Probability Test

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Probability Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The probability that at least one of the events A and B occurs is 0.6.If A and B occur simultaneously with probability 0.2,then $$P(\overline{A})+P(\overline{B})$$ is
(Here $$\overline{A}$$ and $$\overline{B}$$ are complements of A and B, respectively.)

A
0.4

B
0.8

C
1.2

D
none

Solution
Question 2
1 / -0

If A and B are two events such that P(A)$$=\dfrac{1}{2}$$, P(B)$$\dfrac{1}{3}$$, P(A/B) $$= \dfrac{1}{4}$$, then P(A' $$\cap$$ B') equals

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

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

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

D
$$\dfrac{3}{16}$$

Solution

Here, $$P(A)=\dfrac{1}{2}$$, $$P(B)\dfrac{1}{3}$$, and $$P(A/B) = \frac{1}{4}$$

$$\because P(A / B) = \dfrac{P(A \cap B)}{P(B)}$$

$$\Rightarrow P(A' \cap B') = P(A / B) \cdot = \dfrac{1}{4} - \dfrac{1}{3} = \dfrac{1}{12}$$

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

$$= 1 - [P(A) - P(B) - P(A \cap B) =]$$

$$= 1 - \left [ \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{12}\right ] = 1 - \left [ \dfrac{6 + 4 - 1}{12} \right ]$$

$$= 1 - \dfrac{9}{12} = \dfrac{3}{12} = \dfrac{1}{4}$$
Question 3
1 / -0

Let $$S$$ and $$T$$ are two events defined on a sample space with probabilities $$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$ Event $$S$$ and $$T$$ are

A
mutually exculsive

B
independent

C
mutually exclusive and independent

D
neither mutually exclusive nor independent

Solution

Given, $$S$$ and $$T$$ are two events defined on a sample space with probabilities
$$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$
$$P(S/T)=\dfrac{P(S \cap T)}{P(T)}$$
$$\Rightarrow 0.5=\dfrac{P(S \cap T)}{0.69}$$
$$\Rightarrow P(S \cap T)=0.5 \times 0.69=P(S)P(T)$$
Therefore, $$S$$ and $$T$$ are independent.

$$\therefore P(S \cap T)=P(S)P(T)$$
$$=0.69\times 0.5=0.345$$
As intersection is not zero $$S$$ and $$ T$$ are not mutually exclusive.
Option B is correct.
Question 4
1 / -0

E and F are two independent events,The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2.Then,

A
$$P(E)=1/3,P(F)=1/4$$

B
$$P(E)=1/4,P(F)=1/3$$

C
$$P(E)=1/6,P(F)=1/2$$

D
$$P(E)=1/2,P(F)=1/6$$

Solution
Question 5
1 / -0

A and B are events such that $$P(A) = 0.4, P(B) = 0.3$$ and $$P(A \cup B) = 0.5$$, Then $$P(B' \cap A')$$ equals

A
$$\frac{2}{3}$$

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

C
$$\frac{3}{10}$$

D
$$\frac{1}{5}$$

Solution

Here, $$P(A) = 0.4, P(B) = 0.3$$ and $$P(A \cup B) = 0.5$$
$$\because P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$\Rightarrow P(A \cap B) = 0.4 + 0.3 - 0.5 = 0.2$$
$$P(B' \cup A) = P(A) - P(A \cap B) $$
$$= 0.4 - 0.2 = 0.2 = \dfrac{1}{5}$$
Question 6
1 / -0

If A and B are two independent events with P(A) $$= \frac{3}{5}$$ and P(B) $$= \frac{4}{9}$$, then P(A' $$\cap$$ B') equals

A
$$\frac{4}{15}$$

B
$$\frac{8}{45}$$

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

D
$$\frac{2}{9}$$

Solution

$$P(A' \cap B') = 1 - P(A \cup B)$$
$$= 1 - [P(A) + P(B) - P(A \cap B)]$$
$$= 1 - \left [ \dfrac{3}{5} + \dfrac{4}{9} - \dfrac{3}{5} \times \dfrac{4}{9} \right ]$$ $$[P(A \cap B) = P(A) \cdot P(B)]$$
$$= 1 - \left [ \dfrac{27 + 20 - 12}{45} \right ] = 1 - \dfrac{35}{45} = \dfrac{10}{45} = \dfrac{2}{9}$$
Question 7
1 / -0

In a random experiment,it the occurrence of one event prevents the occurrence of other event,is

A
A complementary event

B
a certain event

C
Not mutually exclusive event

D
mutually exclusive event

Solution

In a random experiment if the occurrence of one event prevents the occurrence of other event it is mutually exclusive.

Explanation:

As we know

The two events both of which can not occur at the same time i.e one event prevent the occurrence of the other i.e their intersection is null are called mutually exclusive events.

Hence, In a random experiment if the occurrence of one event prevents the occurrence of other event it is mutually exclusive.
Question 8
1 / -0

If $$P(A) = 0.4, P(B) = 0.8$$ and $$P(B | A) = 0.6$$, then $$P(A \cup B)$$ is equal to

A
0.24

B
0.3

C
0.48

D
0.96

Solution

Here, $$P(A) = 0.4, P(B) = 0.8$$ and $$P(B | A) = 0.6$$
$$\because P(B | A) = \dfrac{P(B \cap A)}{P(A)}$$
$$\Rightarrow P(B \cap A) = P(B/A) \cdot P(A) $$
$$= 0.6 \times 0.4 = 0.24$$
$$\because P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$= 0.4 + 0.8 - 0.24$$
$$= 1.2 - 0.24 = 0.96$$
Question 9
1 / -0

If the events A and B are independent, then $$P(A \cap B)$$ is equal to

A
$$P(A) + P(B)$$

B
$$P(A) - P(B)$$

C
$$P(A).P(B)$$

D
$$P(A) / P(B)$$

Solution

If A and B are independent, then $$P(A \cap B) = P(A) \cdot P(B)$$
Question 10
1 / -0

If $$ E_{1} \cap E_{2} = \phi $$, then $$ E_{1} and E_{2}$$ will be

A
Exclusive

B
Independent

C
Dependent

D
Complementary

Solution

B is correct

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Probability Test - 52

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Probability Test - 52

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Question 1

0.4

0.8

1.2

none

Solution

Question 2

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

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

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

$$\dfrac{3}{16}$$

Solution

Question 3

mutually exculsive

independent

mutually exclusive and independent

neither mutually exclusive nor independent

Solution

Question 4

$$P(E)=1/3,P(F)=1/4$$

$$P(E)=1/4,P(F)=1/3$$

$$P(E)=1/6,P(F)=1/2$$

$$P(E)=1/2,P(F)=1/6$$

Solution

Question 5

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

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

$$\frac{3}{10}$$

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

Solution

Question 6

$$\frac{4}{15}$$

$$\frac{8}{45}$$

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

$$\frac{2}{9}$$

Solution

Question 7

A complementary event

a certain event

Not mutually exclusive event

mutually exclusive event

Solution

Question 8

0.24

0.3

0.48

0.96

Solution

Question 9

$$P(A) + P(B)$$

$$P(A) - P(B)$$

$$P(A).P(B)$$

$$P(A) / P(B)$$

Solution

Question 10

Exclusive

Independent

Dependent

Complementary

Solution

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