A\cap B\cap C \right )$$

Solution

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